Q.(i) Find the minor and cofactor of the second row and second column element of the determinant $\begin{vmatrix}2 & -3 & 5\\6 & 0 & 4\\1 & 5 & -7\end{vmatrix}$. [2 marks]

Main (i): delete row 2/column 2 for the minor, apply sign $(-1)^{2+2}$ for the cofactor. (ii): verify the given matrix identity, then solve for $A^{-1}$. OR: solve the linear system via $X=A^{-1}B$.

Main (i): $\begin{vmatrix}2&-3&5\\6&0&4\\1&5&-7\end{vmatrix}$. The element in row 2, column 2 is $0$. Its minor $M_{22}$ is the determinant left after deleting row 2 and column 2:

$M_{22}=\begin{vmatrix}2&5\\1&-7\end{vmatrix}=2(-7)-5(1)=-14-5=-19$

Cofactor $C_{22}=(-1)^{2+2}M_{22}=(+1)(-19)=-19$.

Main (ii): $A=\begin{bmatrix}3&1\\-1&2\end{bmatrix}$.

$A^2=A\cdot A=\begin{bmatrix}3(3)+1(-1)&3(1)+1(2)\\-1(3)+2(-1)&-1(1)+2(2)\end{bmatrix}=\begin{bmatrix}8&5\\-5&3\end{bmatrix}$

$5A=\begin{bmatrix}15&5\\-5&10\end{bmatrix},\qquad 7I=\begin{bmatrix}7&0\\0&7\end{bmatrix}$

$A^2-5A+7I=\begin{bmatrix}8-15+7&5-5+0\\-5+5+0&3-10+7\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}=0\ \checkmark$

From $A^2-5A+7I=0$, multiply through by $A^{-1}$ (valid since $\det A=6+1=7\ne0$):

$A-5I+7A^{-1}=0 \implies A^{-1}=\frac{5I-A}{7}$

$5I-A=\begin{bmatrix}5-3&0-1\\0-(-1)&5-2\end{bmatrix}=\begin{bmatrix}2&-1\\1&3\end{bmatrix}$

$A^{-1}=\frac17\begin{bmatrix}2&-1\\1&3\end{bmatrix}$

OR: Solve $x+y+z=6$, $y+3z=11$, $x-2y+z=0$ by the matrix method.

Write as $AX=B$ where $A=\begin{bmatrix}1&1&1\\0&1&3\\1&-2&1\end{bmatrix}$, $X=\begin{bmatrix}x\\y\\z\end{bmatrix}$, $B=\begin{bmatrix}6\\11\\0\end{bmatrix}$.

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