Inverse Matrix Questions with Solutions

Definition of the Identity Matrix

The identity matrix \( I_n \) is the square matrix of order \( n \times n \) with all entries on the main diagonal equal to \( 1 \) and all other entries equal to \( 0 \).

Examples of identity matrices:

\[ I_2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} , \quad I_3 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} , \quad I_4 = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \]

Definition of the Inverse of a Matrix

Let \( A \) be a square matrix of order \( n \times n \). If there exists a matrix \( B \) of the same order such that

\( A B = I_n = B A \)

where \( I_n \) is the identity matrix of order \( n \times n \), then \( B \) is called the inverse of \( A \), and \( A \) is the inverse of \( B \).

Find the Inverse Using Row Reduction

This method is also called the Gauss–Jordan elimination method.

We first write the augmented matrix

\[ [ A \mid I ] \]

where \( I \) is the identity matrix of the same order as \( A \), and then use elementary row operations to rewrite it as

\[ [ I \mid A^{-1} ] \]

where \( A^{-1} \) is the inverse of \( A \).

An online calculator to find the inverse of a matrix using row reduction is included.

Find the Inverse Using Minors, Cofactors and Adjugate

This method is explained using a numerical example. Matrix \( A \) is given below:

\( A = \begin{bmatrix} -1 & 0 & 1 \\ 2 & -1 & 2 \\ -1 & 2 & 1 \end{bmatrix} \)

Matrix of Minors

The entry \( M_{i,j} \) of the matrix of minors of \( A \) is the determinant obtained by deleting the \( i^{\text{th}} \) row and \( j^{\text{th}} \) column.

To find \( M_{1,1} \), delete row 1 and column 1:

\( M_{1,1} = \det \begin{bmatrix} -1 & 2 \\ 2 & 1 \end{bmatrix} = (-1)(1) - (2)(2) = -1 - 4 = -5 \)

To find \( M_{1,2} \), delete row 1 and column 2:

\( M_{1,2} = \det \begin{bmatrix} 2 & 2 \\ -1 & 1 \end{bmatrix} = (2)(1) - (2)(-1) = 2 + 2 = 4 \)

To find \( M_{1,3} \), delete row 1 and column 3:

\( M_{1,3} = \det \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} = (2)(2) - (-1)(-1) = 4 - 1 = 3 \)

To find \( M_{2,1} \), delete row 2 and column 1:

\( M_{2,1} = \det \begin{bmatrix} 0 & 1 \\ 2 & 1 \end{bmatrix} = (0)(1) - (1)(2) = -2 \)

The remaining minors are:

\( M_{2,2} = 0, \quad M_{2,3} = -2, \quad M_{3,1} = 1, \quad M_{3,2} = -4, \quad M_{3,3} = 1 \)

The matrix of minors \( M \) is:

\( M = \begin{bmatrix} -5 & 4 & 3 \\ -2 & 0 & -2 \\ 1 & -4 & 1 \end{bmatrix} \)

Matrix of Cofactors

The entries \( C_{i,j} \) of the cofactor matrix are given by:

\( C_{i,j} = (-1)^{i+j} M_{i,j} \)

Evaluating the entries:

\( C_{1,1} = -5,\; C_{1,2} = -4,\; C_{1,3} = 3,\; C_{2,1} = 2,\; C_{2,2} = 0,\; C_{2,3} = -2,\; C_{3,1} = 1,\; C_{3,2} = 4,\; C_{3,3} = 1 \)

Hence:

\( C = \begin{bmatrix} -5 & -4 & 3 \\ 2 & 0 & -2 \\ 1 & 4 & 1 \end{bmatrix} \)

Adjugate of a Matrix

The adjugate (also called the classical adjoint) of \( A \) is the transpose of its cofactor matrix:

\( \operatorname{adj}(A) = C^{T} = \begin{bmatrix} -5 & 2 & 1 \\ -4 & 0 & 4 \\ 3 & -2 & 1 \end{bmatrix} \)

Inverse Matrix

Find the determinant \( D \) of \( A \) using the first row and the minors already found:

\( D = (-1)(-5) - (0)(4) + (1)(3) = 5 + 0 + 3 = 8 \)

