Problems in Mathematics

  • Inverse Matrices
  • An $n\times n$ matrix $A$ is said to be invertible if there exists an $n\times n$ matrix $B$ such that $AB=BA=I$. Such a matrix $B$ is unique and called the inverse matrix of $A$, denoted by $A^{-1}$

Let $A, B$ be $n\times n$ matrices.

  • $A$ is invertible if and only if $\rref([ A \mid I_n])=[ I_n \mid A’]$ for some $n\times n$ matrix $A’$. In this case, $A’=A^{-1}$.
  • $A$ is invertible if and only if $A$ is nonsingular.
  • If $A, B$ are invertible, then $(AB)^{-1}=B^{-1}A^{-1}$
  • A $2\times 2$ matrix $A=\begin{bmatrix} a & b\\ c& d \end{bmatrix}$ is invertible if and only if the determinant $\det(A)=ad-bc \neq 0$. If $A$ is invertible, then the inverse matrix is given by $A^{-1}=\frac{1}{\det(A)}\begin{bmatrix} d & -b\\ -c& a \end{bmatrix}$.
  • If $A$ is invertible, then $A^{\trans}$ is invertible and $(A^{\trans})^{-1}=(A^{-1})^{\trans}$.
  • For each of the following $3\times 3$ matrices $A$, determine whether $A$ is invertible and find the inverse $A^{-1}$ if exists by computing the augmented matrix $[A|I]$, where $I$ is the $3\times 3$ identity matrix. (a) $A=\begin{bmatrix} 1 & 3 & -2 \\ 2 &3 &0 \\ 0 & 1 & -1 \end{bmatrix}$ (b) $A=\begin{bmatrix} 1 & 0 & 2 \\ -1 &-3 &2 \\ 3 & 6 & -2 \end{bmatrix}$.
  • Let A be the matrix \[\begin{bmatrix} 1 & -1 & 0 \\ 0 &1 &-1 \\ 0 & 0 & 1 \end{bmatrix}.\] Is the matrix $A$ invertible? If not, then explain why it isn’t invertible. If so, then find the inverse. ( The Ohio State University )
  • Find the inverse matrix of \[A=\begin{bmatrix} 1 & 1 & 2 \\ 0 &0 &1 \\ 1 & 0 & 1 \end{bmatrix}\] if it exists. If you think there is no inverse matrix of $A$, then give a reason. ( The Ohio State University )
  • Let $A$ be the following $3\times 3$ upper triangular matrix. \[A=\begin{bmatrix} 1 & x & y \\ 0 &1 &z \\ 0 & 0 & 1 \end{bmatrix},\] where $x, y, z$ are some real numbers. Determine whether the matrix $A$ is invertible or not. If it is invertible, then find the inverse matrix $A^{-1}$.
  • For which choice(s) of the constant $k$ is the following matrix invertible? \[A=\begin{bmatrix} 1 & 1 & 1 \\ 1 &2 &k \\ 1 & 4 & k^2 \end{bmatrix}.\] ( Johns Hopkins University )
  • Suppose that $M, P$ are two $n \times n$ non-singular matrix. Prove that there is a matrix $N$ such that $MN = P$.
  • Let $A$ be an $n \times n$ matrix satisfying $A^2+c_1A+c_0I=O$, where $c_0, c_1$ are scalars, $I$ is the $n\times n$ identity matrix, and $O$ is the $n\times n$ zero matrix. Prove that if $c_0\neq 0$, then the matrix $A$ is invertible (nonsingular). How about the converse? Namely, is it true that if $c_0=0$, then the matrix $A$ is not invertible?
  • A square matrix $A$ is called idempotent if $A^2=A$. Show that a square invertible idempotent matrix is the identity matrix.
  • Find the inverse matrix of $A=\begin{bmatrix} 1 & 0 & 1 \\ 1 &0 &0 \\ 2 & 1 & 1 \end{bmatrix}$ if it exists. If you think there is no inverse matrix of $A$, then give a reason. See (a)
  • Find a nonsingular $2\times 2$ matrix $A$ such that $A^3=A^2B-3A^2$, where $B=\begin{bmatrix} 4 & 1\\ 2& 6 \end{bmatrix}$. Verify that the matrix $A$ you obtained is actually a nonsingular matrix. See (b)
  • Determine whether there exists a nonsingular matrix $A$ if $A^2=AB+2A$, where $B$ is the following matrix. If such a nonsingular matrix $A$ exists, find the inverse matrix $A^{-1}$. (a) \[B=\begin{bmatrix} -1 & 1 & -1 \\ 0 &-1 &0 \\ 1 & 2 & -2 \end{bmatrix}\] (b) \[B=\begin{bmatrix} -1 & 1 & -1 \\ 0 &-1 &0 \\ 2 & 1 & -4 \end{bmatrix}.\]
  • Determine whether there exists a nonsingular matrix $A$ if $A^4=ABA^2+2A^3$, where $B$ is the following matrix. $B=\begin{bmatrix} -1 & 1 & -1 \\ 0 &-1 &0 \\ 2 & 1 & -4 \end{bmatrix}.$ If such a nonsingular matrix $A$ exists, find the inverse matrix $A^{-1}$. ( The Ohio State University, Linear Algebra Final Exam Problem )
  • Let $A, B, C$ be the following $3\times 3$ matrices. \[A=\begin{bmatrix} 1 & 2 & 3 \\ 4 &5 &6 \\ 7 & 8 & 9 \end{bmatrix}, B=\begin{bmatrix} 1 & 0 & 1 \\ 0 &3 &0 \\ 1 & 0 & 5 \end{bmatrix}, C=\begin{bmatrix} -1 & 0\ & 1 \\ 0 &5 &6 \\ 3 & 0 & 1 \end{bmatrix}.\] Then compute and simplify the following expression. \[(A^{\trans}-B)^{\trans}+C(B^{-1}C)^{-1}.\] ( The Ohio State University )
  • Let $A$ be the coefficient matrix of the system of linear equations \begin{align*} -x_1-2x_2&=1\\ 2x_1+3x_2&=-1. \end{align*} (a) Solve the system by finding the inverse matrix $A^{-1}$. (b) Let $\mathbf{x}=\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$ be the solution of the system obtained in part (a). Calculate and simplify \[A^{2017}\mathbf{x}.\] ( The Ohio State University )
  • (a) Let $A$ be a $6\times 6$ matrix and suppose that $A$ can be written as $A=BC$, where $B$ is a $6\times 5$ matrix and $C$ is a $5\times 6$ matrix. Prove that the matrix $A$ cannot be invertible. (b) Let $A$ be a $2\times 2$ matrix and suppose that $A$ can be written as $A=BC$, where $B$ is a $ 2\times 3$ matrix and $C$ is a $3\times 2$ matrix. Can the matrix $A$ be invertible?
  • For a real number $a$, consider $2\times 2$ matrices $A, P, Q$ satisfying the following five conditions. (1) $A=aP+(a+1)Q$, (2) $P^2=P$, (3) $Q^2=Q$, (4) $PQ=O$, (5) $QP=O$, where $O$ is the $2\times 2$ zero matrix. Then do the following problems. (a) Prove that $(P+Q)A=A$. (b) Suppose $a$ is a positive real number and let $A=\begin{bmatrix} a & 0\\ 1& a+1 \end{bmatrix}$. Then find all matrices $P, Q$ satisfying conditions (1)-(5). (c) Let $n$ be an integer greater than $1$. For any integer $k$, $2\leq k \leq n$, we define the matrix $A_k=\begin{bmatrix} k & 0\\ 1& k+1 \end{bmatrix}$. Then calculate and simplify the matrix product $A_nA_{n-1}A_{n-2}\cdots A_2$. ( Tokyo University Entrance Exam 2007 )
  • Let $\mathbf{v}$ be a nonzero vector in $\R^n$. Then the dot product $\mathbf{v}\cdot \mathbf{v}=\mathbf{v}^{\trans}\mathbf{v}\neq 0$. Set $a:=\frac{2}{\mathbf{v}^{\trans}\mathbf{v}}$ and define the $n\times n$ matrix $A$ by $A=I-a\mathbf{v}\mathbf{v}^{\trans}$, where $I$ is the $n\times n$ identity matrix. Prove that $A$ is a symmetric matrix and $AA=I$. Conclude that the inverse matrix is $A^{-1}=A$.
  • Consider the system of linear equations \begin{align*} x_1&= 2, \\ -2x_1 + x_2 &= 3, \\ 5x_1-4x_2 +x_3 &= 2 \end{align*} (a) Find the coefficient matrix and its inverse matrix. (b) Using the inverse matrix, solve the system of linear equations. ( The Ohio State University )
  • Consider the following system of linear equations \begin{align*} 2x+3y+z&=-1\\ 3x+3y+z&=1\\ 2x+4y+z&=-2. \end{align*} (a) Find the coefficient matrix $A$ for this system. (b) Find the inverse matrix of the coefficient matrix found in (a) (c) Solve the system using the inverse matrix $A^{-1}$.
  • Consider the matrix \[A=\begin{bmatrix} 1 & 2 & 1 \\ 2 &5 &4 \\ 1 & 1 & 0 \end{bmatrix}.\] (a) Calculate the inverse matrix $A^{-1}$. If you think the matrix $A$ is not invertible, then explain why. (b)  Are the vectors \[ \mathbf{A}_1=\begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}, \mathbf{A}_2=\begin{bmatrix} 2 \\ 5 \\ 1 \end{bmatrix}, \text{ and } \mathbf{A}_3=\begin{bmatrix} 1 \\ 4 \\ 0 \end{bmatrix}\] linearly independent? (c)  Write the vector $\mathbf{b}=\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$ as a linear combination of $\mathbf{A}_1$, $\mathbf{A}_2$, and $\mathbf{A}_3$. ( The Ohio State University, Linear Algebra Exam )
  • A $2 \times 2$ matrix $A$ satisfies $\tr(A^2)=5$ and $\tr(A)=3$. Find $\det(A)$. See (a)
  • Suppose that a real matrix $A$ maps each of the following vectors \[\mathbf{x}_1=\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}, \mathbf{x}_2=\begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}, \mathbf{x}_3=\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \] into the vectors \[\mathbf{y}_1=\begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}, \mathbf{y}_2=\begin{bmatrix} -1 \\ 0 \\ 3 \end{bmatrix}, \mathbf{y}_3=\begin{bmatrix} 3 \\ 1 \\ 1 \end{bmatrix},\] respectively. That is, $A\mathbf{x}_i=\mathbf{y}_i$ for $i=1,2,3$. Find the matrix $A$. ( Kyoto University Exam )
  • Let $A$ and $B$ are $n \times n$ matrices with real entries. Assume that $A+B$ is invertible. Then show that \[A(A+B)^{-1}B=B(A+B)^{-1}A.\] ( University of California, Berkeley Qualifying Exam )
  • Let $A$ be an $n\times n$ invertible matrix. Prove that the inverse matrix of $A$ is uniques.
  • Let $\mathbf{u}$ and $\mathbf{v}$ be vectors in $\R^n$, and let $I$ be the $n \times n$ identity matrix. Suppose that the inner product of $\mathbf{u}$ and $\mathbf{v}$ satisfies \[\mathbf{v}^{\trans}\mathbf{u}\neq -1.\] Define the matrix \[A=I+\mathbf{u}\mathbf{v}^{\trans}.\] Prove that $A$ is invertible and the inverse matrix is given by the formula \[A^{-1}=I-a\mathbf{u}\mathbf{v}^{\trans},\] where \[a=\frac{1}{1+\mathbf{v}^{\trans}\mathbf{u}}.\] This formula is called the Sherman-Woodberry formula .
  • Let $A$ be a singular $2\times 2$ matrix such that $\tr(A)\neq -1$ and let $I$ be the $2\times 2$ identity matrix. Then prove that the inverse matrix of the matrix $I+A$ is given by the following formula: \[(I+A)^{-1}=I-\frac{1}{1+\tr(A)}A.\] Using the formula, calculate the inverse matrix of $\begin{bmatrix} 2 & 1\\ 1& 2 \end{bmatrix}$.
  • Let $A=\begin{bmatrix} a & 0\\ 0& b \end{bmatrix}$. Show that (a) $A^n=\begin{bmatrix} a^n & 0\\ 0& b^n \end{bmatrix}$ for any $n \in \N$. (b) Let $B=S^{-1}AS$, where $S$ be an invertible $2 \times 2$ matrix. Show that $B^n=S^{-1}A^n S$ for any $n \in \N$
  • Let $A$ be an $n\times n$ invertible matrix. Then prove the transpose $A^{\trans}$ is also invertible and that the inverse matrix of the transpose $A^{\trans}$ is the transpose of the inverse matrix $A^{-1}$. Namely, show that $(A^{\trans})^{-1}=(A^{-1})^{\trans}$.
  • A square matrix $A$ is called nilpotent if there exists a positive integer $k$ such that $A^k=O$, where $O$ is the zero matrix. (a) If $A$ is a nilpotent $n \times n$ matrix and $B$ is an $n\times n$ matrix such that $AB=BA$. Show that the product $AB$ is nilpotent. (b) Let $P$ be an invertible $n \times n$ matrix and let $N$ be a nilpotent $n\times n$ matrix. Is the product $PN$ nilpotent? If so, prove it. If not, give a counterexample.
  • (a) Show that if $A$ is invertible, then $A$ is nonsingular. (b) Let $A, B, C$ be $n\times n$ matrices such that $AB=C$. Prove that if either $A$ or $B$ is singular, then so is $C$. (c) Show that if $A$ is nonsingular, then $A$ is invertible.
  • A square matrix $A$ is called nilpotent if some power of $A$ is the zero matrix. Namely, $A$ is nilpotent if there exists a positive integer $k$ such that $A^k=O$, where $O$ is the zero matrix. Suppose that $A$ is a nilpotent matrix and let $B$ be an invertible matrix of the same size as $A$. Is the matrix $B-A$ invertible? If so prove it. Otherwise, give a counterexample.
  • Let $A$ be an $n\times n$ matrix. The $(i, j)$ cofactor $C_{ij}$ of $A$ is defined to be $C_{ij}=(-1)^{ij}\det(M_{ij})$, where $M_{ij}$ is the $(i,j)$ minor matrix obtained from $A$ removing the $i$-th row and $j$-th column. Then consider the $n\times n$ matrix $C=(C_{ij})$, and define the $n\times n$ matrix $\Adj(A)=C^{\trans}$. The matrix $\Adj(A)$ is called the adjoint matrix of $A$. When $A$ is invertible, then its inverse can be obtained by the formula \[A^{-1}=\frac{1}{\det(A)}\Adj(A).\] For each of the following matrices, determine whether it is invertible, and if so, then find the invertible matrix using the above formula. (a) $A=\begin{bmatrix} 1 & 5 & 2 \\ 0 &-1 &2 \\ 0 & 0 & 1 \end{bmatrix}$. (b) $B=\begin{bmatrix} 1 & 0 & 2 \\ 0 &1 &4 \\ 3 & 0 & 1 \end{bmatrix}$.
  • Let $A$ be a real symmetric matrix whose diagonal entries are all positive real numbers. Is it true that the all of the diagonal entries of the inverse matrix $A^{-1}$ are also positive? If so, prove it. Otherwise, give a counterexample.
  • Let $A$ be an $n\times n$ nonsingular matrix with integer entries. Prove that the inverse matrix $A^{-1}$ contains only integer entries if and only if $\det(A)=\pm 1$.
  • Introduction to Matrices
  • Elementary Row Operations
  • Gaussian-Jordan Elimination
  • Solutions of Systems of Linear Equations
  • Linear Combination and Linear Independence
  • Nonsingular Matrices
  • Subspaces in $\R^n$
  • Bases and Dimension of Subspaces in $\R^n$
  • General Vector Spaces
  • Subspaces in General Vector Spaces
  • Linearly Independency of General Vectors
  • Bases and Coordinate Vectors
  • Dimensions of General Vector Spaces
  • Linear Transformation from $\R^n$ to $\R^m$
  • Linear Transformation Between Vector Spaces
  • Orthogonal Bases
  • Determinants of Matrices
  • Computations of Determinants
  • Introduction to Eigenvalues and Eigenvectors
  • Eigenvectors and Eigenspaces
  • Diagonalization of Matrices
  • The Cayley-Hamilton Theorem
  • Dot Products and Length of Vectors
  • Eigenvalues and Eigenvectors of Linear Transformations
  • Jordan Canonical Form

