CBSE Class 12 Mathematics HOTs Determinants Set 03

Question. Find the maximum value of \(\begin{vmatrix} 1 & 1 & 1 \\ 1 & 1+\sin\theta & 1 \\ 1 & 1 & 1+\cos\theta \end{vmatrix}\).
Answer: Let \(\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1+\sin\theta & 1 \\ 1 & 1 & 1+\cos\theta \end{vmatrix}\)
\(\Rightarrow \Delta = 1[(1+\sin\theta)(1+\cos\theta) - 1] - 1(1+\cos\theta - 1) + 1(1 - 1 - \sin\theta)\)
\(= 1 + \cos\theta + \sin\theta + \sin\theta\cos\theta - 1 - \cos\theta - \sin\theta\)
\(= \sin\theta\cos\theta\)
\(\therefore\) Maximum value of \(\Delta\) is \(\frac{1}{2}\).

Question. If \(x \in \mathbb{N}\) and \(\begin{vmatrix} x+3 & -2 \\ -3x & 2x \end{vmatrix} = 8\), then find the value of \(x\).
Answer: Given, \(\begin{vmatrix} x+3 & -2 \\ -3x & 2x \end{vmatrix} = 8\)
\(\Rightarrow (x+3)(2x) - (-2)(-3x) = 8\)
\(\Rightarrow 2x^2 + 6x - 6x = 8 \Rightarrow 2x^2 = 8\)
\(\Rightarrow x^2 = 4 \Rightarrow x = 2\) [since \(x \neq -2\) because \(x \in \mathbb{N}\)].

Question. If \(\begin{vmatrix} x & \sin\theta & \cos\theta \\ -\sin\theta & -x & 1 \\ \cos\theta & 1 & x \end{vmatrix} = 8\), write the value of \(x\).
Answer: Given, \(\begin{vmatrix} x & \sin\theta & \cos\theta \\ -\sin\theta & -x & 1 \\ \cos\theta & 1 & x \end{vmatrix} = 8\)
\(\Rightarrow x(-x^2 - 1) - \sin\theta(-x\sin\theta - \cos\theta) + \cos\theta(-\sin\theta + x\cos\theta) = 8\)
\(\Rightarrow -x^3 - x + x\sin^2\theta + \sin\theta\cos\theta - \sin\theta\cos\theta + x\cos^2\theta = 8\)
\(\Rightarrow -x^3 - x + x(\sin^2\theta + \cos^2\theta) = 8\)
\(\Rightarrow -x^3 - x + x = 8 \Rightarrow x^3 + 8 = 0\)
\(\Rightarrow (x + 2)(x^2 - 2x + 4) = 0 \Rightarrow x+2 = 0\) [since \(x^2 - 2x + 4 > 0\), \(\forall x\)]
\(\Rightarrow x = -2\).

Question. If \(A = \begin{bmatrix} 1 & 2 \\ 3 & -1 \end{bmatrix}\) and \(B = \begin{bmatrix} 1 & 3 \\ -1 & 1 \end{bmatrix}\), write the value of \(|AB|\).
Answer: Given that \(A = \begin{bmatrix} 1 & 2 \\ 3 & -1 \end{bmatrix}\) and \(B = \begin{bmatrix} 1 & 3 \\ -1 & 1 \end{bmatrix}\)
\(\therefore AB = \begin{bmatrix} 1 & 2 \\ 3 & -1 \end{bmatrix} \begin{bmatrix} 1 & 3 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ 4 & 8 \end{bmatrix}\)
\(\therefore |AB| = \begin{vmatrix} -1 & 5 \\ 4 & 8 \end{vmatrix} = (-1)\cdot 8 - 4\cdot 5 = -28\).

Question. If \(\begin{vmatrix} 2x & 5 \\ 8 & x \end{vmatrix} = \begin{vmatrix} 6 & -2 \\ 7 & 3 \end{vmatrix}\), write the value of \(x\).
Answer: Given, \(\begin{vmatrix} 2x & 5 \\ 8 & x \end{vmatrix} = \begin{vmatrix} 6 & -2 \\ 7 & 3 \end{vmatrix}\)
\(\Rightarrow 2x^2 - 40 = 18 + 14\)
\(\Rightarrow 2x^2 = 72 \Rightarrow x^2 = 36\)
\(\Rightarrow x = \pm 6\).

