Question: For the matrices in Example 3, verify that ${\mathbf{MM}}^{\mathbf{-}\mathbf{1}}$ and ${\mathbf{M}}^{\mathbf{-}\mathbf{1}}\mathbf{M}$ both equal a unit matrix. Multiply${\mathbf{M}}^{\mathbf{-}\mathbf{1}}\mathbf{k}$to verify the solution of equations (6.9).

Matrix multiplication is a binary operation that creates a matrix by multiplying two matrices together.

A unit matrix(I) can be defined as a scalar matrix in which all the diagonal elements are equal to 1 and all the other elements are zero.

$$\begin{gathered} \mathrm{I}=\left(100 \\ 010 \\ 001\right) \end{gathered}$$

The given matrices are

$$\begin{gathered} \mathrm{M}=\left(10-1 \\ -230 \\ 1-32\right),{\mathrm{M}}^{-1}=\frac{1}{3}\left(633 \\ 432 \\ 333\right),\mathrm{k}=\left(5 \\ 1 \\ -10\right) \end{gathered}$$

The product of ${\mathrm{MM}}^{-1},{\mathrm{M}}^{-1}\mathrm{M},{\mathrm{M}}^{-1}\mathrm{k}$are to be found.

Find the product of the matrix ${\mathrm{MM}}^{-1}$.

localid="1658990898411" $$\begin{gathered} {\mathrm{MM}}^{-1}=\frac{1}{3}\left(10-1 \\ -230 \\ 1-32\right)\left(633 \\ 432 \\ 333\right) \\ =\frac{1}{3}\left(6-33-33-3 \\ -12+12-6+9-6+6 \\ 6-12+63-9+63-6+6\right) \\ =\frac{1}{3}\left(300 \\ 030 \\ 003\right) \\ =\left(100 \\ 010 \\ 001\right) \\ \mathrm{so},{\mathrm{MM}}^{-1}=\mathrm{I} \end{gathered}$$

Find the product of the matrix ${\mathrm{M}}^{-1}\mathrm{M}$.

$$\begin{gathered} {\mathrm{MM}}^{-1}=\frac{1}{3}\left(633 \\ 432 \\ 333\right)\left(10-1 \\ -230 \\ 1-32\right) \\ =\frac{1}{3}\left(6-6+33-3-6+6 \\ 4-6+29-69-9 \\ 3-6+39-9-3+6\right) \\ =\frac{1}{3}\left(300 \\ 030 \\ 003\right) \\ =\left(100 \\ 010 \\ 001\right) \\ \mathrm{so},{\mathrm{M}}^{-1}\mathrm{M}=\mathrm{I} \end{gathered}$$

$$\begin{gathered} {\mathrm{M}}^{-1}=\frac{1}{3}\left(633 \\ 432 \\ 333\right)\left(5 \\ 1 \\ -10\right) \end{gathered}$$

$$\begin{gathered} {\mathrm{M}}^{-1}\mathrm{k}=\frac{1}{3}\left(\left(6\times 5\right)+\left(3\times 1\right)+\left(3\times \left(-10\right)\right) \\ \left(4\times 5\right)+\left(3\times 1\right)+\left(2\times \left(-10\right)\right) \\ \left(3\times 5\right)+\left(3\times 1\right)+\left(3\times \left(-10\right)\right)\right) \\ =\frac{1}{3}\left(3 \\ 3 \\ -12\right) \\ =\left(1 \\ 1 \\ -4\right) \end{gathered}$$

So, x = 1 , y = 1 , z = -4

Which satisfies equations (6.9).

x - z = 5

-2x + 3y = 1

x - 3y + 2z = -10

Therefore, $$\begin{gathered} {\mathrm{MM}}^{-1}={\mathrm{M}}^{-1}\mathrm{M}=\mathrm{I}\mathrm{and}{\mathrm{M}}^{-1}\mathrm{k}=\left(1 \\ 1 \\ -4\right) \end{gathered}$$