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Mathematics LibreTexts

2.7: Finding the Inverse of a Matrix

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  • Page ID 19845

  • Ken Kuttler
  • Brigham Young University via Lyryx

In Example 2.6.1 , we were given \(A^{-1}\) and asked to verify that this matrix was in fact the inverse of \(A\). In this section, we explore how to find \(A^{-1}\).

Let \[A=\left[ \begin{array}{rr} 1 & 1 \\ 1 & 2 \end{array} \right]\nonumber \] as in Example 2.6.1 . In order to find \(A^{-1}\), we need to find a matrix \(\left[ \begin{array}{rr} x & z \\ y & w \end{array} \right]\) such that \[\left[ \begin{array}{rr} 1 & 1 \\ 1 & 2 \end{array} \right] \left[ \begin{array}{rr} x & z \\ y & w \end{array} \right] =\left[ \begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array} \right]\nonumber \] We can multiply these two matrices, and see that in order for this equation to be true, we must find the solution to the systems of equations, \[\begin{array}{c} x+y=1 \\ x+2y=0 \end{array}\nonumber\] and \[\begin{array}{c} z+w=0 \\ z+2w=1 \end{array}\nonumber \] Writing the augmented matrix for these two systems gives \[\left[ \begin{array}{rr|r} 1 & 1 & 1 \\ 1 & 2 & 0 \end{array} \right] \nonumber \] for the first system and \[\left[ \begin{array}{rr|r} 1 & 1 & 0 \\ 1 & 2 & 1 \end{array} \right] \label{inverse2a}\] for the second.

Let’s solve the first system. Take \(-1\) times the first row and add to the second to get \[\left[ \begin{array}{rr|r} 1 & 1 & 1 \\ 0 & 1 & -1 \end{array} \right]\nonumber \] Now take \(-1\) times the second row and add to the first to get \[\left[ \begin{array}{rr|r} 1 & 0 & 2 \\ 0 & 1 & -1 \end{array} \right]\nonumber \] Writing in terms of variables, this says \(x=2\) and \(y=-1.\)

Now solve the second system, \(\eqref{inverse2a}\) to find \(z\) and \(w.\) You will find that \(z = -1\) and \(w = 1\).

If we take the values found for \(x,y,z,\) and \(w\) and put them into our inverse matrix, we see that the inverse is \[A^{-1} = \left[ \begin{array}{rr} x & z \\ y & w \end{array} \right] = \left[ \begin{array}{rr} 2 & -1 \\ -1 & 1 \end{array} \right]\nonumber \]

After taking the time to solve the second system, you may have noticed that exactly the same row operations were used to solve both systems. In each case, the end result was something of the form \(\left[ I|X\right]\) where \(I\) is the identity and \(X\) gave a column of the inverse. In the above, \[\left[ \begin{array}{c} x \\ y \end{array} \right]\nonumber \] the first column of the inverse was obtained by solving the first system and then the second column \[\left[ \begin{array}{c} z \\ w \end{array} \right]\nonumber \]

To simplify this procedure, we could have solved both systems at once! To do so, we could have written \[\left[ \begin{array}{rr|rr} 1 & 1 & 1 & 0 \\ 1 & 2 & 0 & 1 \end{array} \right]\nonumber \]

and row reduced until we obtained \[\left[ \begin{array}{rr|rr} 1 & 0 & 2 & -1 \\ 0 & 1 & -1 & 1 \end{array} \right]\nonumber \] and read off the inverse as the \(2\times 2\) matrix on the right side.