Question. If \(\begin{vmatrix} 3x & 7 \\ -2 & 4 \end{vmatrix} = \begin{vmatrix} 8 & 7 \\ 6 & 4 \end{vmatrix}\), find the value of \(x\).
Answer: Given, \(\begin{vmatrix} 3x & 7 \\ -2 & 4 \end{vmatrix} = \begin{vmatrix} 8 & 7 \\ 6 & 4 \end{vmatrix}\)
\(\Rightarrow 12x + 14 = 32 - 42\)
\(\Rightarrow 12x = -10 - 14 = -24\)
\(\Rightarrow x = -2\).

Question. Write the value of the determinant \(\begin{vmatrix} p & p+1 \\ p-1 & p \end{vmatrix}\).
Answer: \(\begin{vmatrix} p & p+1 \\ p-1 & p \end{vmatrix} = p^2 - (p-1)(p+1) = p^2 - (p^2 - 1) = 1\).

Question. Write the value of \(\begin{vmatrix} 2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86 \end{vmatrix}\).
Answer: Let \(\Delta = \begin{vmatrix} 2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86 \end{vmatrix}\)
\(\Rightarrow \Delta = 2[8(86) - 9(75)] - 7[3(86) - 5(75)] + 65[3(9) - 5(8)]\)
\(= 2(688 - 675) - 7(258 - 375) + 65(27 - 40)\)
\(= 2(13) - 7(-117) + 65(-13)\)
\(= 26 + 819 - 845 = 0\).

Question. If \(\begin{vmatrix} x+1 & x-1 \\ x-3 & x+2 \end{vmatrix} = \begin{vmatrix} 4 & -1 \\ 1 & 3 \end{vmatrix}\), then write the value of \(x\).
Answer: Given, \(\begin{vmatrix} x+1 & x-1 \\ x-3 & x+2 \end{vmatrix} = \begin{vmatrix} 4 & -1 \\ 1 & 3 \end{vmatrix}\)
\(\Rightarrow (x+1)(x+2) - (x-3)(x-1) = 4 \times 3 - 1 \times (-1)\)
\(\Rightarrow x^2 + x + 2x + 2 - (x^2 - 3x - x + 3) = 12 + 1\)
\(\Rightarrow x^2 + 3x + 2 - x^2 + 4x - 3 = 13\)
\(\Rightarrow 7x = 13 + 1 \Rightarrow x = 2\).

Question. If \(\begin{vmatrix} 2x & x+3 \\ 2(x+1) & x+1 \end{vmatrix} = \begin{vmatrix} 1 & 5 \\ 3 & 3 \end{vmatrix}\), then write the value of \(x\).
Answer: Given, \(\begin{vmatrix} 2x & x+3 \\ 2(x+1) & x+1 \end{vmatrix} = \begin{vmatrix} 1 & 5 \\ 3 & 3 \end{vmatrix}\)
\(\Rightarrow 2x(x+1) - 2(x+1)(x+3) = 3 - 15\)
\(\Rightarrow 2x^2 + 2x - 2(x^2 + 4x + 3) = -12\)
\(\Rightarrow 2x^2 + 2x - 2x^2 - 8x - 6 = -12\)
\(\Rightarrow -6x = -12 + 6 \Rightarrow x = \frac{-6}{-6} = 1\).

Question. Evaluate: \(\begin{vmatrix} \cos 15^\circ & \sin 15^\circ \\ \sin 75^\circ & \cos 75^\circ \end{vmatrix}\).
Answer: \(\begin{vmatrix} \cos 15^\circ & \sin 15^\circ \\ \sin 75^\circ & \cos 75^\circ \end{vmatrix} = \cos 15^\circ \cos 75^\circ - \sin 75^\circ \sin 15^\circ = \cos(15^\circ + 75^\circ) = \cos 90^\circ = 0\).

Question. If \(A = \begin{bmatrix} 3 & 4 \\ 1 & 2 \end{bmatrix}\), find the value of \(3|A|\).
Answer: Here, \(A = \begin{bmatrix} 3 & 4 \\ 1 & 2 \end{bmatrix}\)
\(\Rightarrow |A| = \begin{vmatrix} 3 & 4 \\ 1 & 2 \end{vmatrix} = 3 \times 2 - 1 \times 4 = 2\)
\(\Rightarrow 3|A| = 3 \times 2 = 6\).