The inverse of \( A \) is:

\( A^{-1} = \dfrac{1}{D} C^{T} = \dfrac{1}{8} \begin{bmatrix} -5 & 2 & 1 \\ -4 & 0 & 4 \\ 3 & -2 & 1 \end{bmatrix} = \begin{bmatrix} -\tfrac{5}{8} & \tfrac{1}{4} & \tfrac{1}{8} \\ -\tfrac{1}{2} & 0 & \tfrac{1}{2} \\ \tfrac{3}{8} & -\tfrac{1}{4} & \tfrac{1}{8} \end{bmatrix} \)

Formula for the Inverse of a 2 by 2 Matrix

Using either of the two methods above, the inverse of

\( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \)

is given by

\( A^{-1} = \dfrac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \)

provided \( ad - bc \neq 0 \).

Properties of Inverse Matrices

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

1. If \( A \) is invertible, its inverse is unique.
2. \( A A^{-1} = A^{-1} A = I \)
3. If \( A \) and \( B \) are invertible, then \( (AB)^{-1} = B^{-1} A^{-1} \).
4. A square matrix is invertible if and only if its determinant is not equal to zero.
5. A matrix whose determinant is not equal to zero is called nonsingular.
6. \( (A^{T})^{-1} = (A^{-1})^{T} \)
7. \( \det(A^{-1}) = \dfrac{1}{\det(A)} \)
8. \( (A^{-1})^{-1} = A \)

Questions on Inverse Matrices

1. Question 1
   Use the row reduction method to find the inverse of the following matrices:

   \( A = \begin{bmatrix} -1 & -1 & 1 \\ 2 & 0 & -2 \\ 1 & 1 & 1 \end{bmatrix}, \quad B = \begin{bmatrix} 1 & 0 & 1 & 2 \\ -1 & 1 & 2 & 0 \\ -2 & 0 & 1 & 2 \\ 0 & 0 & 0 & 1 \end{bmatrix} \)

2. Question 2
   Use the method of cofactors to find the inverse of \( A = \begin{bmatrix} -1 & 0 & 3 \\ 3 & 2 & 2 \\ 0 & 0 & 1 \end{bmatrix} \).
3. Question 3
   \( A, B, C \) are 2 by 2 matrices with \( B = \begin{bmatrix} -1 & -1 \\ -2 & 1 \end{bmatrix}, \; C = \begin{bmatrix} 2 & -1 \\ -2 & 2 \end{bmatrix} \). Find matrix \( A \) such that \( AB = C \).
4. Question 4
   For what value(s) of \( k \) is each matrix invertible?

   a) \( \begin{bmatrix} k & -1 & 4 \\ 2 & 0 & 1 \\ -1 & 0 & -1 \end{bmatrix} , \quad \) b) \( \begin{bmatrix} k & -1 \\ -1 & 3 \end{bmatrix}, \quad \) c) \( \begin{bmatrix} k & -1 & 4 \\ 0 & k+1 & 1 \\ 0 & 0 & k-3 \end{bmatrix} \)

5. Question 5
   Square invertible matrices \( P, Q, R, S \) of the same dimension satisfy \( P = Q R^{-1} S \). Express \( R \) in terms of \( P, Q, S \) or their inverses.
6. 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 \( A^{-1} \) assuming none of \( a, b, c, d \) is zero.
7. Question 7
   Use the inverse matrix to solve \[ \begin{bmatrix} 1 & 0 & 1 & 2 \\ -1 & 1 & 2 & 0 \\ -2 & 0 & 1 & 2 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ -1 \\ 2 \end{bmatrix} \]
8. Question 8
   What is the most efficient method to solve the systems \( A X_1 = B_1, \; A X_2 = B_2, \; A X_3 = B_3, \ldots, A X_i = B_i \)?
9. Question 9
   Invertible matrices \( A \) and \( B \) of the same dimension satisfy \( A^{-1} = A B \). Find \( B \) in terms of \( A \) or its inverse.
10. Question 10
    