This exploration motivates the following important algorithm.

Algorithm \(\PageIndex{1}\): Matrix Inverse Algorithm

Suppose \(A\) is an \(n\times n\) matrix. To find \(A^{-1}\) if it exists, form the augmented \(n\times 2n\) matrix \[\left[ A|I\right]\nonumber \] If possible do row operations until you obtain an \(n\times 2n\) matrix of the form \[\left[ I|B\right]\nonumber \] When this has been done, \(B=A^{-1}.\) In this case, we say that \(A\) is invertible . If it is impossible to row reduce to a matrix of the form \(\left[ I|B\right] ,\) then \(A\) has no inverse.

This algorithm shows how to find the inverse if it exists. It will also tell you if \(A\) does not have an inverse.

Consider the following example.

Example \(\PageIndex{1}\): Finding the Inverse

Let \(A=\left[ \begin{array}{rrr} 1 & 2 & 2 \\ 1 & 0 & 2 \\ 3 & 1 & -1 \end{array} \right]\). Find \(A^{-1}\) if it exists.

Set up the augmented matrix \[\left[ A|I\right] = \left[ \begin{array}{rrr|rrr} 1 & 2 & 2 & 1 & 0 & 0 \\ 1 & 0 & 2 & 0 & 1 & 0 \\ 3 & 1 & -1 & 0 & 0 & 1 \end{array} \right]\nonumber \]

Now we row reduce, with the goal of obtaining the \(3 \times 3\) identity matrix on the left hand side. First, take \(-1\) times the first row and add to the second followed by \(-3\) times the first row added to the third row. This yields \[ \ \left[ \begin{array}{rrr|rrr} 1 & 2 & 2 & 1 & 0 & 0 \\ 0 & -2 & 0 & -1 & 1 & 0 \\ 0 & -5 & -7 & -3 & 0 & 1 \end{array} \right]\nonumber \] Then take 5 times the second row and add to -2 times the third row. \[\left[ \begin{array}{rrr|rrr} 1 & 2 & 2 & 1 & 0 & 0 \\ 0 & -10 & 0 & -5 & 5 & 0 \\ 0 & 0 & 14 & 1 & 5 & -2 \end{array} \right]\nonumber \] Next take the third row and add to \(-7\) times the first row. This yields \[\left[ \begin{array}{rrr|rrr} -7 & -14 & 0 & -6 & 5 & -2 \\ 0 & -10 & 0 & -5 & 5 & 0 \\ 0 & 0 & 14 & 1 & 5 & -2 \end{array} \right]\nonumber \] Now take \(-\frac{7}{5}\) times the second row and add to the first row. \[\left[ \begin{array}{rrr|rrr} -7 & 0 & 0 & 1 & -2 & -2 \\ 0 & -10 & 0 & -5 & 5 & 0 \\ 0 & 0 & 14 & 1 & 5 & -2 \end{array} \right]\nonumber \] Finally divide the first row by -7, the second row by -10 and the third row by 14 which yields \[\left[ \begin{array}{rrr|rrr} 1 & 0 & 0 & - \ \frac{1}{7} & \ \frac{2}{7} & \ \frac{2}{7} \\ 0 & 1 & 0 & \ \frac{1}{2} & - \ \frac{1}{2} & 0 \\ 0 & 0 & 1 & \ \frac{1}{14} & \ \frac{5}{14} & - \ \frac{1}{7} \end{array} \right]\nonumber \] Notice that the left hand side of this matrix is now the \(3 \times 3\) identity matrix \(I_3\). Therefore, the inverse is the \(3 \times 3\) matrix on the right hand side, given by \[\left[ \begin{array}{rrr} - \ \frac{1}{7} & \ \frac{2}{7} & \ \frac{2}{7} \\ \ \frac{1}{2} & - \ \frac{1}{2} & 0 \\ \ \frac{1}{14} & \ \frac{5}{14} & - \ \frac{1}{7} \end{array} \right]\nonumber \]

It may happen that through this algorithm, you discover that the left hand side cannot be row reduced to the identity matrix. Consider the following example of this situation.

Example \(\PageIndex{2}\): A Matrix Which Has No Inverse

Let \(A=\left[ \begin{array}{rrr} 1 & 2 & 2 \\ 1 & 0 & 2 \\ 2 & 2 & 4 \end{array} \right]\). Find \(A^{-1}\) if it exists.

Write the augmented matrix \(\left[ A|I\right]\) \[\left[ \begin{array}{rrr|rrr} 1 & 2 & 2 & 1 & 0 & 0 \\ 1 & 0 & 2 & 0 & 1 & 0 \\ 2 & 2 & 4 & 0 & 0 & 1 \end{array} \right]\nonumber \] and proceed to do row operations attempting to obtain \(\left[ I|A^{-1}\right] .\) Take \(-1\) times the first row and add to the second. Then take \(-2\) times the first row and add to the third row. \[\left[ \begin{array}{rrr|rrr} 1 & 2 & 2 & 1 & 0 & 0 \\ 0 & -2 & 0 & -1 & 1 & 0 \\ 0 & -2 & 0 & -2 & 0 & 1 \end{array} \right]\nonumber \] Next add \(-1\) times the second row to the third row. \[\left[ \begin{array}{rrr|rrr} 1 & 2 & 2 & 1 & 0 & 0 \\ 0 & -2 & 0 & -1 & 1 & 0 \\ 0 & 0 & 0 & -1 & -1 & 1 \end{array} \right]\nonumber \] At this point, you can see there will be no way to obtain \(I\) on the left side of this augmented matrix. Hence, there is no way to complete this algorithm, and therefore the inverse of \(A\) does not exist. In this case, we say that \(A\) is not invertible.

If the algorithm provides an inverse for the original matrix, it is always possible to check your answer. To do so, use the method demonstrated in Example 2.6.1 . Check that the products \(AA^{-1}\) and \(A^{-1}A\) both equal the identity matrix. Through this method, you can always be sure that you have calculated \(A^{-1}\) properly!

One way in which the inverse of a matrix is useful is to find the solution of a system of linear equations. Recall from Definition 2.2.4 that we can write a system of equations in matrix form, which is of the form \(AX=B\). Suppose you find the inverse of the matrix \(A^{-1}\). Then you could multiply both sides of this equation on the left by \(A^{-1}\) and simplify to obtain \[\begin{array}{c} \left( A^{-1} \right) AX =A^{-1}B \\ \left(A^{-1}A\right) X = A^{-1}B \\ IX = A^{-1}B \\ X = A^{-1}B \end{array}\nonumber \] Therefore we can find \(X\), the solution to the system, by computing \(X=A^{-1}B\). Note that once you have found \(A^{-1}\), you can easily get the solution for different right hand sides (different \(B\)). It is always just \(A^{-1}B\).

We will explore this method of finding the solution to a system in the following example.

Example \(\PageIndex{3}\): Using the Inverse to Solve a System of Equations

Consider the following system of equations. Use the inverse of a suitable matrix to give the solutions to this system. \[\begin{array}{c} x+z=1 \\ x-y+z=3 \\ x+y-z=2 \end{array}\nonumber \]

First, we can write the system of equations in matrix form \[AX = \left[ \begin{array}{rrr} 1 & 0 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array} \right] \left[ \begin{array}{r} x \\ y \\ z \end{array} \right] =\left[ \begin{array}{r} 1 \\ 3 \\ 2 \end{array} \right] = B \label{inversesystem1}\]

The inverse of the matrix \[A = \left[ \begin{array}{rrr} 1 & 0 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array} \right]\nonumber \] is \[A^{-1} = \left[ \begin{array}{rrr} 0 & \ \frac{1}{2} & \ \frac{1}{2} \\ 1 & -1 & 0 \\ 1 & - \ \frac{1}{2} & - \ \frac{1}{2} \end{array} \right]\nonumber \]

Verifying this inverse is left as an exercise.

From here, the solution to the given system \(\eqref{inversesystem1}\) is found by \[\left[ \begin{array}{r} x \\ y \\ z \end{array} \right] = A^{-1}B = \left[ \begin{array}{rrr} 0 & \ \frac{1}{2} & \ \frac{1}{2} \\ 1 & -1 & 0 \\ 1 & - \ \frac{1}{2} & - \ \frac{1}{2} \end{array} \right] \left[ \begin{array}{r} 1 \\ 3 \\ 2 \end{array} \right] =\left[ \begin{array}{r} \ \frac{5}{2} \\ -2 \\ - \ \frac{3}{2} \end{array} \right]\nonumber \]

What if the right side, \(B\), of \(\eqref{inversesystem1}\) had been \(\left[ \begin{array}{r} 0 \\ 1 \\ 3 \end{array} \right] ?\) In other words, what would be the solution to \[\left[ \begin{array}{rrr} 1 & 0 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array} \right] \left[ \begin{array}{r} x \\ y \\ z \end{array} \right] =\left[ \begin{array}{r} 0 \\ 1 \\ 3 \end{array} \right] ?\nonumber \] By the above discussion, the solution is given by \[\left[ \begin{array}{r} x \\ y \\ z \end{array} \right] = A^{-1}B = \left[ \begin{array}{rrr} 0 & \ \frac{1}{2} & \ \frac{1}{2} \\ 1 & -1 & 0 \\ 1 & - \ \frac{1}{2} & - \ \frac{1}{2} \end{array} \right] \left[ \begin{array}{r} 0 \\ 1 \\ 3 \end{array} \right] =\left[ \begin{array}{r} 2 \\ -1 \\ -2 \end{array} \right]\nonumber \] This illustrates that for a system \(AX=B\) where \(A^{-1}\) exists, it is easy to find the solution when the vector \(B\) is changed.

We conclude this section with some important properties of the inverse.

Theorem \(\PageIndex{1}\): Inverses of Transposes and Products

Let \(A, B\), and \(A_i\) for \(i=1,...,k\) be \(n \times n\) matrices.

