NCERT Solutions Class-12-Chapter-4-Determinants
Miscellaneous Exercises on Chapter 4
Note: This exercise covers advanced determinant properties, matrix inverses, and system of equations. Recall that \( (AB)^{-1} = B^{-1}A^{-1} \) and the range of trigonometric functions when evaluating determinants involving them.
Q1
Prove that the determinant \( \begin{vmatrix} x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x \end{vmatrix} \) is independent of \( \theta \).
Solution:
Expanding along the first row:
\[ \begin{aligned} \Delta &= x(-x^2 - 1) - \sin \theta (-x \sin \theta - \cos \theta) + \cos \theta (-\sin \theta + x \cos \theta) \\ &= -x^3 - x + x \sin^2 \theta + \sin \theta \cos \theta - \sin \theta \cos \theta + x \cos^2 \theta \\ &= -x^3 - x + x(\sin^2 \theta + \cos^2 \theta) \\ &= -x^3 - x + x(1) \\ &= -x^3 \end{aligned} \]
Since the result \( -x^3 \) does not contain \( \theta \), the determinant is independent of \( \theta \).
Q2
Evaluate the determinant: \( \begin{vmatrix} \cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\ -\sin \beta & \cos \beta & 0 \\ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha \end{vmatrix} \)
Solution:
Expanding along the third column (since it has a zero):
\[ \begin{aligned} \Delta &= -\sin \alpha (-\sin \beta \sin \alpha \sin \beta - \cos \beta \sin \alpha \cos \beta) - 0 + \cos \alpha (\cos \alpha \cos \beta \cos \beta - \cos \alpha \sin \beta (-\sin \beta)) \\ &= -\sin \alpha [-\sin \alpha (\sin^2 \beta + \cos^2 \beta)] + \cos \alpha [\cos \alpha (\cos^2 \beta + \sin^2 \beta)] \\ &= -\sin \alpha (-\sin \alpha (1)) + \cos \alpha (\cos \alpha (1)) \\ &= \sin^2 \alpha + \cos^2 \alpha \\ &= 1 \end{aligned} \]
Q3
If \( A^{-1} = \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix} \), find \( (AB)^{-1} \).
Solution:
We know that \( (AB)^{-1} = B^{-1} A^{-1} \).
Step 1: Find \( B^{-1} \)
\( |B| = 1(3-0) - 2(-1-0) - 2(2-0) = 3 + 2 - 4 = 1 \).
Cofactors of B:
- \( C_{11}=3, C_{12}=1, C_{13}=2 \)
- \( C_{21}=2, C_{22}=1, C_{23}=2 \)
- \( C_{31}=6, C_{32}=2, C_{33}=5 \)
\( \text{adj } B = \begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix} \). So \( B^{-1} = \begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix} \).
Step 2: Multiply \( B^{-1} A^{-1} \)
\[ (AB)^{-1} = \begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix} \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix} = \begin{bmatrix} 9-30+30 & -3+12-12 & 3-10+12 \\ 3-15+10 & -1+6-4 & 1-5+4 \\ 6-30+25 & -2+12-10 & 2-10+10 \end{bmatrix} \]
\[ = \begin{bmatrix} 9 & -3 & 5 \\ -2 & 1 & 0 \\ 1 & 0 & 2 \end{bmatrix} \]
Q4
Let \( A = \begin{bmatrix} 1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5 \end{bmatrix} \). Verify that
(i) \( [\text{adj } A]^{-1} = \text{adj } (A^{-1}) \)
(ii) \( (A^{-1})^{-1} = A \)
(i) \( [\text{adj } A]^{-1} = \text{adj } (A^{-1}) \)
(ii) \( (A^{-1})^{-1} = A \)
Solution:
Step 1: Find \( |A| \)
\( |A| = 1(15-1) - (-2)(-10-1) + 1(-2-3) = 14 - 22 - 5 = -13 \neq 0 \).
Step 2: Find \( \text{adj } A \)
Cofactor matrix \( C = \begin{bmatrix} 14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1 \end{bmatrix} \). Since A is symmetric, \( \text{adj } A = C \).
(i) Calculate \( [\text{adj } A]^{-1} \) by finding determinant and adjoint of \( \text{adj } A \). Then calculate \( A^{-1} \) and its adjoint. They will be equal.
(ii) We know the property \( (A^{-1})^{-1} = A \). Calculating the inverse of \( A^{-1} \) will return matrix A.
