CUET Mathematics 2024 - If A, B and C are three singular matrices given by A=[arrayll1 & 4 \\ 3 & 2 aarray], B=[arrayll3 b & 5 \\ a & 2array] and C=[arraycca+b+c & c+1 \\ a+c & carray], then the value of a b c is : | PYQs + Solutions

CUET Mathematics 2024

Algebra

Matrices & Determinants

Medium

If $A, B$ and $C$ are three singular matrices given by $A=\left[\begin{array}{ll}1 & 4 \\ 3 & 2 a\end{array}\right], B=\left[\begin{array}{ll}3 b & 5 \\ a & 2\end{array}\right]$ and $C=\left[\begin{array}{cc}a+b+c & c+1 \\ a+c & c\end{array}\right]$ , then the value of $a b c$ is :

✅ Correct Option: 3

To find the value of

abc

, we need to find the values of

a

b

, and

c

using the fact that all three matrices are singular.

A matrix is singular when its determinant equals zero.

For matrix

A = \begin{bmatrix} 1 & 4 \\ 3 & 2a \end{bmatrix}

\det(A) = 1 \times 2a - 4 \times 3 = 0

2a - 12 = 0

a = 6

For matrix

B = \begin{bmatrix} 3b & 5 \\ a & 2 \end{bmatrix}

, using

a = 6

\det(B) = 3b \times 2 - 5 \times 6 = 0

6b - 30 = 0

b = 5

For matrix

C = \begin{bmatrix} a+b+c & c+1 \\ a+c & c \end{bmatrix}

, substituting

a = 6

and

b = 5

\det(C) = (11+c)c - (c+1)(6+c) = 0

11c + c^2 - 6c - 6 - c^2 - c = 0

4c - 6 = 0

c = \dfrac{6}{4} = \dfrac{3}{2}

Therefore:

abc = 6 \times 5 \times \dfrac{3}{2} = 30 \times \dfrac{3}{2} = 45