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CUET Mathematics 2025 PYQs29 May Shift 2. Free, no login required.

Algebra

past year questions, CUET Mathematics 2025 29 May Shift 2

Matrices & Determinants

past year questions, Algebra, CUET Mathematics 2025 29 May Shift 2

Medium

core

If A=[124−3]A = \begin{bmatrix} 1 & 2 \\ 4 & -3 \end{bmatrix} and f(x)=2x2−4x+5f(x) = 2x^2 - 4x + 5, the f(A)f(A) is equal to

Solution

✅ Correct Option: 1

Given A=[124−3]A = \begin{bmatrix} 1 & 2 \\ 4 & -3 \end{bmatrix} and f(x)=2x2−4x+5f(x) = 2x^2 - 4x + 5

When replacing xx with matrix AA, each constant term becomes a scalar multiple of the identity matrix.

f(A)=2A2−4A+5If(A) = 2A^2 - 4A + 5I

A2=A×AA^2 = A \times A

=[124−3]×[124−3]= \begin{bmatrix} 1 & 2 \\ 4 & -3 \end{bmatrix} \times \begin{bmatrix} 1 & 2 \\ 4 & -3 \end{bmatrix}

First row, first column: (1)(1)+(2)(4)=1+8=9(1)(1) + (2)(4) = 1 + 8 = 9

First row, second column: (1)(2)+(2)(−3)=2−6=−4(1)(2) + (2)(-3) = 2 - 6 = -4

Second row, first column: (4)(1)+(−3)(4)=4−12=−8(4)(1) + (-3)(4) = 4 - 12 = -8

Second row, second column: (4)(2)+(−3)(−3)=8+9=17(4)(2) + (-3)(-3) = 8 + 9 = 17

A2=[9−4−817]A^2 = \begin{bmatrix} 9 & -4 \\ -8 & 17 \end{bmatrix}

2A2=2[9−4−817]2A^2 = 2\begin{bmatrix} 9 & -4 \\ -8 & 17 \end{bmatrix}

=[18−8−1634]= \begin{bmatrix} 18 & -8 \\ -16 & 34 \end{bmatrix}

4A=4[124−3]4A = 4\begin{bmatrix} 1 & 2 \\ 4 & -3 \end{bmatrix}

=[4816−12]= \begin{bmatrix} 4 & 8 \\ 16 & -12 \end{bmatrix}

5I=5[1001]5I = 5\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}

=[5005]= \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}

f(A)=2A2−4A+5If(A) = 2A^2 - 4A + 5I

=[18−8−1634]−[4816−12]+[5005]= \begin{bmatrix} 18 & -8 \\ -16 & 34 \end{bmatrix} - \begin{bmatrix} 4 & 8 \\ 16 & -12 \end{bmatrix} + \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}

Position (1,1): 18−4+5=1918 - 4 + 5 = 19

Position (1,2): −8−8+0=−16-8 - 8 + 0 = -16

Position (2,1): −16−16+0=−32-16 - 16 + 0 = -32

Position (2,2): 34−(−12)+5=34+12+5=5134 - (-12) + 5 = 34 + 12 + 5 = 51

f(A)=[19−16−3251]f(A) = \begin{bmatrix} 19 & -16 \\ -32 & 51 \end{bmatrix}

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Consider a determinant Δ=∣3a−a+b−a+c−b+a3b−b+c−c+a−c+b3c∣\Delta=\begin{vmatrix}3a & -a+b & -a+c\\ -b+a & 3b & -b+c\\ -c+a & -c+b & 3c\end{vmatrix}, then

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Choose the correct answer from the options given below:

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