Question

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

Short answer

The formula ${\mathrm{M}}^{-1}=\frac{1}{\det \mathrm{M}}{\mathrm{C}}^{\mathrm{T}}$is verified from the product of matrices${\mathrm{MC}}^{\mathrm{T}}$.

Step-by-step solution

Definition of Product of a Matrix:

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 ).

Given Parameters:

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.

Determinant and Cofactors of Matrix:

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,

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

Inverse of Matrix:

Find the inverse of matrix from the formula,

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

Product of two Matrices:

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.