CUET Mathematics - For = vmatrix 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c vmatrix where a, b, c are roots of equation x^3 + px + q = 0, then Match the LIST-I with LIST-II | | LIST-I | | LIST-II | |---|---|---|---| | A. | a + b + c | I. | p - q | | B. | | II. | -q | | C. | ab + bc + ca | III. | 0 | | D. | abc | IV. | p | Choose the correct answer from the options given below: | PYQs + Solutions

CUET Mathematics 2026 15th May Shift 1

Algebra

Hard

core

A-III, B-I, C-II, D-IV

A-III, B-I, C-IV, D-II

A-I, B-III, C-IV, D-II

A-I, B-III, C-II, D-IV

✅ Correct Option: 2

Related questions:

23 Aug Shift 1

Let the matrix $A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}$ and $AB = \begin{bmatrix} 5 & 6 & 7 \\ 8 & 9 & 10 \end{bmatrix}$ then order of B is :

17 Aug Shift 2

Identify the correct statements.

(A) If $A$ is a non-singular matrix, then $A^{-1} = \frac{|A|}{(\text{adj } A)}$

(B) If $A$ is an invertible matrix then $\frac{1}{|A^{-1}|} = |A|$

(C) If $A$ and $B$ are two invertible matrices of the same order then $AB$ is also invertible matrix and $(BA)^{-1} = A^{-1}B^{-1}$

(D) If $A$ is an invertible matrix, then $A^T$ is also invertible and $(A^T)^{-1} = \frac{1}{(A^{-1})^T}$

Choose the correct answer from the options given below :

22nd May Shift 1

If matrix $A=\begin{bmatrix}1 & 4 & 5\\3 & 2 & 6\\0 & 1 & 0\end{bmatrix}$ , then

Match List-I with List-II

List-I	List-II
(A) $\left\vert A \right\vert =$	(I) $\dfrac{1}{9}$
(B) $\left\vert A^{-1} \right\vert =$	(II) $81$
(C) $\left\vert \text{adj } A \right\vert =$	(III) $9$
(D) $\left\vert \text{adj } (\text{adj } A) \right\vert =$	(IV) $6561$

Choose the correct answer from the options given below:

CUET Mathematics 2026 15th May Shift 1 PYQs

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