If represents the value of the determinant vmatrix 1 & k & 1 \\ - k & 1 & k \\ -1 & - k & 1 vmatrix, where k is any real number then the value of belongs to Option 1: R (set of real numbers) Option 2: (2, 4) Option 3: (-3,-1)(2,4) Option 4: [2, 4]

Answer: [2, 4]. Solution: Let s = k. Expanding along the first row: = (1+s^2) - s(-s+s) + (s^2+1) = 2 + 2s^2. Since s^2 ranges over [0,1], ranges over [2,4], endpoints included.

Algebra

Matrices & Determinants

Medium

core

If $\Delta$ represents the value of the determinant $\begin{vmatrix} 1 & \sin k\theta & 1 \\ -\sin k\theta & 1 & \sin k\theta \\ -1 & -\sin k\theta & 1 \end{vmatrix}$ , where $k$ is any real number then the value of $\Delta$ belongs to

R (set of real numbers)

(2, 4)

$(-3,-1)\cup(2,4)$

[2, 4]

✅ Correct Option: 4

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

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

Since $s^2$ ranges over $[0,1]$ , $\Delta$ ranges over $[2,4]$ , endpoints included.

2026: 22 May Shift 2

If $A=\begin{bmatrix}k & 2\\2 & k\end{bmatrix}$ such that $\lvert A^3 \rvert = 125$ then which of the following statements are TRUE?

(A) $\lvert k.adj(3A) \rvert = 405$

(B) $\lvert k.adj(3A) \rvert = 2025$

(D) $\lvert adj(3A) \rvert = 45$

Choose the correct answer from the options given below:

2026: 21 May Shift 1

If $A = \begin{bmatrix} 4 & -1 & \lambda \\ 2 & 0 & 6 \\ 1 & -2 & 7 \end{bmatrix}$ , then $\frac{1}{2}A^{-1}$ exist if

2025: 22 May Shift 1

Let A be a non-singular square matrix of order 3 and $|adj A| = 5$ then $|A|$ is equal to

2026: 29 May Shift 1

If $A=\begin{bmatrix}3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2\end{bmatrix}$ and $A(adjA)=xI$ , where $I$ is an identify matrix of order 3, then $x$ is equal to

2025: 21 May Shift 1

Let $A = [a_{ij}]_{3×2}$ and $B = [b_{ij}]_{3×4}$ be two matrices. Then the order of the matrix $(A^T . B)^T$ is:

2026: 22 May Shift 2

Match List-I with List-II

List-I	List-II
(A) If $A=\begin{bmatrix}2 & -5\\3 & 4\end{bmatrix}$ , then adj A =	(I) $\begin{bmatrix}4 & -5\\-2 & -3\end{bmatrix}$
(B) If $A=\begin{bmatrix}2 & 3\\5 & -2\end{bmatrix}$ , then $A^{-1}=$	(II) $\begin{bmatrix}\frac{1}{17} & \frac{-5}{17}\\\frac{3}{17} & \frac{2}{17}\end{bmatrix}$
(C) If $A=\begin{bmatrix}-3 & 5\\2 & 4\end{bmatrix}$ , then adj A =	(III) $\begin{bmatrix}4 & 5\\-3 & 2\end{bmatrix}$
(D) If $A=\begin{bmatrix}2 & 5\\-3 & 1\end{bmatrix}$ , then $A^{-1}=$	(IV) $\begin{bmatrix}\frac{2}{19} & \frac{3}{19}\\\frac{5}{19} & \frac{-2}{19}\end{bmatrix}$

Choose the correct answer from the options given below:

Solution

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