CUET UG 2024 — Mathematics Algebra

To find the value of $abc$abc, we need to find the values of $a$a, $b$b, and $c$c using the fact that all three matrices are singular.

A matrix is singular when its determinant equals zero.

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For matrix $A = \begin{bmatrix} 1 & 4 \\ 3 & 2a \end{bmatrix}$A=[13​42a​]:

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

$$2a - 12 = 0$$2a−12=0

$$a = 6$$a=6

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For matrix $B = \begin{bmatrix} 3b & 5 \\ a & 2 \end{bmatrix}$B=[3ba​52​], using $a = 6$a=6:

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

$$6b - 30 = 0$$6b−30=0

$$b = 5$$b=5

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For matrix $C = \begin{bmatrix} a+b+c & c+1 \\ a+c & c \end{bmatrix}$C=[a+b+ca+c​c+1c​], substituting $a = 6$a=6 and $b = 5$b=5:

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

$$11c + c^2 - 6c - 6 - c^2 - c = 0$$11c+c2−6c−6−c2−c=0

$$4c - 6 = 0$$4c−6=0

$$c = \dfrac{6}{4} = \dfrac{3}{2}$$c=46​=23​

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Therefore:

$$abc = 6 \times 5 \times \dfrac{3}{2} = 30 \times \dfrac{3}{2} = 45$$abc=6×5×23​=30×23​=45