NCERT – Exercise 4.1
Evaluate the determinants in Exercises 1 and 2.
1. \( \begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix} \)
Solution:
\( \begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix} = 2(-1) – (-5)(4) = -2 + 20 = 18 \).
2. \( \begin{vmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{vmatrix} \)
Solution:
\( \begin{vmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{vmatrix} = \cos \theta \cdot \cos \theta – (\sin \theta)(-\sin \theta) \)
= \( \cos^2 \theta + \sin^2 \theta = 1 \).
3. \( \begin{vmatrix} x^2 – x + 1 & x – 1 \\ x + 1 & x + 1 \end{vmatrix} \)
Solution:
\( \begin{vmatrix} x^2 – x + 1 & x – 1 \\ x + 1 & x + 1 \end{vmatrix} = (x^2 – x + 1)(x + 1) – (x – 1)(x + 1) \)
= \( (x^3 + 1) – (x^2 – 1) = x^3 + 1 – x^2 + 1 = x^3 – x^2 + 2 \).
4. If \( A = \begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix} \), then show that \( |2A| = 4|A| \).
Solution:
Given \( A = \begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix} \), so \( 2A = \begin{bmatrix} 2 & 4 \\ 8 & 4 \end{bmatrix} \).
L.H.S. \( = |2A| = \begin{vmatrix} 2 & 4 \\ 8 & 4 \end{vmatrix} = 8 – 32 = -24 \).
R.H.S. \( = 4|A| = 4 \begin{vmatrix} 1 & 2 \\ 4 & 2 \end{vmatrix} = 4(2 – 8) = -24 \).
Hence, \( |2A| = 4|A| \).
5. If \( A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{bmatrix} \), then show that \( |3A| = 27|A| \).
Solution:
Given \( A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{bmatrix} \), so \( 3A = \begin{bmatrix} 3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12 \end{bmatrix} \).
L.H.S. \( = |3A| = \begin{vmatrix} 3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12 \end{vmatrix} \)
= \( 3 \begin{vmatrix} 3 & 6 \\ 0 & 12 \end{vmatrix} – 0 \begin{vmatrix} 0 & 6 \\ 0 & 12 \end{vmatrix} + 3 \begin{vmatrix} 0 & 3 \\ 0 & 0 \end{vmatrix} \)
= \( 3 \times 36 – 0 + 3 \times 0 = 108 \).
R.H.S. \( = 27|A| = 27 \begin{vmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{vmatrix} \)
= \( 27 \left[ 1 \begin{vmatrix} 1 & 2 \\ 0 & 4 \end{vmatrix} – 0 \begin{vmatrix} 0 & 2 \\ 0 & 4 \end{vmatrix} + 1 \begin{vmatrix} 0 & 1 \\ 0 & 0 \end{vmatrix} \right] \)
= \( 27[1(4) – 0 + 0] = 108 \).
Hence, \( |3A| = 27|A| \).
Evaluate the determinants:
6. (i) \( \begin{vmatrix} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{vmatrix} \)
Solution:
\( \begin{vmatrix} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{vmatrix} = 3(0 – 5) + 1(0 + 3) – 2 \times 0 \)
= \( -15 + 3 = -12 \).
6. (ii) \( \begin{vmatrix} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{vmatrix} \)
Solution:
\( \begin{vmatrix} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{vmatrix} = 3 \begin{vmatrix} 1 & -2 \\ 3 & 1 \end{vmatrix} + 4 \begin{vmatrix} 1 & -2 \\ 2 & 1 \end{vmatrix} + 5 \begin{vmatrix} 1 & 1 \\ 2 & 3 \end{vmatrix} \)
= \( 3(1 + 6) + 4(1 + 4) + 5(3 – 2) = 21 + 20 + 5 = 46 \).
6. (iii) \( \begin{vmatrix} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{vmatrix} \)
Solution:
\( \begin{vmatrix} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{vmatrix} = 0 \begin{vmatrix} 0 & -3 \\ 3 & 0 \end{vmatrix} – 1 \begin{vmatrix} -1 & -3 \\ -2 & 0 \end{vmatrix} + 2 \begin{vmatrix} -1 & 0 \\ -2 & 3 \end{vmatrix} \)
= \( 0 – 1(0 – 6) + 2(-3) = 6 – 6 = 0 \).
6. (iv) \( \begin{vmatrix} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{vmatrix} \)
Solution:
\( \begin{vmatrix} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{vmatrix} = 2 \begin{vmatrix} 2 & -1 \\ -5 & 0 \end{vmatrix} + 1 \begin{vmatrix} 0 & -1 \\ 3 & 0 \end{vmatrix} – 2 \begin{vmatrix} 0 & 2 \\ 3 & -5 \end{vmatrix} \)
= \( 2(0 – 5) + 1(0 + 3) – 2(0 – 6) = -10 + 3 + 12 = 5 \).
7. If \( A = \begin{bmatrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{bmatrix} \), find \( |A| \).
Solution:
Given \( A = \begin{bmatrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{bmatrix} \),
\( |A| = \begin{vmatrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{vmatrix} \)
= \( 1 \begin{vmatrix} 1 & -3 \\ 4 & -9 \end{vmatrix} – 1 \begin{vmatrix} 2 & -3 \\ 5 & -9 \end{vmatrix} – 2 \begin{vmatrix} 2 & 1 \\ 5 & 4 \end{vmatrix} \)
= \( 1(-9 + 12) – 1(-18 + 15) – 2(8 – 5) = 3 + 3 – 6 = 0 \).
8. Find the values of \( x \), if:
(i) \( \begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix} = \begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix} \)
Solution:
\( \begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix} = \begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix} \)
\( \Rightarrow 2 – 20 = 2x^2 – 24 \)
\( \Rightarrow -18 = 2x^2 – 24 \)
\( \Rightarrow 2x^2 = 6 \)
\( \Rightarrow x^2 = 3 \)
\( \Rightarrow x = \pm \sqrt{3} \).
(ii) \( \begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix} = \begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix} \)
Solution:
\( \begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix} = \begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix} \)
\( \Rightarrow 2 \times 5 – 4 \times 3 = 5x – 6x \)
\( \Rightarrow 10 – 12 = -x \)
\( \Rightarrow x = 2 \).
9. 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:
\( \begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix} = \begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix} \)
\( \Rightarrow x^2 – 36 = 36 – 36 \)
\( \Rightarrow x^2 = 36 \)
\( \Rightarrow x = \pm 6 \).
Hence, the correct answer is B.