Verify formula (6.13). Hint: Consider the product of the matrices ${\mathbf{MC}}^{\mathbf{T}}$. Use Problem 3.8.

Matrix multiplication is a binary operation that creates a matrix by multiplying two matrices together. For matrix multiplication to work, the number of columns in the first matrix must equal the number of rows in the second matrix.

For example, if matrix A and B is defined as $\left(3\times 2\right)$and $\left(3\times 2\right)$then, the product of the matrices A and B is meaningless as the columns in the first matrix (here, 2 ) is not equals to the number of rows in the second matrix (here, 3 ).

Write the matrix from problem 3.8, M = $\left(\begin{array}{ccc}1& 0& -1\\ -2& 3& 0\\ 1& -3& 2\end{array}\right)$.

Find the cofactors of matrix M.

Find determinant of matrix,

$$\begin{gathered} \det \mathrm{M}=\left|\begin{array}{ccc}1& 0& -1\\ -2& 3& 0\\ 1& -3& 2\end{array}\right| \\ =1\left(3\right)\left(2\right)-\left(0\right)(-3))-0+(-1\left)\right((-2)(-3)-\left(3\right)\left(1\right)) \\ =3 \end{gathered}$$

Find cofactors of matrix,

$$\begin{gathered} \left|\begin{array}{cc}3& 0\\ -3& 2\end{array}\right|=6 \\ -\left|\begin{array}{cc}-2& 0\\ 1& 2\end{array}\right|=4 \\ \left|\begin{array}{cc}-2& 3\\ 1& -3\end{array}\right|=3 \\ -\left|\begin{array}{cc}0& -1\\ -3& 2\end{array}\right|=3 \end{gathered}$$

Solve further,

$$\begin{gathered} \left|\begin{array}{cc}1& -1\\ 1& 2\end{array}\right|=3 \\ -\left|\begin{array}{cc}1& 0\\ 1& -3\end{array}\right|=3 \\ \left|\begin{array}{cc}0& -1\\ 3& 0\end{array}\right|=3 \\ -\left|\begin{array}{cc}1& -1\\ -2& 0\end{array}\right|=3 \end{gathered}$$

Solve further,

$\left|\begin{array}{cc}1& 0\\ -2& 3\end{array}\right|=3$

Write transpose of cofactors from the above elements,

$$\begin{gathered} \mathrm{C}=\left(\begin{array}{ccc}6& 4& 3\\ 3& 3& 3\\ 3& 2& 3\end{array}\right) \\ {\mathrm{C}}^{\mathrm{T}}=\left(\begin{array}{ccc}6& 3& 3\\ 4& 3& 2\\ 3& 3& 3\end{array}\right) \end{gathered}$$

Find the inverse of matrix from the formula,

$$\begin{gathered} M^{-1}=\frac{1}{detM}{C}^T \\ =\frac{1}{3}\left(\begin{array}{ccc}6& 4& 3\\ 3& 3& 3\\ 3& 2& 3\end{array}\right) \\ =\left(\begin{array}{ccc}2& 1& 1\\ \frac{4}{3}& 1& \frac{2}{3}\\ 1& 1& 1\end{array}\right) \end{gathered}$$

Find the product of matrices M and ${\mathrm{C}}^{\mathrm{T}}$,

$$\begin{gathered} {\mathrm{MC}}^{\mathrm{T}}=\left(\begin{array}{ccc}1& 0& -1\\ -2& 3& 0\\ 1& -3& 2\end{array}\right)\left(\begin{array}{ccc}6& 3& 3\\ 4& 3& 2\\ 3& 3& 3\end{array}\right) \\ =\left(\begin{array}{ccc}3& 0& 0\\ 0& 3& 0\\ 0& 0& 3\end{array}\right) \\ =3\mathrm{l} \\ =(\det \mathrm{M})\mathrm{l} \end{gathered}$$

Divide and multiply ${\mathrm{M}}^{-1}$on both sides,

$$\begin{gathered} \frac{1}{\det \mathrm{M}}{\mathrm{MC}}^{\mathrm{T}}=\mathrm{l} \\ {\mathrm{M}}^{-1}\mathrm{l}={\mathrm{M}}^{-1}\left(\mathrm{M}\frac{1}{\det \mathrm{M}}{\mathrm{C}}^{\mathrm{T}}\right) \\ {\mathrm{M}}^{-1}\mathrm{l}=\mathrm{l}\left(\frac{1}{\det \mathrm{M}}{\mathrm{C}}^{\mathrm{T}}\right) \\ {\mathrm{M}}^{-1}=\frac{1}{\det \mathrm{M}}{\mathrm{C}}^{\mathrm{T}} \end{gathered}$$

Therefore, the formula (6.13) ${\mathrm{M}}^{-1}=\frac{1}{\det \mathrm{M}}{\mathrm{C}}^{\mathrm{T}}$is verified.