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CUET Mathematics 2025 PYQs3 June Shift 2. Free, no login required.

Algebra

past year questions, CUET Mathematics 2025 3 June Shift 2

Matrices & Determinants

past year questions, Algebra, CUET Mathematics 2025 3 June Shift 2

Medium

core

If A=[15712],B=[9178]A = \begin{bmatrix} 1 & 5 \\ 7 & 12 \end{bmatrix}, B = \begin{bmatrix} 9 & 1 \\ 7 & 8 \end{bmatrix} and C are three matrices such that 3A+5B+2C=03A + 5B + 2C = 0, then the matrix C is equal to

Solution

✅ Correct Option: 2

Given:

Matrix A=[15712]A = \begin{bmatrix} 1 & 5 \\ 7 & 12 \end{bmatrix}

Matrix B=[9178]B = \begin{bmatrix} 9 & 1 \\ 7 & 8 \end{bmatrix}

Equation: 3A+5B+2C=03A + 5B + 2C = 0

From the equation 3A+5B+2C=03A + 5B + 2C = 0:

2C=−3A−5B2C = -3A - 5B

C=−3A−5B2C = \frac{-3A - 5B}{2}

Calculate 3A3A:

3A=3×[15712]3A = 3 \times \begin{bmatrix} 1 & 5 \\ 7 & 12 \end{bmatrix}

=[3152136]= \begin{bmatrix} 3 & 15 \\ 21 & 36 \end{bmatrix}

Calculate 5B5B:

5B=5×[9178]5B = 5 \times \begin{bmatrix} 9 & 1 \\ 7 & 8 \end{bmatrix}

=[4553540]= \begin{bmatrix} 45 & 5 \\ 35 & 40 \end{bmatrix}

Calculate 3A+5B3A + 5B:

3A+5B=[3152136]+[4553540]3A + 5B = \begin{bmatrix} 3 & 15 \\ 21 & 36 \end{bmatrix} + \begin{bmatrix} 45 & 5 \\ 35 & 40 \end{bmatrix}

=[48205676]= \begin{bmatrix} 48 & 20 \\ 56 & 76 \end{bmatrix}

Calculate −3A−5B-3A - 5B:

−3A−5B=−[48205676]-3A - 5B = -\begin{bmatrix} 48 & 20 \\ 56 & 76 \end{bmatrix}

=[−48−20−56−76]= \begin{bmatrix} -48 & -20 \\ -56 & -76 \end{bmatrix}

Calculate CC:

C=12[−48−20−56−76]C = \frac{1}{2}\begin{bmatrix} -48 & -20 \\ -56 & -76 \end{bmatrix}

=[−24−10−28−38]= \begin{bmatrix} -24 & -10 \\ -28 & -38 \end{bmatrix}

Therefore, C=[−24−10−28−38]C = \begin{bmatrix} -24 & -10 \\ -28 & -38 \end{bmatrix}

Related questions:

2025: 14 May Shift 1

Let A=[121−132241]A = \begin{bmatrix} 1 & 2 & 1 \\ -1 & 3 & 2 \\ 2&4&1\end{bmatrix} and MijM_{ij}, AijA_{ij} respectively denote the minor, co-factor of an element aija_{ij} of matrix A, then which of the following are true?

(A) M22=−1M_{22} = -1

(B) A23=0A_{23} = 0

(C) A32=3A_{32} = 3

(D) M23=1M_{23} = 1

(E) M32=−3M_{32} = -3

Choose the correct answer from the options given below:

2026: 22 May Shift 1

If A=[aij]3×3=[23−11230−11]A=[a_{ij}]_{3\times3}=\begin{bmatrix}2 & 3 & -1\\1 & 2 & 3\\0 & -1 & 1\end{bmatrix} and B=[bij]3×2=[3−21412]B=[b_{ij}]_{3\times2}=\begin{bmatrix}3 & -2\\1 & 4\\1 & 2\end{bmatrix} then the value of a12b21+a32b31a_{12}b_{21}+a_{32}b_{31} is

2025: 14 May Shift 2

The values of λ\lambda for which the system of equation x+2y+z=14,−x+y+z=10,x+λy+z=2x + 2y + z = 14, - x + y + z = 10, x + \lambda y + z = 2 has unique solution is

2026: 22 May Shift 2

If [3121]P[2132]=[1−101]\begin{bmatrix}3 & 1\\2 & 1\end{bmatrix}P\begin{bmatrix}2 & 1\\3 & 2\end{bmatrix}=\begin{bmatrix}1 & -1\\0 & 1\end{bmatrix} then matrix P is

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2026: 19 May Shift 1

Match List-I with List-II

List-IList-II
(A) [53−71]+[7372]\begin{bmatrix} 5 & 3 \\ -7 & 1 \end{bmatrix} + \begin{bmatrix} 7 & 3 \\ 7 & 2 \end{bmatrix}(I) Identity matrix
(B) [53−71]+[−4−3−60]\begin{bmatrix} 5 & 3 \\ -7 & 1 \end{bmatrix} + \begin{bmatrix} -4 & -3 \\ -6 & 0 \end{bmatrix}(II) Upper Triangular Matrix
(C) [5−3−2−1]+[−5321]\begin{bmatrix} 5 & -3 \\ -2 & -1 \end{bmatrix} + \begin{bmatrix} -5 & 3 \\ 2 & 1 \end{bmatrix}(III) Zero Matrix
(D) [513−26]+[−4−132−5]\begin{bmatrix} 5 & 13 \\ -2 & 6 \end{bmatrix} + \begin{bmatrix} -4 & -13 \\ 2 & -5 \end{bmatrix}(IV) Lower Triangular Matrix

Choose the correct answer from the options given below:

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