  • If \(A\) is an invertible matrix, then \((A^{T})^{-1} = (A^{-1})^{T}\)
  • If \(A\) and \(B\) are invertible matrices, then \(AB\) is invertible and \((AB)^{-1} = B^{-1}A^{-1}\)
  • If \(A_1, A_2, ..., A_k\) are invertible, then the product \(A_1A_2 \cdots A_k\) is invertible, and \((A_1A_2 \cdots A_k)^{-1} = A_k^{-1}A_{k-1}^{-1} \cdots A_2^{-1}A_1^{-1}\)

Consider the following theorem.

Theorem \(\PageIndex{2}\): Properties of the Inverse

Let \(A\) be an \(n \times n\) matrix and \(I\) the usual identity matrix.

  • \(I\) is invertible and \(I^{-1} = I\)
  • If \(A\) is invertible then so is \(A^{-1}\), and \((A^{-1})^{-1} = A\)
  • If \(A\) is invertible then so is \(A^k\), and \((A^k)^{-1} = (A^{-1})^k\)
  • If \(A\) is invertible and \(p\) is a nonzero real number, then \(pA\) is invertible and \((pA)^{-1} = \frac{1}{p}A^{-1}\)

Inverse Matrix Questions

Inverse matrix questions and solutions are given here to help students learn how to find the inverse of different matrices using different formulas and techniques. As we know, matrices are one of the most scoring concepts for students. Finding the inverse matrix is simple for 2×2 matrices. However, we can easily find the inverse matrix for 3×3 and 4×4 matrices using some simple rules. In this article, you will learn how to find the inverse of a given matrix using a suitable method.

What is an inverse matrix?

Suppose A is a non-singular square matrix of order n×n, and there is a matrix B of the same order, such that AB = BA = I, then B is called the inverse matrix of A, and I is the identity matrix.

Inverse matrix formula for 2×2 matrix

|A| = ad – bc

Therefore, A -1 = adjA/|A|

That means,

Inverse matrix formula for 3×3 or n×n matrix

Step 1: Find the determinant of the given matrix, say A.

Step 2: Find the cofactor matrix C ij = (-1) i+j det (M ij ), where M ij is the (i,j)th minor matrix after removing the ith row and the jth column.

Step 3: Find the transpose of the cofactor matrix to get the adj A.

Step 4: A -1 = adj A/det(A)

Learn more about the inverse matrix here.

Inverse Matrix Questions and Answers

1. Find the inverse of the matrix \(\begin{array}{l}A=\begin{bmatrix}2 & 3 \\1 & 2 \\\end{bmatrix}\end{array} \) .

Let us find the determinant of A.

Here, |A| ≠ 0, so the inverse of A exists.

Now, A -1 = adjA/|A|

2. What is the inverse of the matrix \(\begin{array}{l}A=\begin{bmatrix}1 & 2 \\-3 & 0 \\\end{bmatrix}\end{array} \) ?

Let us calculate the determinant of A.

As we know, A -1 = adjA/|A|

3. If \(\begin{array}{l}A=\begin{bmatrix}cos\ \theta & sin\ \theta \\-sin\ \theta & cos\ \theta \\\end{bmatrix}\end{array} \) and A -1 = A T , find the value of θ.

Also, given that,

⇒ AA -1 = AA T

From the above,

cos 2 θ + sin 2 θ = 1

This is one of the trigonometric identities and is true for all real values of θ.

4. Calculate the inverse of the matrix \(\begin{array}{l}A=\begin{bmatrix}2 & 4 & -6 \\7 & 3 & 5 \\1 & -2 & 4 \\\end{bmatrix}\end{array} \) .

First, find the determinant of matrix A.

= 2(12 + 10) – 4(28 – 5) – 6(-14 – 3)

= 2(22) – 4(23) – 6(-17)

= 44 – 92 + 102

Thus, the inverse matrix exists.

inverse matrix questions Q4

5. If \(\begin{array}{l}A=\begin{bmatrix}2 & 1 \\7 & 2 \\\end{bmatrix}\end{array} \) , show that (A -1 ) -1 = A.

Here, matrix A is non-singular.

Let A -1 = B

= (-⅔)(-⅔) – (⅓)(7/3)

= (4/9) – (7/9)

= (4 – 7)/9

That means B -1 = (A -1 ) -1 = A

6. Find x, y, z if \(\begin{array}{l}A=\begin{bmatrix}0 & 2y & z \\x & y & -z \\x & -y & z \\\end{bmatrix}\end{array} \) satisfies A T = A -1 .

⇒ AA T = AA -1

⇒ AA T = I {since A -1 A = AA -1 = I}

By performing multiplication on the LHS, we get:

inverse matrix questions Q6

By equating the corresponding elements, we have:

4y 2 + z 2 = 1 …(1)

x 2 + y 2 + z 2 = 1 …(2)

2y 2 – z 2 = 0 …(3)

Adding equations (1) and (3), we get:

4y 2 + z 2 + 2y 2 – z 2 = 1 + 0

⇒ y = ±1/√6

Substituting the value of y in equation (3), we get:

z 2 = 2(1/6)

⇒ z = ±1/√3

Substituting the values of y and z in equation (2), we get:

x 2 = 1 – y 2 – z 2

x 2 = 1 – (1/6) – (1/3)

x 2 = (6 – 1 – 2)/6

⇒ x = ±1/√2

Therefore, x = ±1/√2, y = ±1/√6 and z = ±1/√3.

7. Find the value of x for which the matrix \(\begin{array}{l}A=\begin{bmatrix}2 & 0 & 10 \\ 0 & x+7 & -3 \\0 & 4 & x \\\end{bmatrix}\end{array} \) is invertible.

Let us find the determinant of the given matrix.

= 2[(x + 7)x – (-3)(4)] – 0 + 10(0 – 0)

= 2(x 2 + 7x + 12)

We know that a matrix is invertible if and only if its determinant is not equal to 0.

Let |A| = 0

2(x 2 + 7x + 12) = 0

⇒ x 2 + 7x + 12 = 0

⇒ x 2 + 3x + 4x + 12 = 0

⇒ x(x + 3) + 4(x + 3) = 0

⇒ (x + 3)(x + 4) = 0

⇒ x + 3 = 0, x + 4 = 0

⇒ x = -3, x = -4

Thus, for x = -3 and -4, the given matrix is invertible.

8. Find the inverse of \(\begin{array}{l}A=\begin{bmatrix}1 & 2 & 3 \\2 & 4 & 5 \\3 & 5 & 6 \\\end{bmatrix}\end{array} \) using row operations.

Let us write the augmented matrix [A | I ] such that I is a square matrix of the order same as A.

inverse matrix questions Q8 (i)

R 3 → R 3 – 3R 1

inverse matrix questions Q8 (ii)

R 2 → R 2 – 2R 1

inverse matrix questions Q8 (iii)

Now, interchange R 2 and R 3 .

inverse matrix questions Q8 (iv)

R 2 → (-1).R 2 and R 3 → (-1).R 3

inverse matrix questions Q8 (v)

R 2 → R 2 – 3R 3

inverse matrix questions Q8 (vi)

R 1 → R 1 – 2R 2

inverse matrix questions Q8 (vii)

R 1 → R 1 – 3R 3

inverse matrix questions Q8 (viii)

This is of the form [ I | B].

Here, B is the inverse of A.

9. Determine the formula for the inverse of matrix \(\begin{array}{l}A=\begin{bmatrix}p & 0 & 0 & 0 \\0 & q & 0 & 0 \\0 & 0 & r & 0 \\0 & 0 & 0 & s \\\end{bmatrix}\end{array} \) , where p, q, r, s ≠ 0.

Let us write the augmented matrix [A | I ].

R 1 → (1/p) R 1

R 2 → (1/q) R 2

R 3 → (1/r) R 3

R 4 → (1/s) R 4

Hence, the inverse of A is:

10. If A is 3 × 3 invertible matrix, then show that for any scalar k (non-zero), kA is invertible and (kA) –1 = (1/k)A –1 .

= [k (1/k)] (A A -1 )

= 1. (AA -1 )

= I {since AA -1 = A -1 A = I}

That means kA is the inverse of (1/k)A -1 .

Therefore, (kA) -1 = (1/k) A -1

Practice Questions on Inverse Matrix

  • Find the inverse of the matrix \(\begin{array}{l}A=\begin{bmatrix}3 & 5 \\-2 & 4 \\\end{bmatrix}\end{array} \) .
  • If \(\begin{array}{l}A=\begin{bmatrix}2 & 1 & 1\\1 & 2 & 1\\1 & 1 & 2\\\end{bmatrix}\end{array} \) , find A -1 exists.
  • Using elementary row operations, find the inverse of the matrix \(\begin{array}{l}\begin{bmatrix}-4 & 8 & 4\\-1 & 2 & 1\\-3 & 6 & 3\\\end{bmatrix}\end{array} \) .
  • Calculate the inverse matrix for \(\begin{array}{l}B=\begin{bmatrix}4+3i & -i \\i &4-3i \\\end{bmatrix}\end{array} \) .
  • What is the inverse of the matrix \(\begin{array}{l}A=\begin{bmatrix}1 & 0 & 1 & 2 \\-1 & 1 & 2 & 0 \\-2 & 0 & 1 & 2 \\0 & 0 & 0 & 1 \\\end{bmatrix}\end{array} \) ?

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Inverse Matrix Questions with Solutions

Tutorials including examples and questions with detailed solutions on how to find the inverse of square matrices using the method of the row echelon form and the method of cofactors. The properties of inverse matrices are discussed and various questions, including some challenging ones, related to inverse matrices are included along with their detailed solutions.

Page Content

Definition of the identity matrix, definition of the inverse of a matrix, find the inverse of a square matrix using the row reduction method, find the inverse of a square matrix using minors, cofactors and adjugate, formula for the inverse of a 2 by 2 matrix, properties of inverse matrices, questions on inverse matrices.

