Inverse Matrices

Verify if the following are inverse matrices or not.

a.

$A=\left[\begin{array}{cc}2& 2\\ -1& 4\end{array}\right]andB=\left[\begin{array}{cc}\frac{1}{2}& \frac{1}{2}\\ -1& \frac{1}{4}\end{array}\right]$

b.

$M=\left[\begin{array}{cc}3& 4\\ 1& 2\end{array}\right]andN=\left[\begin{array}{cc}1& -2\\ -\frac{1}{2}& \frac{3}{2}\end{array}\right]$

Solution:

a. find the product between matrix A and B;

$$\begin{gathered} A\times B=\left[\begin{array}{cc}2& 2\\ -1& 4\end{array}\right]\times \left[\begin{array}{cc}\frac{1}{2}& \frac{1}{2}\\ -1& \frac{1}{4}\end{array}\right] \\ A\times B=\left[\begin{array}{cc}(2\times \frac{1}{2})+(2\times (-1\left)\right)& (2\times \frac{1}{2})+(2\times \frac{1}{4})\\ (-1\times \frac{1}{2})+(4\times (-1\left)\right)& (-1\times \frac{1}{2})+(4\times \frac{1}{4})\end{array}\right] \\ A\times B=\left[\begin{array}{cc}1-2& 1+\frac{1}{2}\\ -\frac{1}{2}-4& -\frac{1}{2}+1\end{array}\right] \\ A\times B=\left[\begin{array}{cc}-1& 1\frac{1}{2}\\ -4\frac{1}{2}& \frac{1}{2}\end{array}\right] \end{gathered}$$

Since the product of matrix A and B fails to give an identity matrix, hence, A is not an inverse of B and vice versa.

b.

$$\begin{gathered} M\times N=\left[\begin{array}{cc}3& 4\\ 1& 2\end{array}\right]\times \left[\begin{array}{cc}1& -2\\ -\frac{1}{2}& \frac{3}{2}\end{array}\right] \\ M\times N=\left[\begin{array}{cc}(3\times 1)+(4\times (-\frac{1}{2}\left)\right)& (3\times (-2\left)\right)+(4\times \frac{3}{2})\\ (1\times 1)+(2\times (-\frac{1}{2})& (1\times (-2\left)\right)+(2\times \frac{3}{2})\end{array}\right] \\ M\times N=\left[\begin{array}{cc}3-2& -6+6\\ 1-1& -2+3\end{array}\right] \\ M\times N=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \end{gathered}$$

Since the product of matrices M and N yields an identity matrix, it means matrix M is the inverse of matrix N.

Find the inverse of matrix B.

$B=\left[\begin{array}{cc}1& 0\\ 2& 3\end{array}\right]$

Solution:

$B=\left[\begin{array}{cc}1& 0\\ 2& 3\end{array}\right]$

Using;

${\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]}^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$

Then;

$$\begin{gathered} B^{-1}=\frac{1}{(1\times 3)-(0\times 2)}\left[\begin{array}{cc}3& 0\\ -2& 1\end{array}\right] \\ {B}^{-1}=\frac{1}{3-0}\left[\begin{array}{cc}3& 0\\ -2& 1\end{array}\right] \\ {B}^{-1}=\frac{1}{3}\left[\begin{array}{cc}3& 0\\ -2& 1\end{array}\right] \end{gathered}$$

or,

$$\begin{gathered} B^{-1}=\frac{1}{3}\left[\begin{array}{cc}3& 0\\ -2& 1\end{array}\right]=\left[\begin{array}{cc}\frac{3}{3}& 0\\ -\frac{2}{3}& \frac{1}{3}\end{array}\right] \\ {B}^{-1}=\left[\begin{array}{cc}1& 0\\ -\frac{2}{3}& \frac{1}{3}\end{array}\right] \end{gathered}$$

Find the inverse of the matrix X.

$X=\left[\begin{array}{ccc}2& 1& -3\\ 5& 3& 0\\ -4& 2& 1\end{array}\right]$

Solution:

This is a 3 by 3 matrix.