Question. What is the value of the determinant \(\begin{vmatrix} 0 & 2 & 0 \\ 2 & 3 & 4 \\ 4 & 5 & 6 \end{vmatrix}\)?
Answer: \(\begin{vmatrix} 0 & 2 & 0 \\ 2 & 3 & 4 \\ 4 & 5 & 6 \end{vmatrix} = -2(12 - 16) = 8\).

Question. What positive value of \(x\) makes the following pair of determinants equal? \(\begin{vmatrix} 2x & 3 \\ 5 & x \end{vmatrix}\), \(\begin{vmatrix} 16 & 3 \\ 5 & 2 \end{vmatrix}\).
Answer: \(\begin{vmatrix} 2x & 3 \\ 5 & x \end{vmatrix} = \begin{vmatrix} 16 & 3 \\ 5 & 2 \end{vmatrix}\)
\(\Rightarrow 2x^2 - 15 = 32 - 15 \Rightarrow x = 4\) [since \(x > 0\)].

Question. Find the value of \(x\) from \(\begin{vmatrix} x & 4 \\ 2 & 2x \end{vmatrix} = 0\).
Answer: \(\begin{vmatrix} x & 4 \\ 2 & 2x \end{vmatrix} = 0 \Rightarrow 2x^2 - 8 = 0 \Rightarrow x = \pm 2\).

Question. Evaluate: \(2 \begin{vmatrix} 7 & -2 \\ -10 & 5 \end{vmatrix}\).
Answer: \(2 \begin{vmatrix} 7 & -2 \\ -10 & 5 \end{vmatrix} = 2(35 - 20) = 2 \times 15 = 30\).

Question. Evaluate: \(\begin{vmatrix} \sqrt{6} & \sqrt{5} \\ \sqrt{20} & \sqrt{24} \end{vmatrix}\).
Answer: \(\begin{vmatrix} \sqrt{6} & \sqrt{5} \\ \sqrt{20} & \sqrt{24} \end{vmatrix} = \sqrt{6} \times \sqrt{24} - \sqrt{5} \times \sqrt{20} = \sqrt{144} - \sqrt{100} = 12 - 10 = 2\).

Question. Evaluate: \(\begin{vmatrix} a+ib & c+id \\ -c+id & a-ib \end{vmatrix}\).
Answer: \(\begin{vmatrix} a+ib & c+id \\ -c+id & a-ib \end{vmatrix} = (a+ib)(a-ib) - (c+id)(-c+id) = (a^2 + b^2) - (-c^2 - d^2) = a^2 + b^2 + c^2 + d^2\) [since \(i^2 = -1\)].

Question. Evaluate: \(\begin{vmatrix} \sin 30^\circ & \cos 30^\circ \\ -\sin 60^\circ & \cos 60^\circ \end{vmatrix}\).
Answer: \(\begin{vmatrix} \sin 30^\circ & \cos 30^\circ \\ -\sin 60^\circ & \cos 60^\circ \end{vmatrix} = \sin 30^\circ \cos 60^\circ + \sin 60^\circ \cos 30^\circ = \sin(30^\circ + 60^\circ) = \sin 90^\circ = 1\).

Question. Evaluate: \(\begin{vmatrix} 2\cos\theta & -2\sin\theta \\ \sin\theta & \cos\theta \end{vmatrix}\).
Answer: \(\begin{vmatrix} 2\cos\theta & -2\sin\theta \\ \sin\theta & \cos\theta \end{vmatrix} = 2\cos^2\theta + 2\sin^2\theta = 2\).

Question. If \(\begin{vmatrix} x & x \\ 1 & x \end{vmatrix} = \begin{vmatrix} 3 & 4 \\ 1 & 2 \end{vmatrix}\), find the value of \(x\).
Answer: \(\begin{vmatrix} x & x \\ 1 & x \end{vmatrix} = \begin{vmatrix} 3 & 4 \\ 1 & 2 \end{vmatrix} \Rightarrow x^2 - x = 6 - 4 \Rightarrow x^2 - x - 2 = 0 \Rightarrow (x-2)(x+1) = 0 \Rightarrow x = 2, x = -1\).

Question. Write the value of \(\Delta = \begin{vmatrix} x+y & y+z & z+x \\ z & x & y \\ -3 & -3 & -3 \end{vmatrix}\).
Answer: Here, \(\Delta = \begin{vmatrix} x+y & y+z & z+x \\ z & x & y \\ -3 & -3 & -3 \end{vmatrix}\)
Applying \(R_1 \rightarrow R_1 + R_2\), we get:
\(\Delta = \begin{vmatrix} x+y+z & x+y+z & x+y+z \\ z & x & y \\ -3 & -3 & -3 \end{vmatrix}\)
Taking \(-3\) common from \(R_3\) and \(x+y+z\) common from \(R_1\), we get:
\(\Delta = -3(x+y+z) \begin{vmatrix} 1 & 1 & 1 \\ z & x & y \\ 1 & 1 & 1 \end{vmatrix} = -3(x+y+z) \cdot 0 = 0\) [since \(R_1\) and \(R_3\) are identical].