  • Solutions to the Questions

Example 2 Find the inverse of matrix A given by \[ A = \begin{bmatrix} 1&1 \\ 2&4 \end{bmatrix} \] if it exists. Solution Write the augmented matrix \( [ A | I )\) \[ \begin{bmatrix} 1&1&|&1&0\\2&4&|&0&1 \end{bmatrix} \] step 1 \[ \color{red}{\begin{matrix} \\ R_2 - 2 \times R_1 \end{matrix} } \begin{bmatrix} 1&1&|&1&0\\0&2&|&-2&1 \end{bmatrix} \] step 2 \[ \color{red}{\begin{matrix} \\ (1/2)R_2 \end{matrix} } \begin{bmatrix} 1&1&|&1&0\\0&1&|&-1&1/2 \end{bmatrix} \] step 3 \[ \color{red}{\begin{matrix} R_1 - R_2 \\ \\ \end{matrix} } \begin{bmatrix} 1&0&|&2&-1/2\\0&1&|&-1&1/2 \end{bmatrix} \] The inverse of A is the 2 × 2 matrix on the right side given by \[ A^{-1} = \begin{bmatrix} 2&-1/2\\-1&1/2 \end{bmatrix} \]

Example 3 Find the inverse of matrix A given by \[ A = \begin{bmatrix}-2&2&0 \\ 2&1&3\\ -2&4&-2\end{bmatrix} \] if it exists. Solution Write the augmented matrix \( [ A | I )\) \[ \begin{bmatrix} -2&2&0&|&1&0&0\\ 2&1&3&|&0&1&0 \\ -2 & 4 & -2 &|& 0 & 0 & 1 \end{bmatrix} \] step 1 \[ \color{red}{ \begin{matrix} \\ R_2 + R_1 \\ R_3 - R_1 \end{matrix} } \begin{bmatrix} -2&2&0&|&1&0&0\\ 0&3&3&|&1&1&0 \\ 0 & 2 & -2 &|& -1 & 0 & 1 \end{bmatrix} \] step 2 \[ \color{red}{ \begin{matrix} \\ \\ R_3 - (2/3) R_2 \\ \end{matrix} } \begin{bmatrix} -2&2&0&|&1&0&0\\ 0&3&3&|&1&1&0 \\ 0 & 0 & - 4&|& -5/3 & -2/3 & 1 \end{bmatrix} \] step 3 \[ \color{red}{ \begin{matrix} \\ \\ (-1/4)R_3 \\ \end{matrix}} \begin{bmatrix} -2&2&0&|&1&0&0\\ 0&3&3&|&1&1&0 \\ 0 & 0 & 1&|& 5/12 & 1/6 & -1/4 \end{bmatrix} \] step 4 \[ \color{red}{ \begin{matrix} \\ R_2 - 3\times R_3 \\ \\ \end{matrix} } \begin{bmatrix} -2&2&0&|&1&0&0\\ 0&3&0&|&-1/4&1/2&3/4 \\ 0 & 0 & 1&|& 5/12 & 1/6 & -1/4 \end{bmatrix} \] step 5 \[ \color{red}{ \begin{matrix} \\ (1/3) R_2 \\ \\ \end{matrix}} \begin{bmatrix} -2&2&0&|&1&0&0\\ 0&1&0&|&-1/12&1/6&1/4 \\ 0 & 0 & 1&|& 5/12 & 1/6 & -1/4 \end{bmatrix} \] step 6 \[ \color{red}{ \begin{matrix} R_1- 2\times R_2 \\ \\ \\ \end{matrix} } \begin{bmatrix} -2&0&0&|&7/6&-1/3&-1/2\\ 0&1&0&|&-1/12&1/6&1/4 \\ 0 & 0 & 1&|& 5/12 & 1/6 & -1/4 \end{bmatrix} \] step 7 \[ \color{red}{ \begin{matrix} (-1/2) R_1 \\ \\ \\ \end{matrix} } \begin{bmatrix} 1&0&0&|&-7/12&1/6&1/4\\ 0&1&0&|&-1/12&1/6&1/4 \\ 0 & 0 & 1&|& 5/12 & 1/6 & -1/4 \end{bmatrix} \] Hence \[ A^{-1} = \begin{bmatrix} -7/12&1/6&1/4\\ -1/12&1/6&1/4 \\ 5/12 & 1/6 & -1/4 \end{bmatrix} \] More examples on how to find matrix inverse using row operations are included.

This method is explained using a numerical example. Matrix A is given below. \[ A = \begin{bmatrix} -1&0&1\\ 2&-1&2 \\ -1 & 2 & 1 \end{bmatrix} \] a) Find the matrices of minors and cofactors, the adjugate and the inverse of A. Matrix of Minors The entry \( M_{i,j} \) of the matrix of minors of matrix A is given by the determinant obtained by deleting the \( i^{th}\) row and the \( j^{th}\) column. To find \( M_{1,1} \), delete row 1 and column 1 from matrix A and find the determinant of the remaining 2 by 2 matrix as follows: \( M_{1,1} = Det \begin{bmatrix} .&.&.\\ .&-1&2 \\ .& 2 & 1 \end{bmatrix} = -1 - 4 = -5\) To find \( M_{1,2} \), delete row 1 and column 2 from matrix A and find the determinant of the remaining 2 by 2 matrix as follows: \( M_{1,2} = Det \begin{bmatrix} .&.&.\\ 2&.&2 \\ -1 & . & 1 \end{bmatrix} = 2 -(-2) = 4 \) To find \( M_{1,3} \), delete row 1 and column 3 from matrix A and find the determinant of the remaining 2 by 2 matrix as follows: \( M_{1,3} = Det \begin{bmatrix} .&.&.\\ 2&-1&. \\ -1 & 2 & . \end{bmatrix} = 4 - 1 = 3 \) To find \( M_{2,1} \), delete row 2 and column 1 from matrix A and find the determinant of the remaining 2 by 2 matrix as follows: \( M_{2,1} = Det \begin{bmatrix} .&0&1\\ .&.&. \\ . & 2 & 1 \end{bmatrix} = 0 - 2 = - 2 \) ... ... The remaining entries are given by: \( M_{2,2} = 0 \) , \( M_{2,3} = -2 \) , \( M_{3,1} = 1\) , \( M_{3,2} = -4\) , \( M_{3,3} = 1\). The matrix of minors M is given by \( M = \begin{bmatrix} -5&4&3\\ -2&0&-2\\ 1&-4&1 \end{bmatrix} \) Matrix of Cofactors The entries \( C_{i,j} \) of the matrix of cofactors C of matrix A are given by \( C_{i,j} = (-1)^{i+j}M{i,j} \) An evaluation of the entries \( C_{i,j} \) gives: \( C_{1,1} = (-1)^{1+1} M_{1,1} = -5 \) \( C_{1,2} = (-1)^{1+2} M_{1,2} = - 4 \) \( C_{1,3} = (-1)^{1+3} M_{1,3} = 3 \) \( C_{2,1} = (-1)^{2+1} M_{2,1} = 2 \) \( C_{2,2} = (-1)^{2+2} M_{2,2} = 0 \) \( C_{3,1} = (-1)^{3+1} M_{3,1} = 1 \) \( C_{3,2} = (-1)^{3+2} M_{3,2} = 4 \) \( C_{3,3} = (-1)^{3+3} M_{3,1} = 1 \) Hence the matrix C of cofactors is given by \( C = \begin{bmatrix} -5&-4&3\\ 2&0&2\\ 1&4&1 \end{bmatrix} \) Adjugate (or adjunct) of a Matrix The adjugate (or adjunct) of matrix A is the transpose of its matrix of cofactors C. \( Adjugate(A) = C^T = \begin{bmatrix} -5&2&1\\ -4&0&4\\ 3&2&1 \end{bmatrix} \) Inverse Matrix We now need to find the determinant D of matrix A. Using the first row of matrix A and the corresponding minors already found, D is given by \( D = det\begin{bmatrix} -1&0&1\\ 2&-1&2 \\ -1 & 2 & 1 \end{bmatrix} = A_{11}M_{1,1} - A_{1,2}M_{1,2} + A_{1,3}M_{1,3} = 8\) The inverse of \( A \) is given by \( A^{-1} = \dfrac{1}{D} C^T = \dfrac{1}{8} \begin{bmatrix} -5&2&1\\ -4&0&4\\ 3&2&1 \end{bmatrix} = \begin{bmatrix} -\dfrac{5}{8}&\dfrac{1}{4}&\dfrac{1}{8}\\ -\dfrac{1}{2}&0&\dfrac{1}{2}\\ \dfrac{3}{8}&\dfrac{1}{4}&\dfrac{1}{8}\end{bmatrix}\)

Using any of the two methods described above, it can be shown that the inverse of matrix A given by \[ A = \begin{bmatrix} a & b\\ c & d \\ \end{bmatrix} \] is given by \[ A^{-1} = \dfrac{1}{ad - bc}\begin{bmatrix} d & -b\\ -c & a \\ \end{bmatrix} \]

A matrix that has an inverse is called an invertible matrix.