    1. Give an example of 2 by 2 matrices \( A \) and \( B \) such that neither is invertible yet \( A + B \) is invertible.
    2. Give an example of 2 by 2 matrices \( A \) and \( B \) such that neither is invertible yet \( A - B \) is invertible.
11. Question 11
    Use any of the two methods to find a formula for the inverse of a 2 by 2 matrix.
12. Question 12
    The system \( A X = B \) has the solutions \[ X_1 = \begin{bmatrix} -1 \\ 2 \\ 3 \end{bmatrix} \text{ for } B_1 = \begin{bmatrix} 2 \\ 13 \\ 3 \end{bmatrix}, \quad X_2 = \begin{bmatrix} 0 \\ -1 \\ 1 \end{bmatrix} \text{ for } B_2 = \begin{bmatrix} 4 \\ 2 \\ 2 \end{bmatrix}, \quad X_3 = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \text{ for } B_3 = \begin{bmatrix} 4 \\ 5 \\ 3 \end{bmatrix}. \] Find \( X \) for \( B = \begin{bmatrix} 1 \\ -9 \\ -1 \end{bmatrix} \).

Solutions to the Above Questions

1. Solution to Question 1

   Inverse of matrix \( A \)

   Write the augmented matrix \( [ A \mid I ] \):

   \( \begin{bmatrix} -1 & -1 & 1 & | & 1 & 0 & 0 \\ 2 & 0 & -2 & | & 0 & 1 & 0 \\ 1 & 1 & 1 & | & 0 & 0 & 1 \end{bmatrix} \)

   Step 1 — \( R_1 \leftarrow -R_1 \):

   \( \begin{bmatrix} 1 & 1 & -1 & | & -1 & 0 & 0 \\ 2 & 0 & -2 & | & 0 & 1 & 0 \\ 1 & 1 & 1 & | & 0 & 0 & 1 \end{bmatrix} \)

   Step 2 — \( R_2 \leftarrow R_2 - 2R_1 \):

   \( \begin{bmatrix} 1 & 1 & -1 & | & -1 & 0 & 0 \\ 0 & -2 & 0 & | & 2 & 1 & 0 \\ 1 & 1 & 1 & | & 0 & 0 & 1 \end{bmatrix} \)

   Step 3 — \( R_3 \leftarrow R_3 - R_1 \):

   \( \begin{bmatrix} 1 & 1 & -1 & | & -1 & 0 & 0 \\ 0 & -2 & 0 & | & 2 & 1 & 0 \\ 0 & 0 & 2 & | & 1 & 0 & 1 \end{bmatrix} \)

   Step 4 — \( R_2 \leftarrow -\tfrac{1}{2}R_2 \), \( R_3 \leftarrow \tfrac{1}{2}R_3 \):

   \( \begin{bmatrix} 1 & 1 & -1 & | & -1 & 0 & 0 \\ 0 & 1 & 0 & | & -1 & -\tfrac{1}{2} & 0 \\ 0 & 0 & 1 & | & \tfrac{1}{2} & 0 & \tfrac{1}{2} \end{bmatrix} \)

   Step 5 — \( R_1 \leftarrow R_1 + R_3 \):

   \( \begin{bmatrix} 1 & 1 & 0 & | & -\tfrac{1}{2} & 0 & \tfrac{1}{2} \\ 0 & 1 & 0 & | & -1 & -\tfrac{1}{2} & 0 \\ 0 & 0 & 1 & | & \tfrac{1}{2} & 0 & \tfrac{1}{2} \end{bmatrix} \)

   Step 6 — \( R_1 \leftarrow R_1 - R_2 \):

   \( \begin{bmatrix} 1 & 0 & 0 & | & \tfrac{1}{2} & \tfrac{1}{2} & \tfrac{1}{2} \\ 0 & 1 & 0 & | & -1 & -\tfrac{1}{2} & 0 \\ 0 & 0 & 1 & | & \tfrac{1}{2} & 0 & \tfrac{1}{2} \end{bmatrix} \)

   \( A^{-1} = \begin{bmatrix} \tfrac{1}{2} & \tfrac{1}{2} & \tfrac{1}{2} \\ -1 & -\tfrac{1}{2} & 0 \\ \tfrac{1}{2} & 0 & \tfrac{1}{2} \end{bmatrix} \)

