If bmatrix3 & 1\\2 & 1bmatrixPbmatrix2 & 1\\3 & 2bmatrix=bmatrix1 & -1\\0 & 1bmatrix then matrix P is Option 1: bmatrix-8 & 5\\19 & 12bmatrix Option 2: bmatrix19 & -12\\8 & -5bmatrix Option 3: bmatrix8 & -5\\-19 & 12bmatrix Option 4: bmatrix5 & -8\\-12 & 19bmatrix

Answer: bmatrix8 & -5\\-19 & 12bmatrix. Solution: Write APB=C, so P=A^(-1)CB^(-1). Both A and B have determinant 1, so A^(-1)=bmatrix1&-1\\-2&3bmatrix and B^(-1)=bmatrix2&-1\\-3&2bmatrix. A^(-1)C=bmatrix1&-2\\-2&5bmatrix, and multiplying by B^(-1) gives P=bmatrix8&-5\\-19&12bmatrix.

Algebra

Matrices & Determinants

Hard

core

If $\begin{bmatrix}3 & 1\\2 & 1\end{bmatrix}P\begin{bmatrix}2 & 1\\3 & 2\end{bmatrix}=\begin{bmatrix}1 & -1\\0 & 1\end{bmatrix}$ then matrix P is

$\begin{bmatrix}-8 & 5\\19 & 12\end{bmatrix}$

$\begin{bmatrix}19 & -12\\8 & -5\end{bmatrix}$

$\begin{bmatrix}8 & -5\\-19 & 12\end{bmatrix}$

$\begin{bmatrix}5 & -8\\-12 & 19\end{bmatrix}$

✅ Correct Option: 3

Write $APB=C$ , so $P=A^{-1}CB^{-1}$ .

Both $A$ and $B$ have determinant $1$ , so $A^{-1}=\begin{bmatrix}1&-1\\-2&3\end{bmatrix}$ and $B^{-1}=\begin{bmatrix}2&-1\\-3&2\end{bmatrix}$ .

$A^{-1}C=\begin{bmatrix}1&-2\\-2&5\end{bmatrix}$ , and multiplying by $B^{-1}$ gives $P=\begin{bmatrix}8&-5\\-19&12\end{bmatrix}$ .

2025: 26 May Shift 2

Match List-I with List-II

List-I	List-II
Matrix/equations	Values
(A) $\begin{bmatrix} 2x+1 & 3y \\ 0 & y^2-5y \end{bmatrix} = \begin{bmatrix} x+3 & y^2+2 \\ 0 & -6 \end{bmatrix}$	(I) $x = 2, y = -1$
(B) $\begin{bmatrix} 1 & 2 & -1 \\ x & 0 & 3 \\ y & 3 & 4 \end{bmatrix}$ is symmetric	(II) $x = 2, y = 2$
(C) $[x \ \ 1]\begin{bmatrix} 1 & 0 \\ -2 & -3 \end{bmatrix}\begin{bmatrix} 5 & 2 \\ 0 & y \end{bmatrix} = O$	(III) $x = -2, y = 2$
(D) $\begin{bmatrix} x & 0 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} x & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ -1 & y/2 \end{bmatrix}$	(IV) $x = 2, y = 0$

Choose the correct answer from the options given below:

2025: 30 May Shift 2

If $A = \begin{bmatrix}2 & 3 & 1 \\ 2 & -1 & 0\end{bmatrix}$ and $B^T = \begin{bmatrix}4 & 4 \\ 6 & -2 \\ 2 & 0\end{bmatrix}$ , then $4A + B$ is

2024: 16 May Shift 1

If $\mathrm{A}=\left[\begin{array}{ll}2 & 4 \\ 4 & 3\end{array}\right], \mathrm{X}=\left[\begin{array}{l}\mathrm{n} \\ 1\end{array}\right], \mathrm{B}=\left[\begin{array}{c}8 \\ 11\end{array}\right]$ and $A X=B$ , then the value of $n$ will be :

2022: 30 Aug Shift 1

The system of equations $3x + 4y = 5$ , $6x + 7y = -8$ is written in matrix form as

2025: 16 May Shift 1

Match List-I with List-II

Let $A$ be any invertible square matrix. Then

List-I	List-II
(A) $A - A^T$	(I) $\vert A\vert A^{-1}$
(B) $A A^T$	(II) Skew-symmetric
(C) $\det (A^{-1})$	(III) Symmetric
(D) $\text{adj} A$	(IV) $[\det(A)]^{-1}$

Choose the correct answer from the options given below:

2025: 2 June Shift 1

Which of the given values of $x$ and $y$ make the following pair of matrices equal ?

$\begin{bmatrix}2x-1 & 4\\y-1 & 3+2x\end{bmatrix}$ and $\begin{bmatrix}0 & y-2\\5 & 4\end{bmatrix}$

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}[32​11​]P[23​12​]=[10​−11​] then matrix P is

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

If $\begin{bmatrix}3 & 1\\2 & 1\end{bmatrix}P\begin{bmatrix}2 & 1\\3 & 2\end{bmatrix}=\begin{bmatrix}1 & -1\\0 & 1\end{bmatrix}$ then matrix P is