  • If A is an invertible matrix, its inverse is unique.
  • \( A A^{-1} = A^{-1} A = I \)
  • If matrices A and B are invertible, then:\( (AB)^{-1} = B^{-1}A^{-1} \)
  • A matrix is invertible if and only if its determinant is not equal to zero.
  • A matrix whose determinant is not equal to zero is called nonsingular.
  • \( (A^T)^{-1} = (A^{-1})^T \)
  • \( Det(A^{-1}) = \dfrac{1}{Det(A)} \)
  • \( (A^{-1})^{-1} = A \)
  • Question 1 Use row reduction method to find the inverse of the following matrices: \( A = \begin{bmatrix} -1&-1&1\\ 2&0&-2 \\ 1 & 1 & 1 \end{bmatrix} \) , \( B = \begin{bmatrix} 1&0&1&2\\ -1& 1 & 2 & 0 \\ -2& 0 & 1 & 2 \\ 0 & 0 & 0 & 1 \end{bmatrix} \)
  • Question 2 Use the method of cofactors to find the inverse of the following matrix. \( A = \begin{bmatrix} -1&0&3\\ 3&2&2 \\ 0& 0 & 1 \end{bmatrix} \)
  • Question 3 A, B and C are 2 by 2 matrices. Matrices B and C are given by: \[ B = \begin{bmatrix} -1&-1\\ -2& 1 \end{bmatrix} , C = \begin{bmatrix} 2 & -1\\ -2 & 2 \end{bmatrix} \] Find matrix A such that AB = C.
  • Question 4 For what value(s) of k is each of the matrices given below invertible? a) \( \begin{bmatrix} k & -1 & 4\\ 2 & 0 & 1\\ -1 & 0 & -1 \end{bmatrix} \) , b) \( \begin{bmatrix} k & -1 \\ -1 & 3 \end{bmatrix} \) , c) \( \begin{bmatrix} k & -1 & 4\\ 0 & k + 1 & 1\\ 0 & 0 & k -3 \end{bmatrix} \)
  • Question 6 Matrix A is given by \( A = \begin{bmatrix} a & 0 & 0 & 0\\ 0 & b & 0 & 0 \\ 0 & 0 & c & 0\\ 0 & 0 & 0 & d \end{bmatrix} \) Find a formula for the inverse of matrix A if none of the parameters a, b, c and d is equal to zero.
  • Question 7 Use the inverse matrix to solve the system of equations \( \begin{bmatrix} 1&0&1&2\\ -1& 1 & 2 & 0 \\ -2& 0 & 1 & 2 \\ 0 & 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} x_1\\ x_2 \\ x_3\\ x_4 \end{bmatrix} = \begin{bmatrix} 0\\ 1 \\ -1\\ 2 \end{bmatrix} \)
  • Question 8 What is the most efficient method to solve the following systems of equations? \( A X_1 = B_1 \) , \( A X_2 = B_2 \) , \( A X_3 = B_3 \) ... \( A X_i = B_i \)
  • Question 9 A and B are invertible matrices of the same dimension related by: \( A^{-1} = A B \). Find B in terms of A or its inverse.
  • Question 10 1) Give an example of 2 by 2 matrices A and B such that neither A nor B are invertible yet A + B is invertible. 2) Give an example of 2 by 2 matrices A and B such that neither A nor B are invertible yet A - B is invertible.
  • Question 11 Use any of the two methods to find a formula for the inverse of a 2 by 2 matrix.(It is already given above without proof).
  • Question 12 The system of equations in matrix form \( A X = B \) has the following solutions: \( X_1 = \begin{bmatrix} -1\\ 2 \\ 3 \end{bmatrix} \) for \( B_1 = \begin{bmatrix} 2\\ 13 \\ 3 \end{bmatrix} \) , \( X_2 = \begin{bmatrix} 0\\ -1 \\ 1 \end{bmatrix} \) for \( B_2 = \begin{bmatrix} 4\\ 2 \\ 2 \end{bmatrix} \) , \( X_3 = \begin{bmatrix} 1\\ 1 \\ 1 \end{bmatrix} \) for \( B_3 = \begin{bmatrix} 4\\ 5 \\ 3 \end{bmatrix} \). Find X for \( B = \begin{bmatrix} 1\\ -9 \\ -1 \end{bmatrix} \).

Solutions to the Above Questions

  • Solution to Question 2 We first find the minors. \( M_{1,1} = Det \begin{bmatrix} .&.&.\\ .&2&2 \\ .& 0 & 1 \end{bmatrix} = 2\) , \( M_{1,2} = Det \begin{bmatrix} .&.&.\\ 3&.&2 \\ 0& . & 1 \end{bmatrix} = 3\) , \( M_{1,3} = Det \begin{bmatrix} .&.&.\\ 3&2&. \\ 0& 0 & . \end{bmatrix} = 0 \) \( M_{2,1} = Det \begin{bmatrix} .&0&3\\ .&.&. \\ .& 0 & 1 \end{bmatrix} = 0\) , \( M_{2,2} = Det \begin{bmatrix} -1&.&3\\ . & . & . \\ 0& . & 1 \end{bmatrix} = -1\) , \( M_{2,3} = Det \begin{bmatrix} -1&0&.\\ . & . & .\\ 0& 0 & . \end{bmatrix} = 0\) \( M_{3,1} = Det \begin{bmatrix} .&0&3\\ .&2&2 \\ . & . & . \end{bmatrix} = - 6\) , \( M_{3,2} = Det \begin{bmatrix} -1&.&3\\ 3&.&2 \\ . & . & . \end{bmatrix} = - 11\) , \( M_{3,3} = Det \begin{bmatrix} -1&0& .\\ 3&2& . \\ . & . & . \end{bmatrix} = - 2\) Matrix C of cofactors whose entries defined are defined as \( C_{i,j} = (-1)^{i+j} M_{i,j} \) \[ C = \begin{bmatrix} 2&-3&0\\ 0&-1&0 \\ - 6 & 11 & -2 \end{bmatrix} \] We need to find D the determinant of A using the third row (it has 2 zeros!) \( D = A_{3,3} M_{3,3} = - 2 \) The inverse of A is given by \( A^{-1} = \dfrac{1}{D} C^T = -\dfrac{1}{2} \begin{bmatrix} 2&0&-6\\ -3&-1&11\\ 0&0&-2 \end{bmatrix} = \begin{bmatrix} -1&0&3\\ \dfrac{3}{2}& \dfrac{1}{2} & -\dfrac{11}{2}\\ 0&0&1\end{bmatrix}\)
  • Solution to Question 3 Given \( A B = C \) Right multiply both sides by \( B^{-1} \) \( A B B^{-1} = C B^{-1}\) Use associativity on the left side \( A (B B^{-1}) = C B^{-1} \) Simplify \( A I = C B^{-1} \) \( A = C B^{-1} \) Use the formula for the inverse of a 2 by 2 matrix to find the inverse of B. \( Det(B) = -3 \) \( B^{-1} = - \dfrac{1}{3} \begin{bmatrix} 1&1\\ 2& -1 \end{bmatrix} \) \( A = C B^{-1} = \begin{bmatrix} 2 & -1\\ -2 & 2 \end{bmatrix} (- \dfrac{1}{3}) \begin{bmatrix} 1&1\\ 2& -1 \end{bmatrix} = \begin{bmatrix} 0&-1\\ -2/3& 4/3 \end{bmatrix}\) Note: you may check the answer for matrix A by substituting in the equation \( A B = C \)
  • Solution to Question 4 A matrix is invertible if its determinant is not equal to zero. a) Using the second column, Det\( \begin{bmatrix} k & -1 & 4\\ 2 & 0 & 1\\ -1 & 0 & -1 \end{bmatrix} = - 1\) The matrix is invertible for any k real b) Det\( \begin{bmatrix} k & -1 \\ -1 & 3 \end{bmatrix} = 3k - 1\) \( 3k - 1 \ne 0 \) \( k \ne 1/3 \) The matrix in part b) is invertible for all real values of k not equal to 1/3. c) The given matrix is an upper triangular matrix and its determinant is equal to the product of the terms in the diagonal left to right. Det \( \begin{bmatrix} k & -1 & 4\\ 0 & k + 1 & 1\\ 0 & 0 & k -3 \end{bmatrix} k(k+1)(k-3)\) \( k(k+1)(k-3) \ne 0 \) The given matrix is invertible if k is not equal to 0, - 1 or 3.
  • Solution to Question 5 Right multiply the two sides of the equation by \( S^{-1} \) \( P S^{-1} = Q R^{-1} S S^{-1} \) simplify \( P S^{-1} = Q R^{-1} I \) \( P S^{-1} = Q R^{-1} \) Left multiply the two sides of the equation by \( Q^{-1} \) \( Q^{-1} P S^{-1} = Q^{-1} Q R^{-1} \) simplify \( Q^{-1} P S^{-1} = I R^{-1}\) \( Q^{-1} P S^{-1} = R^{-1} \) Take the inverse of both sides \( (Q^{-1} P S^{-1})^{-1} = (R^{-1})^{-1} \) Simplify \( R = S P^{-1} Q \)
  • Solution to Question 6 Write the augmented matrix \( [ A | I ]\) \( \begin{bmatrix} a & 0 & 0 & 0&|&1&0&0&0\\ 0 & b & 0 & 0&|&0&1&0&0 \\ 0 & 0 & c & 0 &|& 0 & 0 & 1 & 0\\ 0 & 0 & 0 & d &|& 0 & 0 & 0 & 1 \end{bmatrix} \) Multiply row (1) by 1/a, row (2) by 1/b, row (3) by 1/c and row (4) by 1/d and simplify \( \begin{bmatrix} 1 & 0 & 0 & 0&|&1/a&0&0&0\\ 0 & 1 & 0 & 0&|&0&1/b&0&0 \\ 0 & 0 & 1 & 0 &|& 0 & 0 & 1/c & 0\\ 0 & 0 & 0 & 1 &|& 0 & 0 & 0 & 1/d \end{bmatrix} \) The inverse of the given matrix is \( A^{-1} = \begin{bmatrix} 1/a&0&0&0\\ 0&1/b&0&0 \\ 0 & 0 & 1/c & 0\\ 0 & 0 & 0 & 1/d \end{bmatrix} \)
  • Solution to Question 7 The system is of the form A X = B with A = \( \begin{bmatrix} 1&0&1&2\\ -1& 1 & 2 & 0 \\ -2& 0 & 1 & 2 \\ 0 & 0 & 0 & 1 \end{bmatrix} \) , \( B =\begin{bmatrix} 0\\ 1 \\ -1\\ 2 \end{bmatrix} \) and \( X = \begin{bmatrix} x_1\\ x_2 \\ x_3\\ x_4 \end{bmatrix} \) Right multiply both sides of the equation by \( A^{-1} \) and simplify. \( A^{-1} A X = A^{-1} B \) \( I_3 X = A^{-1} B , I_3 \) is the 3 by 3 identity matrix Simplify the above \( X = A^{-1} B \) The inverse of matrix A was calculated in question 1 and is given by (it is matrix B in question 1) \( A^{-1} = \begin{bmatrix} 1/3&0&-1/3&0\\ -1&1&-1&4 \\ 2/3 & 0 & 1/3 & -2\\ 0 & 0 & 0 & 1 \end{bmatrix} \) \( \begin{bmatrix} x_1\\ x_2 \\ x_3\\ x_4 \end{bmatrix} = A^{-1} B = \begin{bmatrix} 1/3&0&-1/3&0\\ -1&1&-1&4 \\ 2/3 & 0 & 1/3 & -2\\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 0\\ 1 \\ -1\\ 2 \end{bmatrix} = \begin{bmatrix}\dfrac{1}{3}\\ 10\\ -\dfrac{13}{3}\\ 2\end{bmatrix}\)
  • Solution to Question 8 Since matrix A is common to all the given systems, the most efficient method solving systems of equations of the form \( A X_1 = B_1 \) , \( A X_2 = B_2 \) , \( A X_3 = B_3 \) ... \( A X_2 = B_i \) is to find the inverse of matrix A and solve as follows (see question 7 above) \( X_1 = A^{-1} B_1 \) , \( X_2 = A^{-1} B_2 \) , \( X_3 = A^{-1} B_3 \) ... \( X_i = A^{-1} B_i \)
  • Solution to Question 10 There are many possible answers to both parts of this question. 1) \(A = \begin{bmatrix} 1 & 1\\ 0 & 0 \\ \end{bmatrix} , B = \begin{bmatrix} 0 & 0\\ - 1 & 1 \\ \end{bmatrix} \) , \(A + B = \begin{bmatrix} 1 & 1\\ - 1 & 0 \\ \end{bmatrix} \) 2) \(A = \begin{bmatrix} 3 & 1\\ 0 & 0 \\ \end{bmatrix} , B = \begin{bmatrix} 0 & 0\\ - 1 & 4 \\ \end{bmatrix} \) , \(A - B = \begin{bmatrix} 3 & 1\\ 1 & - 4 \\ \end{bmatrix} \) Check that the determinant of matrices A and B are equal to zero and therefore non invertible. Check that the determinants of A + B and A - B are not equal to zero and therefore invertible.
  • Solution to Question 11 Let \(A = \begin{bmatrix} a & b\\ c & d \\ \end{bmatrix} \) We shall use the method of cofactors. We first calculate the minors \( M_{1,1} = d\) , \( M_{1,2} = c\) , \( M_{2,1} = b\) , \( M_{2,2} = a\) Then the cofactors using the formula: \( C_{i,j} = (-1)^{i+j}M_{i,j} \) \( C_{1,1} = d\) , \( C_{1,2} = - c\) , \( C_{2,1} = - b\) , \( C_{2,2} = a\) The determinant of A is \( D = a d - b c \) \( A^{-1} = \dfrac{1}{a d - b c} \begin{bmatrix} d & - c\\ - d & a \\ \end{bmatrix}^T = \dfrac{1}{a d - b c} \begin{bmatrix} d & - d\\ - c & a \\ \end{bmatrix} \)
  • Solution to Question 12 The solutions may be written in matrix form as follows \( A \begin{bmatrix} -1 & 0 & 1 \\ 2 & - 1 & 1\\ 3 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 5 & 4 & 4\\ 17 & 2 & 5\\ 5 & 2 & 3 \end{bmatrix} \) which gives \( A = \begin{bmatrix} 5 & 4 & 4\\ 17 & 2 & 5\\ 5 & 2 & 3 \end{bmatrix} \begin{bmatrix} -1 & 0 & 1 \\ 2 & - 1 & 1\\ 3 & 1 & 1 \end{bmatrix}^{-1} \) which gives \( A^{-1} = \begin{bmatrix} -1 & 0 & 1 \\ 2 & - 1 & 1\\ 3 & 1 & 1 \end{bmatrix} \begin{bmatrix} 5 & 4 & 4\\ 17 & 2 & 5\\ 5 & 2 & 3 \end{bmatrix}^{-1} \) The solution X is given by \( X = A^{-1} B = \begin{bmatrix} -1 & 0 & 1 \\ 2 & - 1 & 1\\ 3 & 1 & 1 \end{bmatrix} \begin{bmatrix} 5 & 4 & 4\\ 17 & 2 & 5\\ 5 & 2 & 3 \end{bmatrix}^{-1} \begin{bmatrix} 1\\ -9 \\ -1 \end{bmatrix} = \begin{bmatrix} 1\\ -2 \\ -1 \end{bmatrix} \)

