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.

Find the Inverse of a Square Matrix Using Minors, Cofactors and Adjugate

This method is explained using a numerical example. Matrix A is given below.
\[ A = \begin{bmatrix} -1&0&1\\ 2&-1&2 \\ -1 & 2 & 1 \end{bmatrix} \] 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

Formula for the Inverse of a 2 by 2 Matrix

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

Properties of Inverse Matrices

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

1. If A is an invertible matrix, its inverse is unique.
   
2. \( A A^{-1} = A^{-1} A = I \)
   
3. If matrices A and B are invertible, then:\( (AB)^{-1} = B^{-1}A^{-1} \)
   
4. A 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

• 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 5
  The square matrices P, Q, R and S are of the same dimension and invertible such that \[ P = Q R^{-1} S \] Express (or determine) \( R\) in terms of \( P \), \( Q\) and \( S \) or/and their inverses.
  
  
• 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 1
  Inverse of matrix A
  Write the augmented matrix \( [ A | I )\) \[ \begin{bmatrix} -1&-1&1&|&1&0&0\\ 2&0&-2&|&0&1&0 \\ 1 & 1 & 1 &|& 0 & 0 & 1 \end{bmatrix} \]
  step 1 \[ \color{red}{ \begin{matrix} -R_1\\ \\ \\ \end{matrix} } \begin{bmatrix} 1&1&-1&|&-1&0&0\\ 2&0&-2&|&0&1&0 \\ 1 & 1 & 1 &|& 0 & 0 & 1 \end{bmatrix} \]
  step 2 \[ \color{red}{ \begin{matrix} \\ R_2-2R_1 \\ \\ \end{matrix} } \begin{bmatrix} 1&1&-1&|&-1&0&0\\ 0&-2&0&|&2&1&0 \\ 1 & 1 & 1 &|& 0 & 0 & 1 \end{bmatrix} \]
  step 3 \[ \color{red}{ \begin{matrix} \\ \\ R_3-R_1\\ \end{matrix} } \begin{bmatrix} 1&1&-1&|&-1&0&0\\ 0&-2&0&|&2&1&0 \\ 0 & 0 & 2 &|& 1 & 0 & 1 \end{bmatrix} \]
  step 4 \[ \color{red}{ \begin{matrix} \\ R_2=(-1/2)R_2\\ R_3=(1/2)R_3\\ \end{matrix} } \begin{bmatrix} 1&1&-1&|&-1&0&0\\ 0&1&0&|&-1&-1/2&0 \\ 0 & 0 & 1 &|& 1/2 & 0 & 1/2 \end{bmatrix} \]
  step 5 \[ \color{red}{ \begin{matrix} R_1+R_3\\ \\ \\ \end{matrix} } \begin{bmatrix} 1&1&0&|&-1/2&0&1/2\\ 0&1&0&|&-1&-1/2&0 \\ 0 & 0 & 1 &|& 1/2 & 0 & 1/2 \end{bmatrix} \]
  step 6 \[ \color{red}{ \begin{matrix} R_1-R_2\\ \\ \\ \end{matrix} } \begin{bmatrix} 1&0&0&|&1/2&1/2&1/2\\ 0&1&0&|&-1&-1/2&0 \\ 0 & 0 & 1 &|& 1/2 & 0 & 1/2 \end{bmatrix} \]
  The inverse of matrix A is the 3 by 3 matrix on the right side. Hence
  \[ A^{-1} = \begin{bmatrix} 1/2&1/2&1/2\\ -1&-1/2&0 \\ 1/2 & 0 & 1/2 \end{bmatrix} \]
  
  Inverse of matrix B
  Write the augmented matrix \( [ A | I )\) \[ \begin{bmatrix} 1&0&1&2 & |&1&0&0&0\\ -1& 1 & 2 & 0 &|&0&1&0&0 \\ -2& 0 & 1 & 2 &|& 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 & |& 0 & 0 & 0 & 1 \end{bmatrix} \]
  step 1 \[ \color{red}{ \begin{matrix} \\ R_2+R_1\\ R_3+2R_1\\ \\ \end{matrix} } \begin{bmatrix} 1&0&1&2 & |&1&0&0&0\\ 0& 1 & 3 & 2 &|&1&1&0&0 \\ 0& 0 & 3 & 6 &|& 2 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 & |& 0 & 0 & 0 & 1 \end{bmatrix} \]
  
  step 2 \[ \color{red}{ \begin{matrix} \\ \\ (1/3) R_3\\ \\ \end{matrix} } \begin{bmatrix} 1&0&1&2 & |&1&0&0&0\\ 0& 1 & 3 & 2 &|&1&1&0&0 \\ 0& 0 & 1 & 2 &|& 2/3 & 0 & 1/3 & 0\\ 0 & 0 & 0 & 1 & |& 0 & 0 & 0 & 1 \end{bmatrix} \]
  step 3 \[ \color{red}{ \begin{matrix} R_1-2R_4\\ R_2-2R_4\\ R_3-2R_4\\ \\ \end{matrix} } \begin{bmatrix} 1&0&1&0 & |&1&0&0&-2\\ 0& 1 & 3 & 0 &|&1&1&0&-2 \\ 0& 0 & 1 & 0 &|& 2/3 & 0 & 1/3 & -2\\ 0 & 0 & 0 & 1 & |& 0 & 0 & 0 & 1 \end{bmatrix} \]
  step 4 \[ \color{red}{ \begin{matrix} R_1-R_3\\ R_2-3R_3\\ \\ \\ \end{matrix} } \begin{bmatrix} 1&0&1&0 & |&1/3&0&-1/3&0\\ 0& 1 & 0 & 0 &|&-1&1&-1&4 \\ 0& 0 & 1 & 0 &|& 2/3 & 0 & 1/3 & -2\\ 0 & 0 & 0 & 1 & |& 0 & 0 & 0 & 1 \end{bmatrix} \]
  The inverse of matrix B is the 4 by 4 matrix on the right side. Hence
  \[ B^{-1} = \begin{bmatrix} 1/3&0&-1/3&0\\ -1&1&-1&4 \\ 2/3 & 0 & 1/3 & -2\\ 0 & 0 & 0 & 1 \end{bmatrix} \]
  
• 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 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.
  Right multiply the equation by \( A^{-1} \)
  \( A^{-1} A^{-1} = A^{-1} A B \)
  simplify
  \( A^{-1} A^{-1} = I B \)
  simplify to
  \( B = A^{-2}\)
  
• 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.