Q.Evaluate the determinant $\begin{vmatrix}4&3\\2&5\end{vmatrix}$.

Solving $\begin{vmatrix} x & 2 \\ 8 & 5 \end{vmatrix}=0$ gives $x=\dfrac{16}{5}$.

The value of a $2\times2$ determinant $\begin{vmatrix} a & b \\ c & d \end{vmatrix}$ is $ad-bc$. Here:

$\begin{vmatrix} x & 2 \\ 8 & 5 \end{vmatrix}=(x)(5)-(2)(8)=5x-16.$

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Expand the $3\times3$ determinant along the first row, get $-(x^2-x)=0$, and solve to get $x=0,\ 1$.

Expand along the first row $(0,\ 1,\ 0)$; only the middle entry contributes:

$\begin{vmatrix} 0 & 1 & 0 \\ x & 2 & x \\ 1 & 3 & x \end{vmatrix} = -\,1\cdot\begin{vmatrix} x & x \\ 1 & x \end{vmatrix}$

The sign is $(-1)^{1+2}=-1$ for the $(1,2)$ position. Now evaluate the $2\times2$ minor:

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Delete row 3 and column 3, evaluate the remaining minor, and apply the $(-1)^{3+3}=+1$ sign to get cofactor $=18$.

The entry $-7$ is in the third row and third column, so $i=3,\ j=3$. Its cofactor is

$C_{33} = (-1)^{3+3}\, M_{33} = (+1)\begin{vmatrix} 2 & -3 \\ 6 & 0 \end{vmatrix}$

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Evaluating the $2\times2$ determinant gives $x^2-(x+2)(x-2)=x^2-(x^2-4)=4$, so its value is $4$.

For a $2\times2$ determinant $\begin{vmatrix} a & b \\ c & d \end{vmatrix}=ad-bc$. Here $a=x,\;b=x+2,\;c=x-2,\;d=x$:

$\begin{vmatrix} x & x+2 \\ x-2 & x \end{vmatrix} = (x)(x) - (x+2)(x-2).$ …