More References and Links to Matrices

  • Matrices with Examples and Questions with Solutions .
  • Find the Inverse of a Matrix Using Row Reduction
  • multiplication of matrices using an applet .
  • Find Inverse Matrix - Calculator.
  • Find Inverse of 3 by 3 Matrix - Calculator.
  • Step by Step Solver to Find the Inverse of a 3 by 3 Matrix .
  • Step by Step Solver to Calculate the Determinant of a 3 by 3 Matrix .

Solving Systems with Inverses

Solving a system of linear equations using the inverse of a matrix.

Solving a system of linear equations using the inverse of a matrix requires the definition of two new matrices: [latex]X[/latex] is the matrix representing the variables of the system, and [latex]B[/latex] is the matrix representing the constants. Using matrix multiplication , we may define a system of equations with the same number of equations as variables as

To solve a system of linear equations using an inverse matrix , let [latex]A[/latex] be the coefficient matrix , let [latex]X[/latex] be the variable matrix, and let [latex]B[/latex] be the constant matrix. Thus, we want to solve a system [latex]AX=B[/latex]. For example, look at the following system of equations.

From this system, the coefficient matrix is

The variable matrix is

And the constant matrix is

Then [latex]AX=B[/latex] looks like

Recall the discussion earlier in this section regarding multiplying a real number by its inverse, [latex]\left({2}^{-1}\right)2=\left(\frac{1}{2}\right)2=1[/latex]. To solve a single linear equation [latex]ax=b[/latex] for [latex]x[/latex], we would simply multiply both sides of the equation by the multiplicative inverse (reciprocal) of [latex]a[/latex]. Thus,

The only difference between a solving a linear equation and a system of equations written in matrix form is that finding the inverse of a matrix is more complicated, and matrix multiplication is a longer process. However, the goal is the same—to isolate the variable.

We will investigate this idea in detail, but it is helpful to begin with a [latex]2\times 2[/latex] system and then move on to a [latex]3\times 3[/latex] system.

A General Note: Solving a System of Equations Using the Inverse of a Matrix

Given a system of equations, write the coefficient matrix [latex]A[/latex], the variable matrix [latex]X[/latex], and the constant matrix [latex]B[/latex]. Then

Multiply both sides by the inverse of [latex]A[/latex] to obtain the solution.

If the coefficient matrix does not have an inverse, does that mean the system has no solution?

No, if the coefficient matrix is not invertible, the system could be inconsistent and have no solution, or be dependent and have infinitely many solutions.

Example 7: Solving a 2 × 2 System Using the Inverse of a Matrix

Solve the given system of equations using the inverse of a matrix.

Write the system in terms of a coefficient matrix, a variable matrix, and a constant matrix.

First, we need to calculate [latex]{A}^{-1}[/latex]. Using the formula to calculate the inverse of a 2 by 2 matrix, we have:

Now we are ready to solve. Multiply both sides of the equation by [latex]{A}^{-1}[/latex].

The solution is [latex]\left(-1,1\right)[/latex].

Can we solve for [latex]X[/latex] by finding the product [latex]B{A}^{-1}?[/latex]

No, recall that matrix multiplication is not commutative, so [latex]{A}^{-1}B\ne B{A}^{-1}[/latex]. Consider our steps for solving the matrix equation.

Notice in the first step we multiplied both sides of the equation by [latex]{A}^{-1}[/latex], but the [latex]{A}^{-1}[/latex] was to the left of [latex]A[/latex] on the left side and to the left of [latex]B[/latex] on the right side. Because matrix multiplication is not commutative, order matters.

Example 8: Solving a 3 × 3 System Using the Inverse of a Matrix

Solve the following system using the inverse of a matrix.

Write the equation [latex]AX=B[/latex].

First, we will find the inverse of [latex]A[/latex] by augmenting with the identity.

Multiply row 1 by [latex]\frac{1}{5}[/latex].

Multiply row 1 by 4 and add to row 2.

Add row 1 to row 3.

Multiply row 2 by −3 and add to row 1.

Multiply row 3 by 5.

Multiply row 3 by [latex]\frac{1}{5}[/latex] and add to row 1.

Multiply row 3 by [latex]-\frac{19}{5}[/latex] and add to row 2.

Multiply both sides of the equation by [latex]{A}^{-1}[/latex]. We want [latex]{A}^{-1}AX={A}^{-1}B:[/latex]

The solution is [latex]\left(1,2,0\right)[/latex].

Solve the system using the inverse of the coefficient matrix.

How To: Given a system of equations, solve with matrix inverses using a calculator.

  • Save the coefficient matrix and the constant matrix as matrix variables [latex]\left[A\right][/latex] and [latex]\left[B\right][/latex].
  • Enter the multiplication into the calculator, calling up each matrix variable as needed.
  • If the coefficient matrix is invertible, the calculator will present the solution matrix; if the coefficient matrix is not invertible, the calculator will present an error message.

Example 9: Using a Calculator to Solve a System of Equations with Matrix Inverses

Solve the system of equations with matrix inverses using a calculator

On the matrix page of the calculator, enter the coefficient matrix as the matrix variable [latex]\left[A\right][/latex], and enter the constant matrix as the matrix variable [latex]\left[B\right][/latex].

On the home screen of the calculator, type in the multiplication to solve for [latex]X[/latex], calling up each matrix variable as needed.

Evaluate the expression.