   Verification: \( A A^{-1} = \begin{bmatrix} -1 & -1 & 1 \\ 2 & 0 & -2 \\ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} 1/2 & 1/2 & 1/2 \\ -1 & -1/2 & 0 \\ 1/2 & 0 & 1/2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I_3 \)

   Inverse of matrix \( B \)

   Write the augmented matrix \( [ B \mid I ] \):

   \( \begin{bmatrix} 1 & 0 & 1 & 2 & | & 1 & 0 & 0 & 0 \\ -1 & 1 & 2 & 0 & | & 0 & 1 & 0 & 0 \\ -2 & 0 & 1 & 2 & | & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & | & 0 & 0 & 0 & 1 \end{bmatrix} \)

   Step 1 — \( R_2 \leftarrow R_2 + R_1 \), \( R_3 \leftarrow R_3 + 2R_1 \):

   \( \begin{bmatrix} 1 & 0 & 1 & 2 & | & 1 & 0 & 0 & 0 \\ 0 & 1 & 3 & 2 & | & 1 & 1 & 0 & 0 \\ 0 & 0 & 3 & 6 & | & 2 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & | & 0 & 0 & 0 & 1 \end{bmatrix} \)

   Step 2 — \( R_3 \leftarrow \tfrac{1}{3}R_3 \):

   \( \begin{bmatrix} 1 & 0 & 1 & 2 & | & 1 & 0 & 0 & 0 \\ 0 & 1 & 3 & 2 & | & 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 2 & | & \tfrac{2}{3} & 0 & \tfrac{1}{3} & 0 \\ 0 & 0 & 0 & 1 & | & 0 & 0 & 0 & 1 \end{bmatrix} \)

   Step 3 — \( R_1 \leftarrow R_1 - 2R_4 \), \( R_2 \leftarrow R_2 - 2R_4 \), \( R_3 \leftarrow R_3 - 2R_4 \):

   \( \begin{bmatrix} 1 & 0 & 1 & 0 & | & 1 & 0 & 0 & -2 \\ 0 & 1 & 3 & 0 & | & 1 & 1 & 0 & -2 \\ 0 & 0 & 1 & 0 & | & \tfrac{2}{3} & 0 & \tfrac{1}{3} & -2 \\ 0 & 0 & 0 & 1 & | & 0 & 0 & 0 & 1 \end{bmatrix} \)

   Step 4 — \( R_1 \leftarrow R_1 - R_3 \), \( R_2 \leftarrow R_2 - 3R_3 \):

   \( \begin{bmatrix} 1 & 0 & 0 & 0 & | & \tfrac{1}{3} & 0 & -\tfrac{1}{3} & 0 \\ 0 & 1 & 0 & 0 & | & -1 & 1 & -1 & 4 \\ 0 & 0 & 1 & 0 & | & \tfrac{2}{3} & 0 & \tfrac{1}{3} & -2 \\ 0 & 0 & 0 & 1 & | & 0 & 0 & 0 & 1 \end{bmatrix} \)

   \( B^{-1} = \begin{bmatrix} \tfrac{1}{3} & 0 & -\tfrac{1}{3} & 0 \\ -1 & 1 & -1 & 4 \\ \tfrac{2}{3} & 0 & \tfrac{1}{3} & -2 \\ 0 & 0 & 0 & 1 \end{bmatrix} \)

   Verification (partial): Check that \(B B^{-1}\) equals \(I_4\). Row 1 of \(B\) times column 1 of \(B^{-1}\): \(1(1/3)+0(-1)+1(2/3)+2(0) = 1\).
   Row 1 of \(B\) times column 4 of \(B^{-1}\): \(1(0)+0(4)+1(-2)+2(1) = 0\).
2. Solution to Question 2

   Find the minors of \( A = \begin{bmatrix} -1 & 0 & 3 \\ 3 & 2 & 2 \\ 0 & 0 & 1 \end{bmatrix} \):

   \( M_{1,1} = \det \begin{bmatrix} 2 & 2 \\ 0 & 1 \end{bmatrix} = 2, \quad M_{1,2} = \det \begin{bmatrix} 3 & 2 \\ 0 & 1 \end{bmatrix} = 3, \quad M_{1,3} = \det \begin{bmatrix} 3 & 2 \\ 0 & 0 \end{bmatrix} = 0 \)

