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 and all the other elements are zero.

$\mathrm{I}=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)$

The given matrices are

$\mathrm{M}=\left(\begin{array}{ccc}1& 0& -1\\ -2& 3& 0\\ 1& -3& 2\end{array}\right),{\mathrm{M}}^{-1}=\frac{1}{3}\left(\begin{array}{ccc}6& 3& 6\\ 4& 3& 2\\ 3& 3& 3\end{array}\right),\mathrm{k}=\left(\begin{array}{c}5\\ 1\\ -10\end{array}\right)$

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${\mathrm{MM}}^{-1}.$

$$\begin{gathered} {\mathrm{MM}}^{-1}=\frac{1}{3}\left(\begin{array}{ccc}1& 0& -1\\ -2& 3& 0\\ 1& -3& 2\end{array}\right)\left(\begin{array}{ccc}6& 3& 3\\ 4& 3& 2\\ 3& 3& 3\end{array}\right) \\ =\frac{1}{3}\left(\begin{array}{ccc}6-3& 3-3& 3-3\\ -12+12& -6+9& -6+6\\ 6-12+6& 3-9+6& 3-6+6\end{array}\right) \\ =\frac{1}{3}\left(\begin{array}{ccc}3& 0& 0\\ 0& 3& 0\\ 0& 0& 3\end{array}\right) \\ =\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right) \\ \mathrm{So},{\mathrm{MM}}^{-1}=\mathrm{I} \\ \mathrm{Find}\mathrm{the}\mathrm{product}\mathrm{of}\mathrm{the}\mathrm{matrix}{\mathrm{M}}^{-1}\mathrm{M}. \\ {\mathrm{M}}^{-1}\mathrm{M}=\frac{1}{3}\left(\begin{array}{ccc}6& 3& 3\\ 4& 3& 2\\ 3& 3& 3\end{array}\right)\left(\begin{array}{ccc}1& 0& -1\\ -2& 3& 0\\ 1& -3& 2\end{array}\right) \\ =\frac{1}{3}\left(\begin{array}{ccc}6-6+3& 3-3& -6+6\\ 4-6+2& 9-6& 9-9\\ 3-6+3& 9-9& -3+6\end{array}\right) \\ =\frac{1}{3}\left(\begin{array}{ccc}3& 0& 0\\ 0& 3& 0\\ 0& 0& 3\end{array}\right) \\ =\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right) \\ \mathrm{So},{\mathrm{M}}^{-1}\mathrm{M}=\mathrm{I} \end{gathered}$$

$$\begin{gathered} {\mathrm{M}}^{-1}\mathrm{k}=\frac{1}{3}\left(\begin{array}{ccc}6& 3& \begin{array}{c}3\end{array}\\ 4& 3& 2\\ 3& 3& 3\end{array}\right)\left(\begin{array}{c}5\\ 1\\ -10\end{array}\right) \\ {\mathrm{M}}^{-1}\mathrm{k}=\frac{1}{3}\left(\begin{array}{ccc}\left(6\times 5\right)+& \left(3\times 1\right)+& \left(3\times \left(-10\right)\right)\\ \left(4\times 5\right)+& \left(3\times 1\right)+& \left(2\times \left(-10\right)\right)\\ \left(3\times 5\right)+& \left(3\times 1\right)+& \left(3\times \left(-10\right)\right)\end{array}\right) \\ =\frac{1}{3}\left(\begin{array}{c}3\\ 3\\ -12\end{array}\right) \\ =\left(\begin{array}{c}1\\ 1\\ -4\end{array}\right) \\ \mathrm{So},\mathrm{x}=1,\mathrm{y}=1,\mathrm{z}=-4 \end{gathered}$$

Which satisfies equations (6.9).

$$\begin{gathered} \mathrm{x}-\mathrm{z}=5 \\ -2\mathrm{x}+3\mathrm{y}=1 \\ \mathrm{x}-3\mathrm{y}+2\mathrm{z}=-10 \end{gathered}$$

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