Exercise 4.1, Question 1
\(\begin{vmatrix}2 & 4 \\ -5 & -1\end{vmatrix}\)
Show solution
- \((2)(-1) - (4)(-5) = -2 + 20\).
Answer: 18
NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.1
Exercise 4.1: Evaluating determinants. \(\begin{vmatrix}a & b \\ c & d\end{vmatrix} = ad - bc\). For order 3, expand along any row or column with signs \(+ - +\); choose the row or column with the most zeros. \(|kA| = k^n|A|\) for an \(n \times n\) matrix.
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Exercise 4.1 questions and solutions
Exercise 4.1, Question 2
Evaluate:
(i) \(\begin{vmatrix}\cos{\left(\theta \right)} & - \sin{\left(\theta \right)} \\ \sin{\left(\theta \right)} & \cos{\left(\theta \right)}\end{vmatrix}\)
Show solution
- \(\cos^2\theta + \sin^2\theta\).
Answer: 1
(ii) \(\begin{vmatrix}x^{2} - x + 1 & x - 1 \\ x + 1 & x + 1\end{vmatrix}\)
Show solution
- \[(x^2 - x + 1)(x + 1) - (x - 1)(x + 1) = (x^3 + 1) - (x^2 - 1)\]
Answer: \(x^3 - x^2 + 2\)
Exercise 4.1, Question 3
\(A = \begin{bmatrix}1 & 2 \\ 4 & 2\end{bmatrix}\). Show \(|2A| = 4|A|\).
Show solution
- \(|A| = 2 - 8 = -6\).
- \[2A = \begin{bmatrix}2 & 4 \\ 8 & 4\end{bmatrix}\], \(|2A| = 8 - 32 = -24 = 4(-6)\).
Answer: \(|2A| = -24 = 4|A|\)
Exercise 4.1, Question 4
\(A = \begin{bmatrix}1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4\end{bmatrix}\). Show \(|3A| = 27|A|\).
Show solution
- A is upper triangular: \(|A| = 1 \cdot 1 \cdot 4 = 4\).
- \(3A\) is upper triangular with diagonal \(3, 3, 12\): \(|3A| = 108 = 27 \cdot 4\).
Answer: \(|3A| = 108 = 27|A|\)
Exercise 4.1, Question 5
Evaluate:
(i) \(\begin{vmatrix}3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0\end{vmatrix}\)
Show solution
- Expand along row 2 (two zeros): only \(a_{23} = -1\) contributes, with sign \((-1)^{2+3} = -1\).
- \[-(-1)\begin{vmatrix}3 & -1 \\ 3 & -5\end{vmatrix} = -15 + 3\]
Answer: \(-12\)
(ii) \(\begin{vmatrix}3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1\end{vmatrix}\)
Show solution
- Along row 1: \[3(1 + 6) + 4(1 + 4) + 5(3 - 2) = 21 + 20 + 5\]
Answer: 46
(iii) \(\begin{vmatrix}0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0\end{vmatrix}\)
Show solution
- Along row 1: \(0 - 1(0 - 6) + 2(-3 - 0) = 6 - 6\). (It is skew symmetric of odd order, so the value must be 0.)
Answer: 0
(iv) \(\begin{vmatrix}2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0\end{vmatrix}\)
Show solution
- Along row 1: \[2(0 - 5) + 1(0 + 3) - 2(0 - 6) = -10 + 3 + 12\]
Answer: 5
Exercise 4.1, Question 6
\(A = \begin{bmatrix}1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9\end{bmatrix}\). Find \(|A|\).
Show solution
- Along row 1: \[1(-9 + 12) - 1(-18 + 15) - 2(8 - 5) = 3 + 3 - 6\]
Answer: 0
Exercise 4.1, Question 7
Find x.
(i) \(\begin{vmatrix}2 & 4 \\ 5 & 1\end{vmatrix} = \begin{vmatrix}2 x & 4 \\ 6 & x\end{vmatrix}\)
Show solution
- \(2 - 20 = 2x^2 - 24 \Rightarrow 2x^2 = 6\).
Answer: \(x = \pm\sqrt3\)
(ii) \(\begin{vmatrix}2 & 3 \\ 4 & 5\end{vmatrix} = \begin{vmatrix}x & 3 \\ 2 x & 5\end{vmatrix}\)
Show solution
- \(10 - 12 = 5x - 6x \Rightarrow -2 = -x\).
Answer: \(x = 2\)
Exercise 4.1, Question 8
If \(\begin{vmatrix}x & 2 \\ 18 & x\end{vmatrix} = \begin{vmatrix}6 & 2 \\ 18 & 6\end{vmatrix}\), x is: (A) 6 (B) \(\pm6\) (C) \(-6\) (D) 0
Show solution
- \(x^2 - 36 = 36 - 36 = 0\).
Answer: (B) \(\pm6\)
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Determinants in our sample papers: Sample paper 1 (questions 3, 4, 7, 32) · Sample paper 2 (questions 4, 6, 7, 35) · Sample paper 3 (questions 4, 7, 32) · Sample paper 4 (questions 4, 5, 7, 35) · Sample paper 5 (questions 4, 6, 7, 32).
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