   \( M_{2,1} = \det \begin{bmatrix} 0 & 3 \\ 0 & 1 \end{bmatrix} = 0, \quad M_{2,2} = \det \begin{bmatrix} -1 & 3 \\ 0 & 1 \end{bmatrix} = -1, \quad M_{2,3} = \det \begin{bmatrix} -1 & 0 \\ 0 & 0 \end{bmatrix} = 0 \)

   \( M_{3,1} = \det \begin{bmatrix} 0 & 3 \\ 2 & 2 \end{bmatrix} = -6, \quad M_{3,2} = \det \begin{bmatrix} -1 & 3 \\ 3 & 2 \end{bmatrix} = -11, \quad M_{3,3} = \det \begin{bmatrix} -1 & 0 \\ 3 & 2 \end{bmatrix} = -2 \)

   Matrix of cofactors \( C_{i,j} = (-1)^{i+j}M_{i,j} \):

   \( C = \begin{bmatrix} 2 & -3 & 0 \\ 0 & -1 & 0 \\ -6 & 11 & -2 \end{bmatrix} \)

   Determinant using the third row (two zeros):

   \( D = A_{3,3} M_{3,3} = (1)(-2) = -2 \)

   \( A^{-1} = \dfrac{1}{D} C^{T} = -\dfrac{1}{2} \begin{bmatrix} 2 & 0 & -6 \\ -3 & -1 & 11 \\ 0 & 0 & -2 \end{bmatrix} = \begin{bmatrix} -1 & 0 & 3 \\ \tfrac{3}{2} & \tfrac{1}{2} & -\tfrac{11}{2} \\ 0 & 0 & 1 \end{bmatrix} \)

   Verification (partial): Row 1 of \(A\) times column 3 of \(A^{-1}\): \((-1)(3)+(0)(-11/2)+(3)(1) = -3+3 = 0\).
   Row 2 of \(A\) times column 1 of \(A^{-1}\): \(3(-1)+2(3/2)+2(0) = -3+3 = 0\).
3. Solution to Question 3

   Given \( A B = C \), right-multiply both sides by \( B^{-1} \):

   \( A B B^{-1} = C B^{-1} \Rightarrow A = C B^{-1} \)

   Compute the inverse of \( B \) using the 2 by 2 formula. Since \( \det(B) = (-1)(1) - (-1)(-2) = -1 - 2 = -3 \):

   \( B^{-1} = -\dfrac{1}{3} \begin{bmatrix} 1 & 1 \\ 2 & -1 \end{bmatrix} \)

   Therefore:

   \( A = C B^{-1} = \begin{bmatrix} 2 & -1 \\ -2 & 2 \end{bmatrix} \left(-\dfrac{1}{3}\right) \begin{bmatrix} 1 & 1 \\ 2 & -1 \end{bmatrix} = \begin{bmatrix} 0 & -1 \\ -\tfrac{2}{3} & \tfrac{4}{3} \end{bmatrix} \)

   \( A = \begin{bmatrix} 0 & -1 \\ -\tfrac{2}{3} & \tfrac{4}{3} \end{bmatrix} \)

   Verification: \( A B = \begin{bmatrix} 0 & -1 \\ -2/3 & 4/3 \end{bmatrix} \begin{bmatrix} -1 & -1 \\ -2 & 1 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ -2 & 2 \end{bmatrix} = C \).
4. Solution to Question 4

   A matrix is invertible if and only if its determinant is not zero.

   a) Expand along the second column (which has two zeros):

   \( \det \begin{bmatrix} k & -1 & 4 \\ 2 & 0 & 1 \\ -1 & 0 & -1 \end{bmatrix} = -(-1) \cdot \det \begin{bmatrix} 2 & 1 \\ -1 & -1 \end{bmatrix} = (1)\bigl((2)(-1) - (1)(-1)\bigr) = (1)(-2+1) = -1 \)

   The determinant is \(-1 \neq 0\) for every real \( k \). So the matrix is invertible for all real \( k \).

   b)

   \( \det \begin{bmatrix} k & -1 \\ -1 & 3 \end{bmatrix} = 3k - 1 \)

   Set \( 3k - 1 \neq 0 \Rightarrow k \neq \tfrac{1}{3} \).