Step1: Find the determinant of the given matrix.

$$\begin{gathered} \left|\begin{array}{c}X\end{array}\right|=2\left|\begin{array}{cc}3& 0\\ 2& 1\end{array}\right|-1\left|\begin{array}{cc}5& 0\\ -4& 1\end{array}\right|-3\left|\begin{array}{cc}5& 3\\ -4& 2\end{array}\right| \\ \left|\begin{array}{c}X\end{array}\right|=2(3-0)-1(5-0)-3(10+12) \\ \left|\begin{array}{c}X\end{array}\right|=6-5-66 \\ \left|\begin{array}{c}X\end{array}\right|=-65 \end{gathered}$$

Since the determinant is not equal to 0, it means that the matrix X has an inverse.

Step2: Find the cofactor of the matrix.

The cofactor of 2 which is C₁₁ is

$$\begin{gathered} C_{11}={(-1)}^{1+1}\times \left|\begin{array}{cc}3& 0\\ 2& 1\end{array}\right| \\ {C}_{11}=1(3-0) \\ {C}_{11}=3 \end{gathered}$$

The cofactor of 1 which is C₁₂ is

$$\begin{gathered} C_{12}={(-1)}^{1+2}\times \left|\begin{array}{cc}5& 0\\ -4& 1\end{array}\right| \\ {C}_{12}=-1(5-0) \\ {C}_{12}=-5 \end{gathered}$$

The cofactor of -3 which is C₁₃ is

$$\begin{gathered} C_{13}={(-1)}^{1+3}\times \left|\begin{array}{cc}5& 3\\ -4& 2\end{array}\right| \\ {C}_{13}=1(10+12) \\ {C}_{13}=22 \end{gathered}$$

The cofactor of 5 which is C₂₁ is

$$\begin{gathered} C_{21}={(-1)}^{2+1}\times \left|\begin{array}{cc}1& -3\\ 2& 1\end{array}\right| \\ {C}_{21}=-1(1+6) \\ {C}_{21}=-7 \end{gathered}$$

The cofactor of 3 which is C₂₂ is

$$\begin{gathered} C_{22}={(-1)}^{2+2}\times \left|\begin{array}{cc}2& -3\\ -4& 1\end{array}\right| \\ {C}_{22}=1(2+12) \\ {C}_{22}=14 \end{gathered}$$

The cofactor of 0 which is C₂₃ is

$$\begin{gathered} C_{23}={(-1)}^{2+3}\times \left|\begin{array}{cc}2& 1\\ -4& 2\end{array}\right| \\ {C}_{23}=-1(4+4) \\ {C}_{23}=-8 \end{gathered}$$

The cofactor of -4 which is C₃₁ is

$$\begin{gathered} C_{31}={(-1)}^{3+1}\times \left|\begin{array}{cc}1& -3\\ 3& 0\end{array}\right| \\ {C}_{31}=1(0+9) \\ {C}_{31}=9 \end{gathered}$$

The cofactor of 2 which is C₃₂ is

$$\begin{gathered} C_{32}={(-1)}^{3+2}\times \left|\begin{array}{cc}2& -3\\ 5& 0\end{array}\right| \\ {C}_{32}=-1(0+15) \\ {C}_{32}=-15 \end{gathered}$$

The cofactor of 1 which is C₃₃ is

$$\begin{gathered} C_{33}={(-1)}^{3+3}\times \left|\begin{array}{cc}2& 1\\ 5& 3\end{array}\right| \\ {C}_{33}=1(6-5) \\ {C}_{33}=1 \end{gathered}$$

So the cofactor of the matrix X is

$X^c=\left[\begin{array}{ccc}3& -5& 22\\ -7& 14& -8\\ 9& -15& 1\end{array}\right]$

Step 3: Transpose of the cofactor matrix to give the adjoint of the matrix.

the transpose of X^c is

${\left(X^c\right)}^T=Adj\left(X\right)=\left[\begin{array}{ccc}3& -7& 9\\ -5& 14& -15\\ 22& -8& 1\end{array}\right]$

Step 4: Divide the adjoint matrix by the determinant of the matrix.

