NCERT Solutions Class-12-Chapter-4-Determinants
Excercise-4.1
Note: A determinant is a scalar value that can be computed from the elements of a square matrix. For a \( 2 \times 2 \) matrix \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \), the determinant is given by \( |A| = ad - bc \).
Q1
Evaluate the determinant: \( \begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix} \)
Solution:
\[ \begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix} = (2 \times -1) - (4 \times -5) \]
\[ = -2 - (-20) = -2 + 20 = 18 \]
Q2
Evaluate the determinants:
(i) \( \begin{vmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{vmatrix} \)
(ii) \( \begin{vmatrix} x^2-x+1 & x-1 \\ x+1 & x+1 \end{vmatrix} \)
(i) \( \begin{vmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{vmatrix} \)
(ii) \( \begin{vmatrix} x^2-x+1 & x-1 \\ x+1 & x+1 \end{vmatrix} \)
Solution:
(i)
\[ \begin{aligned} \begin{vmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{vmatrix} &= (\cos \theta)(\cos \theta) - (-\sin \theta)(\sin \theta) \\ &= \cos^2 \theta - (-\sin^2 \theta) \\ &= \cos^2 \theta + \sin^2 \theta \\ &= 1 \end{aligned} \]
(ii)
\[ \begin{aligned} \begin{vmatrix} x^2-x+1 & x-1 \\ x+1 & x+1 \end{vmatrix} &= (x^2-x+1)(x+1) - (x-1)(x+1) \\ &= (x^3 + 1) - (x^2 - 1) \\ &= x^3 + 1 - x^2 + 1 \\ &= x^3 - x^2 + 2 \end{aligned} \]
Q3
If \( A = \begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix} \), then show that \( |2A| = 4|A| \).
Solution:
First, calculate \( 2A \):
\[ 2A = 2 \begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix} = \begin{bmatrix} 2 & 4 \\ 8 & 4 \end{bmatrix} \]
Calculate determinant of \( 2A \):
\[ |2A| = (2 \times 4) - (4 \times 8) = 8 - 32 = -24 \]
Now, calculate determinant of \( A \):
\[ |A| = (1 \times 2) - (2 \times 4) = 2 - 8 = -6 \]
Calculate \( 4|A| \):
\[ 4|A| = 4 \times (-6) = -24 \]
Since \( |2A| = 4|A| \), the statement is verified.
Q4
If \( A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{bmatrix} \), then show that \( |3A| = 27|A| \).
Solution:
First, calculate \( 3A \):
\[ 3A = 3 \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{bmatrix} = \begin{bmatrix} 3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12 \end{bmatrix} \]
Calculate \( |3A| \) (expanding along first column):
\[ |3A| = 3(3 \times 12 - 6 \times 0) - 0 + 0 = 3(36) = 108 \]
Now, calculate \( |A| \) (expanding along first column):
\[ |A| = 1(1 \times 4 - 2 \times 0) - 0 + 0 = 1(4) = 4 \]
Calculate \( 27|A| \):
\[ 27|A| = 27 \times 4 = 108 \]
Since \( |3A| = 27|A| \), the statement is verified.
Q5
Evaluate the determinants:
(i) \( \begin{vmatrix} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{vmatrix} \)
(ii) \( \begin{vmatrix} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{vmatrix} \)
(iii) \( \begin{vmatrix} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{vmatrix} \)
(iv) \( \begin{vmatrix} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{vmatrix} \)
(i) \( \begin{vmatrix} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{vmatrix} \)
(ii) \( \begin{vmatrix} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{vmatrix} \)
(iii) \( \begin{vmatrix} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{vmatrix} \)
(iv) \( \begin{vmatrix} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{vmatrix} \)
Solution:
(i) Expanding along the second row (as it has two zeros):
\[ \Delta = -(-1) \begin{vmatrix} 3 & -1 \\ 3 & -5 \end{vmatrix} = 1 [3(-5) - (-1)(3)] = 1 [-15 + 3] = -12 \]
(ii) Expanding along the first row:
\[ \begin{aligned} \Delta &= 3 \begin{vmatrix} 1 & -2 \\ 3 & 1 \end{vmatrix} - (-4) \begin{vmatrix} 1 & -2 \\ 2 & 1 \end{vmatrix} + 5 \begin{vmatrix} 1 & 1 \\ 2 & 3 \end{vmatrix} \\ &= 3(1 + 6) + 4(1 + 4) + 5(3 - 2) \\ &= 3(7) + 4(5) + 5(1) \\ &= 21 + 20 + 5 = 46 \end{aligned} \]
(iii) Expanding along the first row:
\[ \begin{aligned} \Delta &= 0 - 1 \begin{vmatrix} -1 & -3 \\ -2 & 0 \end{vmatrix} + 2 \begin{vmatrix} -1 & 0 \\ -2 & 3 \end{vmatrix} \\ &= -1(0 - 6) + 2(-3 - 0) \\ &= 6 - 6 = 0 \end{aligned} \]
(iv) Expanding along the second row:
\[ \begin{aligned} \Delta &= -0 + 2 \begin{vmatrix} 2 & -2 \\ 3 & 0 \end{vmatrix} - (-1) \begin{vmatrix} 2 & -1 \\ 3 & -5 \end{vmatrix} \\ &= 2(0 - (-6)) + 1(-10 - (-3)) \\ &= 2(6) + 1(-7) \\ &= 12 - 7 = 5 \end{aligned} \]
Q6
If \( A = \begin{bmatrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{bmatrix} \), find \( |A| \).
Solution:
Expanding along the first row:
\[ \begin{aligned} |A| &= 1 \begin{vmatrix} 1 & -3 \\ 4 & -9 \end{vmatrix} - 1 \begin{vmatrix} 2 & -3 \\ 5 & -9 \end{vmatrix} + (-2) \begin{vmatrix} 2 & 1 \\ 5 & 4 \end{vmatrix} \\ &= 1(-9 - (-12)) - 1(-18 - (-15)) - 2(8 - 5) \\ &= 1(3) - 1(-3) - 2(3) \\ &= 3 + 3 - 6 = 0 \end{aligned} \]
Q7
Find values of x, if
(i) \( \begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix} = \begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix} \)
(ii) \( \begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix} = \begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix} \)
(i) \( \begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix} = \begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix} \)
(ii) \( \begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix} = \begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix} \)
Solution:
(i)
\[ \begin{aligned} (2 \times 1) - (4 \times 5) &= (2x \times x) - (4 \times 6) \\ 2 - 20 &= 2x^2 - 24 \\ -18 &= 2x^2 - 24 \\ 2x^2 &= 6 \\ x^2 &= 3 \Rightarrow x = \pm \sqrt{3} \end{aligned} \]
(ii)
\[ \begin{aligned} (2 \times 5) - (3 \times 4) &= (x \times 5) - (3 \times 2x) \\ 10 - 12 &= 5x - 6x \\ -2 &= -x \\ x &= 2 \end{aligned} \]
Q8
If \( \begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix} = \begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix} \), then x is equal to- 6
- \(\pm 6\)
- -6
- 0
- 6
- \(\pm 6\)
- -6
- 0
Answer: (B)
Solution:
\[ \begin{aligned} x^2 - 36 &= 36 - 36 \\ x^2 - 36 &= 0 \\ x^2 &= 36 \\ x &= \pm 6 \end{aligned} \]