   The matrix is invertible for all real \( k \neq \tfrac{1}{3} \).

   c) The matrix is upper triangular, so its determinant is the product of the diagonal entries:

   \( \det = k(k+1)(k-3) \)

   The matrix is invertible when \( k(k+1)(k-3) \neq 0 \), i.e. \( k \neq 0, -1, 3 \).

5. Solution to Question 5

   Start with \( P = Q R^{-1} S \). Right-multiply by \( S^{-1} \) and simplify:

   \( P S^{-1} = Q R^{-1}S S^{-1}\)

   \( P S^{-1} = Q R^{-1} \)

   Left-multiply by \( Q^{-1} \):

   \( Q^{-1} P S^{-1} = R^{-1} \)

   Take the inverse of both sides (using \( (XYZ)^{-1} = Z^{-1}Y^{-1}X^{-1} \)):

   \( R = (Q^{-1} P S^{-1})^{-1} = S P^{-1} Q \)

6. Solution to Question 6

   Write the augmented matrix \( [ A \mid I ] \):

   \( \begin{bmatrix} a & 0 & 0 & 0 & | & 1 & 0 & 0 & 0 \\ 0 & b & 0 & 0 & | & 0 & 1 & 0 & 0 \\ 0 & 0 & c & 0 & | & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & d & | & 0 & 0 & 0 & 1 \end{bmatrix} \)

   Multiply each row by the reciprocal of its diagonal entry:

   \( \begin{bmatrix} 1 & 0 & 0 & 0 & | & 1/a & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & | & 0 & 1/b & 0 & 0 \\ 0 & 0 & 1 & 0 & | & 0 & 0 & 1/c & 0 \\ 0 & 0 & 0 & 1 & | & 0 & 0 & 0 & 1/d \end{bmatrix} \)

   \( A^{-1} = \begin{bmatrix} 1/a & 0 & 0 & 0 \\ 0 & 1/b & 0 & 0 \\ 0 & 0 & 1/c & 0 \\ 0 & 0 & 0 & 1/d \end{bmatrix} \)

7. Solution to Question 7
   Hint: The coefficient matrix here is the same as matrix \( B \) from Question 1.

   The system is \( A X = B \) where

   \( A = \begin{bmatrix} 1 & 0 & 1 & 2 \\ -1 & 1 & 2 & 0 \\ -2 & 0 & 1 & 2 \\ 0 & 0 & 0 & 1 \end{bmatrix}, \quad B = \begin{bmatrix} 0 \\ 1 \\ -1 \\ 2 \end{bmatrix}, \quad X = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} \)

   Then \( X = A^{-1} B \). From Question 1 (matrix \(B\)),

   \( A^{-1} = \begin{bmatrix} \tfrac{1}{3} & 0 & -\tfrac{1}{3} & 0 \\ -1 & 1 & -1 & 4 \\ \tfrac{2}{3} & 0 & \tfrac{1}{3} & -2 \\ 0 & 0 & 0 & 1 \end{bmatrix} \)

   Therefore:

   \( X = A^{-1} B = \begin{bmatrix} \tfrac{1}{3} & 0 & -\tfrac{1}{3} & 0 \\ -1 & 1 & -1 & 4 \\ \tfrac{2}{3} & 0 & \tfrac{1}{3} & -2 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \\ -1 \\ 2 \end{bmatrix} = \begin{bmatrix} \tfrac{1}{3} \\ 10 \\ -\tfrac{13}{3} \\ 2 \end{bmatrix} \)

   \( x_1 = \tfrac{1}{3}, \; x_2 = 10, \; x_3 = -\tfrac{13}{3}, \; x_4 = 2 \)

   Verification (first equation): \( x_1 + x_3 + 2x_4 = \tfrac{1}{3} - \tfrac{13}{3} + 4 = -4 + 4 = 0\).
8. Solution to Question 8

   Since matrix \( A \) is common to all systems of the form

   \( A X_1 = B_1, \; A X_2 = B_2, \; A X_3 = B_3, \ldots, A X_i = B_i \)

   the most efficient method is to compute \( A^{-1} \) once, then use

   \( X_1 = A^{-1} B_1, \; X_2 = A^{-1} B_2, \; X_3 = A^{-1} B_3, \ldots, X_i = A^{-1} B_i \)

9. Solution to Question 9

   Given \( A^{-1} = A B \), right-multiply both sides by \( A^{-1} \):

   \( A^{-1} A^{-1} = A^{-1} A B \Rightarrow A^{-2} = I B \)

   \( B = A^{-2} \)

10. Solution to Question 10

    Many examples are possible.