Remember the determinant of matrix X is 65. This final stage gives us the inverse of matrix X which is X^-1. Hence, we have

$$\begin{gathered} X^{-1}=\frac{1}{-65}\left[\begin{array}{ccc}3& -7& 9\\ -5& 14& -15\\ 22& -8& 1\end{array}\right] \\ {X}^{-1}=\left[\begin{array}{ccc}-\frac{3}{65}& \frac{7}{65}& -\frac{9}{65}\\ \frac{5}{65}& -\frac{14}{65}& \frac{15}{65}\\ -\frac{22}{65}& \frac{8}{65}& -\frac{1}{65}\end{array}\right] \\ {X}^{-1}=\mathbf{\left[}\begin{array}{ccc}\mathbf{-}\frac{\mathbf{3}}{\mathbf{65}}& \frac{\mathbf{7}}{\mathbf{65}}& \mathbf{-}\frac{\mathbf{9}}{\mathbf{65}}\\ \frac{\mathbf{1}}{\mathbf{13}}& \mathbf{-}\frac{\mathbf{14}}{\mathbf{65}}& \frac{\mathbf{3}}{\mathbf{13}}\\ \mathbf{-}\frac{\mathbf{22}}{\mathbf{65}}& \frac{\mathbf{8}}{\mathbf{65}}& \mathbf{-}\frac{\mathbf{1}}{\mathbf{65}}\end{array}\mathbf{\right]} \end{gathered}$$

Using matrix operations solve for x and y in the following:

$$\begin{gathered} 2x+3y=6 \\ x-2y=-2 \end{gathered}$$

Solution:

This equation can be represented in matrix form as

$\left[\begin{array}{cc}2& 3\\ 1& -2\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}6\\ -2\end{array}\right]$

Let the matrices be represented by P, Q and R respectively such that

$\left[P\right]\times \left[Q\right]=\left[R\right]$

We intend on finding matrix Q since it represents our unknowns x and y. So we make matrix Q the subject of the formula

$$\begin{gathered} {\left[P\right]}^{-1}\times \left[P\right]\times \left[Q\right]={\left[P\right]}^{-1}\times \left[R\right] \\ {\left[P\right]}^{-1}\times \left[P\right]=I \end{gathered}$$

I is an Identity matrix and its determinant is 1.

$$\begin{gathered} I\left[Q\right]=\left[R\right]\times {\left[P\right]}^{-1} \\ \left[Q\right]=\left[R\right]\times {\left[P\right]}^{-1} \end{gathered}$$

$$\begin{gathered} {\left[P\right]}^{-1}={\left[\begin{array}{cc}2& 3\\ 1& -2\end{array}\right]}^{-1} \\ {\left[P\right]}^{-1}=\frac{1}{(-4-3)}\left[\begin{array}{cc}-2& -3\\ -1& 2\end{array}\right] \\ {\left[P\right]}^{-1}=\left[\begin{array}{cc}\frac{2}{7}& \frac{3}{7}\\ \frac{1}{7}& -\frac{2}{7}\end{array}\right] \end{gathered}$$

Then,

$$\begin{gathered} \left[Q\right]=\left[\begin{array}{cc}\frac{2}{7}& \frac{3}{7}\\ \frac{1}{7}& -\frac{2}{7}\end{array}\right]\times \left[\begin{array}{c}6\\ -2\end{array}\right] \\ \left[Q\right]=\left[\begin{array}{c}(\frac{2}{7}\times 6)+(\frac{3}{7}\times -2)\\ (\frac{1}{7}\times 6)+\left(\right(-\frac{2}{7})\times -2)\end{array}\right] \\ \left[Q\right]=\left[\begin{array}{c}\frac{12}{7}-\frac{6}{7}\\ \frac{6}{7}+\frac{4}{7}\end{array}\right] \\ \left[Q\right]=\left[\begin{array}{c}\frac{6}{7}\\ \frac{10}{7}\end{array}\right] \\ \left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}\frac{6}{7}\\ \frac{10}{7}\end{array}\right] \\ x=\frac{\mathbf{6}}{\mathbf{7}} \\ y=\frac{\mathbf{10}}{\mathbf{7}} \end{gathered}$$