Match List - I with List - II. List - I List - II (A) A is a square m… | Maths. Section B1 CUET question

Match List - I with List - II.

List - I	List - II
(A) A is a square matrix of order 3 and \|2A\| = k\|A\|, then k is	(I) 0
(B) Value of $\begin{vmatrix}1 & a & b+c\\1 & b & c+a\\1 & c & a+b\end{vmatrix}$	(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 a₂₃ is	(IV) 7

Choose the correct answer from the options given below :

Answer & explanation

Correct answer: option 1

(A) A is a square matrix of order 3 and |2A| = k|A|, then k is

|2A| = k|A|

$2^3|A|=k|A|⇒k=8$

(B) Value of $\begin{vmatrix}1 & a & b+c\\1 & b & c+a\\1 & c & a+b\end{vmatrix}$

We know that for a 3×3 matrix $A=\begin{bmatrix}a&f&c\\d&e&f\\g&h&i\end{bmatrix}$, the determinant is:

$∣A∣=a(ei−fh)−b(di−fg)+c(dh−eg)$

given matrix is $A=\begin{bmatrix}a&f&c\\d&e&f\\g&h&i\end{bmatrix}$

then det. of A is ⇒ $∣A∣=1[b(a+b)−c(c+a)]−a[1(a+b)−1(c+a)]+(b+c)[1(c)−1(b)]$

$=1(ab+b^2−c^2−ac)−a(a+b−c−a)+(b+c)(c−b)$

$=ab+b^2−c^2−ac−a^2−ab+ac+a^2+bc−b^2+c^2−bc$

$=ab−ab+b^2−b^2−c^2+c^2−ac+ac−a^2+a^2+bc−bc$

$=0$

$\begin{bmatrix}5-x & x+1 \\2 & 4\end{bmatrix}=0$

$⇒20−4x−2x−2=0$

$⇒18−6x=0$

$⇒6x=18$

$⇒x=\frac{18}{6}=3$

(D) If $A = (a_{ij})=\begin{bmatrix}5 & 3 & 8\\2 & 0 & 1\\1 & 2 & 3\end{bmatrix}$ then the minor of the element a₂₃ is

We need the minor of element a₂₃, the row II and column III would get cut.

The determinant of the remaining square matrix, i.e. det$\begin{bmatrix}5&13\\1&2\end{bmatrix}$ is the required minor.

∴ minor of a₂₃ = 5 × 2 − 1 × 3 = 10 − 3 = 7.

So, correct matching is

(A) - (III), (B) - (I), (C) - (II), (D) - (IV)

Option 1 is correct.