Q5
Evaluate the determinant: \( \begin{vmatrix} x & y & x+y \\ y & x+y & x \\ x+y & x & y \end{vmatrix} \)
Solution:
Apply \( R_1 \to R_1 + R_2 + R_3 \):
\[ \Delta = \begin{vmatrix} 2(x+y) & 2(x+y) & 2(x+y) \\ y & x+y & x \\ x+y & x & y \end{vmatrix} = 2(x+y) \begin{vmatrix} 1 & 1 & 1 \\ y & x+y & x \\ x+y & x & y \end{vmatrix} \]
Apply \( C_2 \to C_2 - C_1 \) and \( C_3 \to C_3 - C_1 \):
\[ \Delta = 2(x+y) \begin{vmatrix} 1 & 0 & 0 \\ y & x & x-y \\ x+y & -y & -x \end{vmatrix} \]
Expanding along first row:
\[ \Delta = 2(x+y) [1(-x^2 - (-y)(x-y))] = 2(x+y)(-x^2 + xy - y^2) = -2(x+y)(x^2 - xy + y^2) = -2(x^3 + y^3) \]
Q6
Evaluate the determinant: \( \begin{vmatrix} 1 & x & y \\ 1 & x+y & y \\ 1 & x & x+y \end{vmatrix} \)
Solution:
Apply \( R_2 \to R_2 - R_1 \) and \( R_3 \to R_3 - R_1 \):
\[ \Delta = \begin{vmatrix} 1 & x & y \\ 0 & y & 0 \\ 0 & 0 & x \end{vmatrix} \]
Expanding along \( C_1 \):
\( \Delta = 1(xy - 0) = xy \).
Q7
Solve the system of equations:
\( \frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4 \)
\( \frac{4}{x} - \frac{6}{y} + \frac{5}{z} = 1 \)
\( \frac{6}{x} + \frac{9}{y} - \frac{20}{z} = 2 \)
\( \frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4 \)
\( \frac{4}{x} - \frac{6}{y} + \frac{5}{z} = 1 \)
\( \frac{6}{x} + \frac{9}{y} - \frac{20}{z} = 2 \)
Solution:
Let \( \frac{1}{x} = u, \frac{1}{y} = v, \frac{1}{z} = w \).
The system becomes \( AX = B \):
\[ \begin{bmatrix} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \end{bmatrix} = \begin{bmatrix} 4 \\ 1 \\ 2 \end{bmatrix} \]
Calculate \( |A| = 1200 \). Calculate \( \text{adj } A \).
Find \( A^{-1} = \frac{1}{1200} \begin{bmatrix} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{bmatrix} \).
\[ X = A^{-1}B = \frac{1}{1200} \begin{bmatrix} 300+150+150 \\ 440-100+60 \\ 288+0-48 \end{bmatrix} = \frac{1}{1200} \begin{bmatrix} 600 \\ 400 \\ 240 \end{bmatrix} = \begin{bmatrix} 1/2 \\ 1/3 \\ 1/5 \end{bmatrix} \]
Thus, \( u = \frac{1}{2} \Rightarrow x = 2 \). \( v = \frac{1}{3} \Rightarrow y = 3 \). \( w = \frac{1}{5} \Rightarrow z = 5 \).
Q8
If \( x, y, z \) are nonzero real numbers, then the inverse of matrix \( A = \begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{bmatrix} \) is- \( \begin{bmatrix} x^{-1} & 0 & 0 \\ 0 & y^{-1} & 0 \\ 0 & 0 & z^{-1} \end{bmatrix} \)
- \( xyz \begin{bmatrix} x^{-1} & 0 & 0 \\ 0 & y^{-1} & 0 \\ 0 & 0 & z^{-1} \end{bmatrix} \)
- ...
- ...
- \( \begin{bmatrix} x^{-1} & 0 & 0 \\ 0 & y^{-1} & 0 \\ 0 & 0 & z^{-1} \end{bmatrix} \)
- \( xyz \begin{bmatrix} x^{-1} & 0 & 0 \\ 0 & y^{-1} & 0 \\ 0 & 0 & z^{-1} \end{bmatrix} \)
- ...
- ...
Answer: (A)
Solution:
For a diagonal matrix \( A = \text{diag}(x, y, z) \), the inverse is simply the diagonal matrix with reciprocal elements: \( A^{-1} = \text{diag}(x^{-1}, y^{-1}, z^{-1}) \).
Q9
Let \( A = \begin{bmatrix} 1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1 \end{bmatrix} \), where \( 0 \le \theta \le 2\pi \). Then- Det(A) = 0
- Det(A) \( \in (2, \infty) \)
- Det(A) \( \in (2, 4) \)
- Det(A) \( \in [2, 4] \)
- Det(A) = 0
- Det(A) \( \in (2, \infty) \)
- Det(A) \( \in (2, 4) \)
- Det(A) \( \in [2, 4] \)
Answer: (D)
Solution:
Expanding the determinant:
\[ \begin{aligned} |A| &= 1(1+\sin^2 \theta) - \sin \theta (-\sin \theta + \sin \theta) + 1(\sin^2 \theta + 1) \\ &= 1 + \sin^2 \theta - 0 + \sin^2 \theta + 1 \\ &= 2(1 + \sin^2 \theta) \end{aligned} \]
Range of \( \sin \theta \) is \( [-1, 1] \), so \( \sin^2 \theta \in [0, 1] \).
Range of \( 1 + \sin^2 \theta \in [1, 2] \).
Range of \( 2(1 + \sin^2 \theta) \in [2, 4] \).