Target Exam

CUET

Subject

-- Applied Mathematics - Section B2

Chapter

Algebra

Question:

Match List-I with List-II

List-I (Matrix)

List-II (Determinant)

(A) $\begin{bmatrix}1&7\\-3&5\end{bmatrix}$

(I) 24

(B) $\begin{bmatrix}-2&5\\-3&-3\end{bmatrix}$

(II) 32

(C) $\begin{bmatrix}-12&8\\-16&8\end{bmatrix}$

(III) 21

(D) $\begin{bmatrix}15&9\\-21&-11\end{bmatrix}$

(IV) 26

Choose the correct answer from the options given below:

Options:

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

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

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

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

Correct Answer:

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

Explanation:

The correct answer is Option (2) → (A)-(IV), (B)-(III), (C)-(II), (D)-(I)

List-I (Matrix)

List-II (Determinant)

(A) $\begin{bmatrix}1&7\\-3&5\end{bmatrix}$

(IV) 26

(B) $\begin{bmatrix}-2&5\\-3&-3\end{bmatrix}$

(III) 21

(C) $\begin{bmatrix}-12&8\\-16&8\end{bmatrix}$

(II) 32

(D) $\begin{bmatrix}15&9\\-21&-11\end{bmatrix}$

(I) 24

(A) $\begin{bmatrix}1&7\\-3&5\end{bmatrix}$

$|A|=1\times5-7(-3)=5+21=26$ → (IV)

(B) $\begin{bmatrix}-2&5\\-3&-3\end{bmatrix}$

$|B|=(-2)(-3)-5(-3)=6+15=21$ → (III)

(C) $\begin{bmatrix}-12&8\\-16&8\end{bmatrix}$

$|C|=(-12)(8)-8(-16)=-96+128=32$ → (II)

(D) $\begin{bmatrix}15&9\\-21&-11\end{bmatrix}$

$|D|=15(-11)-9(-21)=-165+189=24$ → (I)

Correct matching: A–IV, B–III, C–II, D–I