  • Precalculus. Authored by : OpenStax College. Provided by : OpenStax. Located at : http://cnx.org/contents/[email protected]:1/Preface . License : CC BY: Attribution

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  • Published: 16 February 2024

Inverse-designed low-index-contrast structures on a silicon photonics platform for vector–matrix multiplication

  • Vahid Nikkhah   ORCID: orcid.org/0000-0003-0666-2614 1 ,
  • Ali Pirmoradi   ORCID: orcid.org/0009-0007-4653-8470 1 ,
  • Farshid Ashtiani   ORCID: orcid.org/0000-0002-8418-9626 1 , 2 ,
  • Brian Edwards   ORCID: orcid.org/0000-0001-9354-125X 1 ,
  • Firooz Aflatouni   ORCID: orcid.org/0000-0001-9314-2486 1 &
  • Nader Engheta   ORCID: orcid.org/0000-0003-3219-9520 1  

Nature Photonics ( 2024 ) Cite this article

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  • Metamaterials
  • Nanophotonics and plasmonics

Inverse-designed silicon photonic metastructures offer an efficient platform to perform analogue computations with electromagnetic waves. However, due to computational difficulties, scaling up these metastructures to handle a large number of data channels is not trivial. Furthermore, a typical inverse-design procedure is limited to a small computational domain and therefore tends to employ resonant features to achieve its objectives. This results in structures that are narrow-bandwidth and highly sensitive to fabrication errors. Here we employ a two-dimensional (2D) inverse-design method based on the effective index approximation with a low-index contrast constraint. This results in compact amorphous lens systems that are generally feed-forward and low-resonance. We designed and experimentally demonstrated a vector–matrix product for a 2 × 2 matrix and a 3 × 3 matrix. We also designed a 10 × 10 matrix using the proposed 2D computational method. These examples demonstrate that these techniques have the potential to enable larger-scale wave-based analogue computing platforms.

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inverse matrix solved problems pdf

Data availability

Data for Figs. 2 – 4 are available via Zenodo at https://doi.org/10.5281/zenodo.10083901 (ref. 39 ).

Code availability

All codes produced during this research are available from the corresponding author upon reasonable request.

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Acknowledgements

This work is supported in part by the US Air Force Office of Scientific Research (AFOSR) Multidisciplinary University Research Initiative (MURI) grant no. FA9550-21-1-0312 (to N.E.) and in part by the US Office of Naval Research (ONR) grant no. N00014-19-1-2248 (to F. Aflatouni.).

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Vahid Nikkhah, Ali Pirmoradi, Farshid Ashtiani, Brian Edwards, Firooz Aflatouni & Nader Engheta

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N.E., F. Aflatouni. and B.E. conceived the idea, envisioned the experiments and supervised the project. V.N. conducted theoretical analysis, numerical simulations and inverse-design of the structures. A.P., F. Ashtiani. and B.E. prepared the designs for nanofabrication by the Advanced Micro Foundry (AMF) and designed the experiments. A.P. performed the experiments and collected data. All authors reviewed, studied and discussed the experimental and numerical simulation results, and discussed the main outcomes of the project. V.N. prepared the first draft of the main text and the Supplementary Information. All authors subsequently worked on the manuscript up to its completion for submission.

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Nikkhah, V., Pirmoradi, A., Ashtiani, F. et al. Inverse-designed low-index-contrast structures on a silicon photonics platform for vector–matrix multiplication. Nat. Photon. (2024). https://doi.org/10.1038/s41566-024-01394-2

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COMMENTS

  1. PDF 2.5 Inverse Matrices

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  5. PDF Worksheet 8: Matrix algebra and inverses

    Note that in the last case, the inverse matrix to the matrix of rotation by ˚ degrees counterclockwise is the matrix of rotation by ˚degrees clockwise. 7. Use the inverse found in exercise 4 to solve the equation 1 2 2 1 ~x= 1 1 : Solution: We have ~x= 1 2 2 1 1 1 1 = 1 3 1 2 2 1 1 1 = 1=3 1=3 : 8. Use invertibility to prove that the equation ...

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  7. PDF 10 Problems: Inverse Matrix

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  8. PDF Matrices: §2.3 The Inverse of Matrices

    Theorem. Suppose A is an invetible matrix. Then, its inverse is unique. This unique inverse is denoted by A 1: Proof. Since A is invertible, it has at least one inverse. Suppose it has two inverses, B and C: By de nition AB = BA = In = AC = CA: So; B = BIn = B(AC) = (BA)C = InC = C: So, B = C: The proof is complete. Example: Computing Inverse

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    Many answers. Ex: 1 2 2 4 18) Give an example of a matrix which is its own inverse (that is, where Many answers. Ex: −10 −11 9 10 Create your own worksheets like this one with Infinite Algebra 2. Free trial available at KutaSoftware.com

  10. PDF Chapter 8 Matrices II: inverses

    A matrix A is said to be invertible if we can nd a matrix B such that AB = I = BA . The matrix B we call it an inverse of A , and we say that the matrix A is invertible . Observe that A has to be square. A matrix that is not invertible is said to be singular . Example 8.1.1. A real number r regarded as a 1 1 matrix is invertible if

  11. PDF Solving Linear Systems, Continued and The Inverse of a Matrix

    Conclusion. Inverse matrices are an elegant way of solving linear systems. They do have some drawbacks: I They are only applicable when the coe square. cient matrix is. Even in the case of a square matrix, an inverse may not exist. They are hard to compute, at least as complicated as doing Gauss-Jordan elimination.

  12. PDF 1.The Inverse of a Matrix

    1 2 1 b = 0 1 1 . By an easy calculation, the matrix S = 1 3 0 " 1 −2 # 0 is a left inverse for A, because 0 1 0 " −2 0 # 1 2 " 1 0 #

  13. PDF Lecture Notes 1: Matrix Algebra Part B: Determinants and Inverses

    Assuming that the m m matrix A has an inverse, we can: 1. construct new rst m equations by premultiplying the old ones by A 1; 2. construct new second n equations by: premultiplying the new rst m equations by the n m matrix C; then subtracting this product from the old second n equations. The result is.

  14. PDF Matrix inversion of a 3matrix

    Here is the matrix A that we saw in the leaflet on finding cofactors and determinants. Alongside, we have assembled the matrix of cofactors of A. A = 7 2 1 0 3 −1 −3 4 −2 C = −2 3 9 8 −11 −34 −5 7 21 In order to find the inverse of A, we first need to use the matrix of cofactors, C, to create the adjoint of matrix A.

  15. PDF Matrix Inverses and Determinants Date Period

    For each matrix state if an inverse exists. 15) Yes 16) Yes Find the inverse of each matrix. 17) 18) Critical thinking questions: 19) For what value(s) of x does the matrix M have an inverse? M x x All values except and 20) Give an example of a 3×3 matrix that has a determinant of .

  16. Inverse Matrices

    For each of the following 3 × 3 matrices A, determine whether A is invertible and find the inverse A − 1 if exists by computing the augmented matrix [A | I], where I is the 3 × 3 identity matrix. (a) A = [1 3 − 2 2 3 0 0 1 − 1] (b) A = [ 1 0 2 − 1 − 3 2 3 6 − 2]. Let A be the matrix [1 − 1 0 0 1 − 1 0 0 1]. Is the matrix A invertible?

  17. PDF Matrix algebra for beginners, Part I matrices, determinants, inverses

    6 Determinants and the inverse matrix 7 7 Solving systems of linear equations 9 8 Properties of determinants 10 9 Gaussian elimination 11 1. ... a more general problem (this is the kind of thing mathematicians love to do) in which we do not know exactly what the coefficients are (ie: 1, 2/3, 1/2, 1800, 1100):

  18. PDF 4.5 Solving Systems Using Inverse Matrices

    Use matrices to solve the linear system in Example 1. º3x+ 4y = 5 Equation 1 2xº y = º10 Equation 2 SOLUTION Begin by writing the linear system in matrix form, as in Example 1. Then find the inverse of matrix A. Aº1= 3º 1 8 4 = Finally, multiply the matrix of constants by Aº1. X= Aº1B= 4 = 1 = The solution of the system is (º7, º4).

  19. 2.7: Finding the Inverse of a Matrix

    Algorithm 2.7.1: Matrix Inverse Algorithm. Suppose A is an n × n matrix. To find A − 1 if it exists, form the augmented n × 2n matrix [A | I] If possible do row operations until you obtain an n × 2n matrix of the form [I | B] When this has been done, B = A − 1. In this case, we say that A is invertible. If it is impossible to row reduce ...

  20. PDF Problems and Solutions in Matrix Calculus

    Problem 5. A square matrix Aover C is called skew-hermitian if A= A. Show that such a matrix is normal, i.e., we have AA = AA. Problem 6. Let Abe an n nskew-hermitian matrix over C, i.e. A = A. Let U be an n n unitary matrix, i.e., U = U 1. Show that B:= U AUis a skew-hermitian matrix. Problem 7. Let A, X, Y be n nmatrices. Assume that XA= I n ...

  21. Inverse Matrix Questions

    Inverse matrix questions and solutions are given here to help students learn how to find the inverse of different matrices using different formulas and techniques. As we know, matrices are one of the most scoring concepts for students. Finding the inverse matrix is simple for 2×2 matrices.

  22. Inverse Matrix Questions with Solutions

    Example 1 Verify that matrices A and B given below are inverses of each other. Solution Let us find the products AB and BA AB = BA = I 2 and therefore A and B are inverse of each other. The inverse of a square matrix A is denoted as A -1 and is unique. Find the Inverse of a Square Matrix Using the Row Reduction Method

  23. Solving a System of Linear Equations Using the Inverse of a Matrix

    Solution Write the system in terms of a coefficient matrix, a variable matrix, and a constant matrix. \displaystyle A=\left [\begin {array} {cc}3& 8\\ 4& 11\end {array}\right],X=\left [\begin {array} {c}x\\ y\end {array}\right],B=\left [\begin {array} {c}5\\ 7\end {array}\right] A =[ 3 4 8 11], X = [ x y], B =[ 5 7] Then

  24. Inverse-designed low-index-contrast structures on a silicon photonics

    Using inverse design, a 3D silicon photonics platform that can be used for the mathematical operation of vector-matrix multiplication with light is demonstrated, potentially enabling large-scale ...