Match List - I with List - II.

List - I	List - II
(A) $A$ is a square matrix of order $3$ and $\lvert 2A \rvert = k\lvert A \rvert$, then $k$ is	(I) $0$
(B) Value of $\begin{vmatrix}1 & a & b+c \\ 1 & b & c+a \\ 1 & c & a+b\end{vmatrix}$ is	(II) $3$
(C) Matrix $\begin{bmatrix}5-x & x+1 \\ 2 & 4\end{bmatrix}$ is singular, then $x =$	(III) $8$
(D) If $A = (a_{ij}) = \begin{bmatrix}5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3\end{bmatrix}$, then the minor of the element $a_{23}$ is	(IV) $7$

Choose the correct answer from the options given below :

(A) For a square matrix of order $n$, the property is $|cA| = c^n|A|$.

Here, $A$ is of order $3$ and $c = 2$:

$|2A| = 2^3|A| = 8|A|$

Comparing with $|2A| = k|A|$:

$k = 8 \longrightarrow$ (III)

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(B)

$\begin{vmatrix}1 & a & b+c \\ 1 & b & c+a \\ 1 & c & a+b\end{vmatrix}$

Apply $C_3 \to C_3 + C_2$:

$= \begin{vmatrix}1 & a & a+b+c \\ 1 & b & a+b+c \\ 1 & c & a+b+c\end{vmatrix}$

Take $(a+b+c)$ common from $C_3$:

$= (a+b+c)\begin{vmatrix}1 & a & 1 \\ 1 & b & 1 \\ 1 & c & 1\end{vmatrix}$

Since $C_1$ and $C_3$ are identical, the determinant $= 0$.

$= (a+b+c) \times 0 = 0 \longrightarrow$ (I)

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(C) A matrix is singular when its determinant $= 0$.

$(5-x)(4) - (2)(x+1) = 0$

$20 - 4x - 2x - 2 = 0$

$18 - 6x = 0$

$x = 3 \longrightarrow$ (II)

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(D) The minor of $a_{ij}$ is the determinant formed by deleting the $i$-th row and $j$-th column.

For $a_{23}$, delete Row 2 and Column 3:

$\text{Minor of } a_{23} = \begin{vmatrix}5 & 3 \\ 1 & 2\end{vmatrix}$

$= (5)(2) - (3)(1) = 10 - 3 = 7 \longrightarrow$ (IV)

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$(A) \to (III), \quad (B) \to (I), \quad (C) \to (II), \quad (D) \to (IV)$