The maximum value of = vmatrix1 & & 1\\ - & 1 & \\ -1 & - & 1vmatrix is ( is real number) Option 1: 2 Option 2: 3 Option 3: 4 Option 4: 10

Answer: 4. Solution: Let s=. Expanding along the first row: = (1+s^2)-s(-s+s)+(s^2+1) = 2+2s^2. Since s^2≤ 1, the maximum is 2+2 = 4 (at = ±()/2).

Algebra

Matrices & Determinants

Medium

core

The maximum value of $\Delta = \begin{vmatrix}1 & \sin\theta & 1\\ -\sin\theta & 1 & \sin\theta\\ -1 & -\sin\theta & 1\end{vmatrix}$ is ( $\theta$ is real number)

✅ Correct Option: 3

Let $s=\sin\theta$ . Expanding along the first row:

$\Delta = (1+s^2)-s(-s+s)+(s^2+1) = 2+2s^2$ .

Since $s^2\leq 1$ , the maximum is $2+2 = 4$ (at $\theta = \pm\frac{\pi}{2}$ ).

2026: 30 May Shift 2

If A and B are invertible matrices of the same order, then

Match List-I with List-II

List-I (Determinant)	List-II (Equal to)
(A) $\left\vert A^{-1}B \right\vert$	(I) $\dfrac{1}{\left\vert A \right\vert \left\vert B \right\vert}$
(B) $\left\vert AB^{-1} \right\vert$	(II) $\dfrac{\left\vert B \right\vert}{\left\vert A \right\vert}$
(C) $\left\vert A^{-1}B^{-1} \right\vert$	(III) $\left\vert A \right\vert \left\vert B \right\vert$
(D) $\left\vert AB \right\vert$	(IV) $\dfrac{\left\vert A \right\vert}{\left\vert B \right\vert}$

Choose the correct answer from the options given below:

2025: 2 June Shift 1

If the matrix $\begin{bmatrix}2 & -1 & 3\\ \lambda & 0 & 7\\-1 & 1 & 4\end{bmatrix}$ is not invertible, then value of $\lambda$ is

2025: 3 June Shift 1

Which of the following statements are correct?

(A) Inverse of a matrix, if it exists, is unique

(B) $(kA)' = -kA'$ (where k is any real number)

(D) For an invertible matrix $A$ , $(A')^{-1} = (A^{-1})'$

Choose the correct answer from the options given below:

2026: 25 May Shift 1

The value of $\Delta = \begin{vmatrix} 2\sin40^\circ & 2\sin50^\circ \\ -3\cos40^\circ & 3\cos50^\circ \end{vmatrix}$ is:

2026: 30 May Shift 1

For what values of $\lambda$ and $\mu$ , the following system of linear equations:

$x+y+2z=1, 2x+y-z=4, x+2y+\lambda z=\mu$

have infinitely many solution?

2025: 26 May Shift 2

If $\begin{bmatrix} 1 & 0 & 0 \\ 0 & y+1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 2x & \\ -2 & \\ z-3 & \end{bmatrix} = \begin{bmatrix} 6 \\ 4 \\ 1 \end{bmatrix}$ then $x + y + z$ is

The maximum value of Δ=∣1sin⁡θ1−sin⁡θ1sin⁡θ−1−sin⁡θ1∣\Delta = \begin{vmatrix}1 & \sin\theta & 1\\ -\sin\theta & 1 & \sin\theta\\ -1 & -\sin\theta & 1\end{vmatrix}Δ=​1−sinθ−1​sinθ1−sinθ​1sinθ1​​ is (θ\thetaθ is real number)

Solution

Related questions:

Keyboard Shortcuts

CUET Mathematics Past Year Questions (Topic-Wise)

The maximum value of $\Delta = \begin{vmatrix}1 & \sin\theta & 1\\ -\sin\theta & 1 & \sin\theta\\ -1 & -\sin\theta & 1\end{vmatrix}$ is ( $\theta$ is real number)