Question. If \(A\) is a \(3 \times 3\) matrix, \(|A| \neq 0\) and \(|3A| = k|A|\), then write the value of \(k\).
Answer: We have, \(|3A| = k|A|\)
\(\Rightarrow 3^3 |A| = k|A|\) [using \(|mA| = m^n|A|\), where \(n\) is the order of \(A\)]
\(\Rightarrow k = 27\).

Question. Let \(A\) be a square matrix of order \(3 \times 3\). Write the value of \(|2A|\), where \(|A| = 4\).
Answer: \(A\) is a \(3 \times 3\) matrix and \(|A| = 4\)
\(\Rightarrow |2A| = 2^3 |A| = 8 \times 4 = 32\).

Question. The value of the determinant of a matrix \(A\) of order \(3 \times 3\) is 4. Find the value of \(|5A|\).
Answer: Since \(A\) is of order \(3 \times 3\) and \(|A| = 4\), we have \(|5A| = 5^3 |A| = 125 \times 4 = 500\).

Question. If the determinant of matrix \(A\) of order \(3 \times 3\) is of value 4, write the value of \(|3A|\).
Answer: Since \(A\) is of order \(3 \times 3\) and \(|A| = 4\), we have \(|3A| = 3^3 |A| = 27 \times 4 = 108\).

Question. If \(A\) is a square matrix of order 3 and \(|3A| = p|A|\), then write the value of \(p\).
Answer: Since \(A\) is of order 3, we have \(|3A| = 3^3 |A| = 27|A|\). Therefore, comparing with \(|3A| = p|A|\), we get \(p = 27\).

Question. Write the value of the determinant \(\begin{vmatrix} 2 & 3 & 4 \\ 5 & 6 & 8 \\ 6x & 9x & 12x \end{vmatrix}\).
Answer: Taking \(3x\) common from \(R_3\), we get \(3x \begin{vmatrix} 2 & 3 & 4 \\ 5 & 6 & 8 \\ 2 & 3 & 4 \end{vmatrix}\). Since \(R_1\) and \(R_3\) are identical, the value of the determinant is zero.

Question. Show that the following determinant vanishes \(\begin{vmatrix} a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c \end{vmatrix}\).
Answer: Let \(\Delta = \begin{vmatrix} a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c \end{vmatrix}\)
Applying \(C_1 \rightarrow C_1 + C_2 + C_3\), we get:
\(\Delta = \begin{vmatrix} a-b+b-c+c-a & b-c & c-a \\ b-c+c-a+a-b & c-a & a-b \\ c-a+a-b+b-c & a-b & b-c \end{vmatrix} = \begin{vmatrix} 0 & b-c & c-a \\ 0 & c-a & a-b \\ 0 & a-b & b-c \end{vmatrix} = 0\) [since all elements of \(C_1\) are \(0\)].

Question. Without expanding, show that \(\begin{vmatrix} \sin\alpha & \cos\alpha & \cos(\alpha+\delta) \\ \sin\beta & \cos\beta & \cos(\beta+\delta) \\ \sin\gamma & \cos\gamma & \cos(\gamma+\delta) \end{vmatrix} = 0\).
Answer: Let \(\Delta = \begin{vmatrix} \sin\alpha & \cos\alpha & \cos(\alpha+\delta) \\ \sin\beta & \cos\beta & \cos(\beta+\delta) \\ \sin\gamma & \cos\gamma & \cos(\gamma+\delta) \end{vmatrix}\)
Applying \(C_3 \rightarrow C_3 + (\sin\delta)C_1 - (\cos\delta)C_2\), we get:
\(\Delta = \begin{vmatrix} \sin\alpha & \cos\alpha & 0 \\ \sin\beta & \cos\beta & 0 \\ \sin\gamma & \cos\gamma & 0 \end{vmatrix}\)
[since \(\cos(a + b) = \cos a \cos b - \sin a \sin b\)]
\(= 0\) [since all elements of \(C_3\) are \(0\)].