    1)

    \( A = \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}, \quad B = \begin{bmatrix} 0 & 0 \\ -1 & 1 \end{bmatrix}, \quad A + B = \begin{bmatrix} 1 & 1 \\ -1 & 1 \end{bmatrix} \)

    Both \( A \) and \( B \) are singular (\( \det = 0 \)), but \( \det(A+B) = 1 - (-1) = 2 \neq 0 \).

    2)

    \( A = \begin{bmatrix} 3 & 1 \\ 0 & 0 \end{bmatrix}, \quad B = \begin{bmatrix} 0 & 0 \\ -1 & 4 \end{bmatrix}, \quad A - B = \begin{bmatrix} 3 & 1 \\ 1 & -4 \end{bmatrix} \)

    Both \( A \) and \( B \) are singular, but \( \det(A-B) = -12 - 1 = -13 \neq 0 \).

11. Solution to Question 11

    Let \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \). Use the cofactor method.

    Minors: \( M_{1,1} = d, \; M_{1,2} = c, \; M_{2,1} = b, \; M_{2,2} = a \).

    Cofactors: \( C_{1,1} = d, \; C_{1,2} = -c, \; C_{2,1} = -b, \; C_{2,2} = a \).

    Determinant: \( D = ad - bc \).

    The adjugate is the transpose of the cofactor matrix:

    \( \operatorname{adj}(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \)

    \( A^{-1} = \dfrac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}, \quad ad - bc \neq 0 \)

12. Solution to Question 12

    The three given solutions can be written together in a single matrix equation:

    \[ A \begin{bmatrix} -1 & 0 & 1 \\ 2 & -1 & 1 \\ 3 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 4 & 4 \\ 13 & 2 & 5 \\ 3 & 2 & 3 \end{bmatrix} \]

    This follows because each column of the left matrix is \(X_i\), and each column of the right matrix is the corresponding \(B_i\), with \(A X_i = B_i\).

    Solving for \(A\) gives

    \[ A = \begin{bmatrix} 2 & 4 & 4 \\ 13 & 2 & 5 \\ 3 & 2 & 3 \end{bmatrix} \begin{bmatrix} -1 & 0 & 1 \\ 2 & -1 & 1 \\ 3 & 1 & 1 \end{bmatrix}^{-1} \]

    Taking inverses on both sides (using \((XY)^{-1} = Y^{-1}X^{-1}\)) gives

    \[ A^{-1} = \begin{bmatrix} -1 & 0 & 1 \\ 2 & -1 & 1 \\ 3 & 1 & 1 \end{bmatrix} \begin{bmatrix} 2 & 4 & 4 \\ 13 & 2 & 5 \\ 3 & 2 & 3 \end{bmatrix}^{-1} \]

    The solution \(X\) for the new right-hand side \(B = \begin{bmatrix} 1 \\ -9 \\ -1 \end{bmatrix}\) is

    \[ X = A^{-1} B = \begin{bmatrix} -1 & 0 & 1 \\ 2 & -1 & 1 \\ 3 & 1 & 1 \end{bmatrix} \begin{bmatrix} 2 & 4 & 4 \\ 13 & 2 & 5 \\ 3 & 2 & 3 \end{bmatrix}^{-1} \begin{bmatrix} 1 \\ -9 \\ -1 \end{bmatrix} = \begin{bmatrix} 1 \\ -2 \\ -\tfrac{11}{6} \end{bmatrix} \]

    \( X = \begin{bmatrix} 1 \\ -2 \\ -\tfrac{11}{6} \end{bmatrix} \)

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