Algebra

Matrices & Determinants

Easy

core

$\Delta=\left|\begin{array}{ccc}1 & \cos x & 1 \\ -\cos x & 1 & \cos x \\ -1 & -\cos x & 1\end{array}\right|$

(A) $\Delta=2\left(1-\cos ^{2} x\right)$

(B) $\Delta=2\left(2-\sin ^{2} x\right)$

(C) Minimum value of $\Delta$ is $2$

(D) Maximum value of $\Delta$ is $4$

Choose the correct answer from the options given below :

(A), (C) and (D) only

(A), (B) and (C) only

(A), (B), (C) and (D)

(B), (C) and (D) only

✅ Correct Option: 4

We need to calculate the determinant $\Delta=\left|\begin{array}{ccc}1 & \cos x & 1 \\ -\cos x & 1 & \cos x \\ -1 & -\cos x & 1\end{array}\right|$

Expanding the determinant using cofactor expansion along the first row:

$\Delta = 1 \cdot \begin{vmatrix} 1 & \cos x \\ -\cos x & 1 \end{vmatrix} - \cos x \cdot \begin{vmatrix} -\cos x & \cos x \\ -1 & 1 \end{vmatrix} + 1 \cdot \begin{vmatrix} -\cos x & 1 \\ -1 & -\cos x \end{vmatrix}$

Calculating each 2×2 determinant:

$\begin{vmatrix} 1 & \cos x \\ -\cos x & 1 \end{vmatrix} = 1 \cdot 1 - \cos x \cdot (-\cos x) = 1 + \cos^2 x$

$\begin{vmatrix} -\cos x & \cos x \\ -1 & 1 \end{vmatrix} = (-\cos x) \cdot 1 - \cos x \cdot (-1) = -\cos x + \cos x = 0$

$\begin{vmatrix} -\cos x & 1 \\ -1 & -\cos x \end{vmatrix} = (-\cos x) \cdot (-\cos x) - 1 \cdot (-1) = \cos^2 x + 1$

Substituting back into the original equation:

$\Delta = 1 \cdot (1 + \cos^2 x) - \cos x \cdot 0 + 1 \cdot (1 + \cos^2 x)$

$\Delta = 1 + \cos^2 x + 1 + \cos^2 x$

$\Delta = 2 + 2\cos^2 x$

Using the identity $\cos^2 x = 1 - \sin^2 x$ :

$\Delta = 2 + 2(1 - \sin^2 x)$

$\Delta = 2 + 2 - 2\sin^2 x$

$\Delta = 4 - 2\sin^2 x$

$\Delta = 2(2 - \sin^2 x)$

Since $0 \leq \sin^2 x \leq 1$ for all real values of $x$ :

Maximum value: when $\sin^2 x = 0$ $sin^{2} x = 0$ , giving $\Delta_{max} = 2(2 - 0) = 4$ $Δ_{ma x} = 2 (2 - 0) = 4$
- Minimum value: when $\sin^2 x = 1$ , giving $\Delta_{min} = 2(2 - 1) = 2$

The correct options are (B), (C), and (D).

2025: 15 May Shift 1

The area of a triangle whose vertices are $(x_1, y_1), (x_2, y_2)$ and $(x_3, y_3)$ is given by the absolute value of

2026: 30 May Shift 1

If $A=\begin{bmatrix}1 & 2\\2 & 1\end{bmatrix}$ and $f(x)=x^2-2x-3$ , then $f(A)$ is equal to

2025: 3 June Shift 2

Which of the following statement is/are correct?

(A) A square matrix $A = [a_{ij}]$ is called a symmetric matrix if $a_{ij} = a_{ji}$ for all $i, j$

(B) $A = [a_{ij}]_{m \times m}$ is a diagonal matrix if $a_{ij} = 0$ when $i = j$

(D) The multiplication of diagonal matrices of same order is commutative

Choose the correct answer from the options given below:

2025: 22 May Shift 2

Let $A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 4 & -6 \\ -2 & 4 \end{bmatrix}$

(A) $\det(A^T) = 1$

(B) $AB = I$ , where $I$ is the identity matrix of order 2.

(D) adj $(B) = \begin{bmatrix} 4 & 2 \\ 6 & 4 \end{bmatrix}$

Choose the correct answer from the options given below:

2026: 19 May Shift 1

If $P = \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix}$ , $Q = \begin{bmatrix} 4 & 0 \\ 1 & 5 \end{bmatrix}$ and $R = \begin{bmatrix} 2 & 0 \\ 1 & -2 \end{bmatrix}$ , then which of the following are TRUE?

(A) $(P^T)^T = P^T$

(B) $(P-Q)R = PR - QR$

(D) $(PQ)^T = Q^TP^T$

Choose the correct answer from the options given below:

2025: 13 May Shift 2

If $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ then $A^{-1}$ is equal to

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