Here are the solutions for Exercise 4.1, which involves evaluating determinants of $2 \times 2$ and $3 \times 3$ matrices and verifying determinant properties.
Table of Contents
1. Evaluate the determinant: $\begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix}$
The determinant of a $2 \times 2$ matrix $\begin{vmatrix} a & b \\ c & d \end{vmatrix}$ is $ad – bc$.
$$\begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix} = (2)(-1) – (4)(-5)$$
$$= -2 – (-20)$$
$$= -2 + 20$$
$$\mathbf{= 18}$$
2. Evaluate the determinants:
(i) $\begin{vmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{vmatrix}$
$$\begin{vmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{vmatrix} = (\cos \theta)(\cos \theta) – (-\sin \theta)(\sin \theta)$$
$$= \cos^2 \theta – (-\sin^2 \theta)$$
$$= \cos^2 \theta + \sin^2 \theta$$
$$\mathbf{= 1}$$
(Using the identity $\cos^2 \theta + \sin^2 \theta = 1$)
(ii) $\begin{vmatrix} x^2 – x + 1 & x – 1 \\ x + 1 & x + 1 \end{vmatrix}$
$$\begin{vmatrix} x^2 – x + 1 & x – 1 \\ x + 1 & x + 1 \end{vmatrix} = (x^2 – x + 1)(x + 1) – (x – 1)(x + 1)$$
Using the identity $(a-b)(a+b) = a^2 – b^2$:
$$= (x^2 – x + 1)(x + 1) – (x^2 – 1^2)$$
Using the identity $(x^2 – x + 1)(x + 1) = x^3 + 1^3$:
$$= (x^3 + 1) – (x^2 – 1)$$
$$= x^3 + 1 – x^2 + 1$$
$$\mathbf{= x^3 – x^2 + 2}$$
3. If $A = \begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}$, then show that $| 2A | = 4 | A |$.
- Calculate $| A |$:$$| A | = \begin{vmatrix} 1 & 2 \\ 4 & 2 \end{vmatrix} = (1)(2) – (2)(4) = 2 – 8 = -6$$$$4 | A | = 4(-6) = \mathbf{-24}$$
- Calculate $| 2A |$:First find $2A$:$$2A = 2 \begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix} = \begin{bmatrix} 2 & 4 \\ 8 & 4 \end{bmatrix}$$Now find the determinant of $2A$:$$| 2A | = \begin{vmatrix} 2 & 4 \\ 8 & 4 \end{vmatrix} = (2)(4) – (4)(8) = 8 – 32 = \mathbf{-24}$$Since $| 2A | = -24$ and $4 | A | = -24$, it is shown that $| 2A | = 4 | A |$.
4. If $A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{bmatrix}$, then show that $| 3A | = 27 | A |$.
This verifies the property: $| kA | = k^n | A |$, where $n$ is the order of the matrix. Here $k=3$ and $n=3$, so $| 3A | = 3^3 | A | = 27 | A |$.
- Calculate $| A |$:Since $A$ is an upper triangular matrix, its determinant is the product of its diagonal elements:$$| A | = (1)(1)(4) = 4$$$$27 | A | = 27(4) = \mathbf{108}$$
- Calculate $| 3A |$:First find $3A$:$$3A = 3 \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{bmatrix} = \begin{bmatrix} 3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12 \end{bmatrix}$$$3A$ is also an upper triangular matrix. Its determinant is the product of its diagonal elements:$$| 3A | = (3)(3)(12) = 9 \times 12 = \mathbf{108}$$Since $| 3A | = 108$ and $27 | A | = 108$, it is shown that $| 3A | = 27 | A |$.
