|
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
\[ A^{-1} =
\begin{bmatrix}
-7/12&1/6&1/4\\
-1/12&1/6&1/4 \\
5/12 & 1/6 & -1/4
\end{bmatrix} \]
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
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
\( A^{-1} = \dfrac{1}{D} C^T = \dfrac{1}{8} \begin{bmatrix} -5&2&1\\ -4&0&4\\ 3&2&1 \end{bmatrix} = \begin{bmatrix} -\dfrac{5}{8}&\dfrac{1}{4}&\dfrac{1}{8}\\ -\dfrac{1}{2}&0&\dfrac{1}{2}\\ \dfrac{3}{8}&\dfrac{1}{4}&\dfrac{1}{8}\end{bmatrix}\)
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.
- 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.
- \( (A^T)^{-1} = (A^{-1})^T \)
- \( Det(A^{-1}) = \dfrac{1}{Det(A)} \)
- \( (A^{-1})^{-1} = A \)
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
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