Question

Show that P^-1HP = 3à —3 matrix

Use cofactor method to find the inverse of P

Answer 1

We shall find the inverse of matrix $p,$ denoted as $p^{-1},$ using the cofactor method

Say the $3$ X $3$ matrix is given by

$p=\begin{bmatrix} 2 & 1 & 0 \\ 1 & -1 & 1 \\ 3 & 2 & 1 \end{bmatrix}$

Next, we find the determinant of $p$ denoted as $\mid p\mid$

$\mid p \mid=2\begin{vmatrix} -1 & 1 \\ 2 & 1 \end{vmatrix}-1\begin{vmatrix} 1 & 1 \\ 3 & 1 \end{vmatrix}+0\begin{vmatrix} 1 & -1 \\ 3 & 2 \end{vmatrix}$

$=2(-3)-1(-2)+0$

$= -6+2$

$=-4$

Thus, we have $\mid p \mid= -4$

Now, We shall find the cofactor which is given by

$= \begin{bmatrix} +\begin{vmatrix} -1 & 1 \\ 2 & 1 \end{vmatrix} & -\begin{vmatrix} 1 & 1 \\ 3 & 1 \end{vmatrix} & + \begin{vmatrix} 1 & -1 \\ 3 & 2 \end{vmatrix}\\ - \begin{vmatrix} 1 & 0 \\ 2 & 1 \end{vmatrix} & +\begin{vmatrix} 2 & 0 \\ 3 & 1 \end{vmatrix} & -\begin{vmatrix} 2 & 1 \\ 3 & 2 \end{vmatrix}\\ + \begin{vmatrix} 1 & 0 \\ -1 & 1 \end{vmatrix} & -\begin{vmatrix} 2 & 0 \\ 1 & 1 \end{vmatrix} & +\begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} \end{bmatrix}$

$=\begin{bmatrix} -3 & 2 & 5 \\ -1 & 2 & -1\\ 1 & -2 & -3 \end{bmatrix}$

Now, by finding the transpose of the above matrix and dividing by the determinant

we have,

$\displaystyle=\frac{1}{-4}\begin{bmatrix} -3 & -1 & 1 \\ 2 & 2 & -2\\ 5 & -1 & -3 \end{bmatrix}=p^{-1}$

Hence, we have

$\displaystyle p^{-1}= \begin{bmatrix} \frac{3}{4} & \frac{1}{4} & \frac{-1}{4} \\ \frac{-1}{2}& \frac{-1}{2} & \frac{1}{2}\\ \frac{-5}{4} & \frac{1}{4} & \frac{3}{4} \end{bmatrix}$

which is the inverse of the matrix by the cofactor method.