5. Evaluate the determinants:
(i) $\begin{vmatrix} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{vmatrix}$
Expand along the second row ($\mathbf{R_2}$), as it has two zeros, simplifying the calculation:
$$D = 0 \cdot C_{21} + 0 \cdot C_{22} + (-1) \cdot C_{23}$$
$$C_{23} = (-1)^{2+3} \begin{vmatrix} 3 & -1 \\ 3 & -5 \end{vmatrix} = (-1) [(3)(-5) – (-1)(3)]$$
$$C_{23} = (-1) [-15 – (-3)] = (-1) [-15 + 3] = (-1)(-12) = 12$$
$$D = (-1) \cdot C_{23} = (-1)(12)$$
$$\mathbf{= -12}$$
(ii) $\begin{vmatrix} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{vmatrix}$
Expand along the first row ($\mathbf{R_1}$):
$$D = 3 \begin{vmatrix} 1 & -2 \\ 3 & 1 \end{vmatrix} – (-4) \begin{vmatrix} 1 & -2 \\ 2 & 1 \end{vmatrix} + 5 \begin{vmatrix} 1 & 1 \\ 2 & 3 \end{vmatrix}$$
$$D = 3 [(1)(1) – (-2)(3)] + 4 [(1)(1) – (-2)(2)] + 5 [(1)(3) – (1)(2)]$$
$$D = 3 [1 – (-6)] + 4 [1 – (-4)] + 5 [3 – 2]$$
$$D = 3 (7) + 4 (5) + 5 (1)$$
$$D = 21 + 20 + 5$$
$$\mathbf{= 46}$$
(iii) $\begin{vmatrix} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{vmatrix}$
Expand along the first row ($\mathbf{R_1}$):
$$D = 0 \begin{vmatrix} 0 & -3 \\ 3 & 0 \end{vmatrix} – 1 \begin{vmatrix} -1 & -3 \\ -2 & 0 \end{vmatrix} + 2 \begin{vmatrix} -1 & 0 \\ -2 & 3 \end{vmatrix}$$
$$D = 0 – 1 [(-1)(0) – (-3)(-2)] + 2 [(-1)(3) – (0)(-2)]$$
$$D = -1 [0 – 6] + 2 [-3 – 0]$$
$$D = -1 (-6) + 2 (-3)$$
$$D = 6 – 6$$
$$\mathbf{= 0}$$
(iv) $\begin{vmatrix} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{vmatrix}$
Expand along the first column ($\mathbf{C_1}$), as it contains a zero:
$$D = 2 \begin{vmatrix} 2 & -1 \\ -5 & 0 \end{vmatrix} – 0 \begin{vmatrix} -1 & -2 \\ -5 & 0 \end{vmatrix} + 3 \begin{vmatrix} -1 & -2 \\ 2 & -1 \end{vmatrix}$$
$$D = 2 [(2)(0) – (-1)(-5)] – 0 + 3 [(-1)(-1) – (-2)(2)]$$
$$D = 2 [0 – 5] + 3 [1 – (-4)]$$
$$D = 2 (-5) + 3 (5)$$
$$D = -10 + 15$$
$$\mathbf{= 5}$$
6. If $A = \begin{bmatrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{bmatrix}$, find $| A |$.
Expand along the first row ($\mathbf{R_1}$):
$$| A | = 1 \begin{vmatrix} 1 & -3 \\ 4 & -9 \end{vmatrix} – 1 \begin{vmatrix} 2 & -3 \\ 5 & -9 \end{vmatrix} + (-2) \begin{vmatrix} 2 & 1 \\ 5 & 4 \end{vmatrix}$$
$$| A | = 1 [(1)(-9) – (-3)(4)] – 1 [(2)(-9) – (-3)(5)] – 2 [(2)(4) – (1)(5)]$$
$$| A | = 1 [-9 – (-12)] – 1 [-18 – (-15)] – 2 [8 – 5]$$
$$| A | = 1 [-9 + 12] – 1 [-18 + 15] – 2 [3]$$
$$| A | = 1 (3) – 1 (-3) – 6$$
$$| A | = 3 + 3 – 6$$
$$\mathbf{= 0}$$
(Since $|A| = 0$, the matrix $A$ is singular.)
7. Find values of $x$, if:
(i) $\begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix} = \begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}$
- Evaluate the determinant of the LHS:$$\begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix} = (2)(1) – (4)(5) = 2 – 20 = -18$$
- Evaluate the determinant of the RHS:$$\begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix} = (2x)(x) – (4)(6) = 2x^2 – 24$$
- Equate the two expressions and solve for $x$:$$-18 = 2x^2 – 24$$$$2x^2 = 24 – 18$$$$2x^2 = 6$$$$x^2 = 3$$$$\mathbf{x = \pm \sqrt{3}}$$
(ii) $\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix} = \begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix}$
- Evaluate the determinant of the LHS:$$\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix} = (2)(5) – (3)(4) = 10 – 12 = -2$$
- Evaluate the determinant of the RHS:$$\begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix} = (x)(5) – (3)(2x) = 5x – 6x = -x$$
- Equate the two expressions and solve for $x$:$$-2 = -x$$$$\mathbf{x = 2}$$
8. If $\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix} = \begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix}$, then $x$ is equal to
(A) 6 (B) $\pm 6$ (C) – 6 (D) 0
Solution:
- Evaluate the determinant of the LHS:$$\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix} = (x)(x) – (2)(18) = x^2 – 36$$
- Evaluate the determinant of the RHS:$$\begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix} = (6)(6) – (2)(18) = 36 – 36 = 0$$
- Equate the two expressions and solve for $x$:$$x^2 – 36 = 0$$$$x^2 = 36$$$$x = \pm 6$$
$$\text{The correct option is } \mathbf{(B)}$$