Exercise 4.1 Practice
Determinants: Evaluation and Properties
Q1
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Evaluate the determinant: \( \begin{vmatrix} 3 & -2 \\ 4 & 5 \end{vmatrix} \)
Formula
For \( \begin{vmatrix} a & b \\ c & d \end{vmatrix} \), value is \( ad - bc \).
Calculation
\( (3 \times 5) - (-2 \times 4) = 15 - (-8) = 15 + 8 = 23 \).
Value = 23
Q2
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Evaluate the determinants:
(i) \( \begin{vmatrix} \sec \theta & \tan \theta \\ \tan \theta & \sec \theta \end{vmatrix} \)
(ii) \( \begin{vmatrix} x^2 - x + 2 & x - 2 \\ x + 2 & x + 2 \end{vmatrix} \)
(i) \( \begin{vmatrix} \sec \theta & \tan \theta \\ \tan \theta & \sec \theta \end{vmatrix} \)
(ii) \( \begin{vmatrix} x^2 - x + 2 & x - 2 \\ x + 2 & x + 2 \end{vmatrix} \)
Part (i) Solution
\( \sec^2 \theta - \tan^2 \theta = 1 \) (Trigonometric Identity).
Part (ii) Solution
\( (x^2 - x + 2)(x + 2) - (x - 2)(x + 2) \)
\( = (x^3 + 2x^2 - x^2 - 2x + 2x + 4) - (x^2 - 4) \)
\( = (x^3 + x^2 + 4) - x^2 + 4 = x^3 + 8 \).
\( = (x^3 + 2x^2 - x^2 - 2x + 2x + 4) - (x^2 - 4) \)
\( = (x^3 + x^2 + 4) - x^2 + 4 = x^3 + 8 \).
Q3
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If \( A = \begin{bmatrix} 1 & 4 \\ 2 & 2 \end{bmatrix} \), show that \( |3A| = 9|A| \).
LHS: |3A|
\( 3A = \begin{bmatrix} 3 & 12 \\ 6 & 6 \end{bmatrix} \).
\( |3A| = 18 - 72 = -54 \).
\( |3A| = 18 - 72 = -54 \).
RHS: 9|A|
\( |A| = 2 - 8 = -6 \).
\( 9|A| = 9(-6) = -54 \).
\( 9|A| = 9(-6) = -54 \).
LHS = RHS. Verified.
Q4
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If \( A = \begin{bmatrix} 2 & 0 & 1 \\ 0 & 1 & 3 \\ 1 & 0 & 2 \end{bmatrix} \), show that \( |2A| = 8|A| \).
LHS: |2A|
\( 2A = \begin{bmatrix} 4 & 0 & 2 \\ 0 & 2 & 6 \\ 2 & 0 & 4 \end{bmatrix} \).
Expand along C2 (contains zeros): \( 2(-1)^{2+2}(16 - 4) = 2(12) = 24 \).
Expand along C2 (contains zeros): \( 2(-1)^{2+2}(16 - 4) = 2(12) = 24 \).
RHS: 8|A|
Expand A along C2: \( 1(4 - 1) = 3 \).
\( 8|A| = 8(3) = 24 \).
\( 8|A| = 8(3) = 24 \).
LHS = RHS. Verified.
Q5
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Evaluate the determinants:
(i) \( \begin{vmatrix} 2 & -1 & 3 \\ 0 & 2 & -1 \\ 3 & -2 & 0 \end{vmatrix} \)
(ii) \( \begin{vmatrix} 3 & -2 & 1 \\ 1 & 2 & 3 \\ -2 & 1 & 0 \end{vmatrix} \)
(i) \( \begin{vmatrix} 2 & -1 & 3 \\ 0 & 2 & -1 \\ 3 & -2 & 0 \end{vmatrix} \)
(ii) \( \begin{vmatrix} 3 & -2 & 1 \\ 1 & 2 & 3 \\ -2 & 1 & 0 \end{vmatrix} \)
Part (i) Expansion (Row 1)
\( 2(0 - 2) - (-1)(0 - (-3)) + 3(0 - 6) \)
\( = 2(-2) + 1(3) + 3(-6) = -4 + 3 - 18 = -19 \).
\( = 2(-2) + 1(3) + 3(-6) = -4 + 3 - 18 = -19 \).
Part (ii) Expansion (Row 1)
\( 3(0 - 3) - (-2)(0 - (-6)) + 1(1 - (-4)) \)
\( = 3(-3) + 2(6) + 1(5) = -9 + 12 + 5 = 8 \).
\( = 3(-3) + 2(6) + 1(5) = -9 + 12 + 5 = 8 \).
Q6
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If \( A = \begin{bmatrix} 1 & 2 & 5 \\ 1 & 1 & 4 \\ 2 & 3 & 9 \end{bmatrix} \), find \( |A| \).
Expansion
Expand along R1:
\( 1(9 - 12) - 2(9 - 8) + 5(3 - 2) \)
\( = 1(-3) - 2(1) + 5(1) = -3 - 2 + 5 = 0 \).
\( 1(9 - 12) - 2(9 - 8) + 5(3 - 2) \)
\( = 1(-3) - 2(1) + 5(1) = -3 - 2 + 5 = 0 \).
|A| = 0
Q7
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Find values of x, if:
(i) \( \begin{vmatrix} 3 & 5 \\ 4 & 1 \end{vmatrix} = \begin{vmatrix} 3x & 6 \\ 2 & x \end{vmatrix} \)
(ii) \( \begin{vmatrix} 3 & 4 \\ 2 & 5 \end{vmatrix} = \begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix} \)
(i) \( \begin{vmatrix} 3 & 5 \\ 4 & 1 \end{vmatrix} = \begin{vmatrix} 3x & 6 \\ 2 & x \end{vmatrix} \)
(ii) \( \begin{vmatrix} 3 & 4 \\ 2 & 5 \end{vmatrix} = \begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix} \)
Part (i)
LHS: \( 3 - 20 = -17 \).
RHS: \( 3x^2 - 12 \).
\( 3x^2 - 12 = -17 \Rightarrow 3x^2 = -5 \Rightarrow x^2 = -5/3 \).
No real solution.
RHS: \( 3x^2 - 12 \).
\( 3x^2 - 12 = -17 \Rightarrow 3x^2 = -5 \Rightarrow x^2 = -5/3 \).
No real solution.
Part (ii)
LHS: \( 15 - 8 = 7 \).
RHS: \( 5x - 6x = -x \).
\( -x = 7 \Rightarrow x = -7 \).
RHS: \( 5x - 6x = -x \).
\( -x = 7 \Rightarrow x = -7 \).
Q8
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If \( \begin{vmatrix} x & 3 \\ 12 & x \end{vmatrix} = \begin{vmatrix} 6 & 3 \\ 12 & 6 \end{vmatrix} \), then x
is equal to:
(A) 6 (B) ±6 (C) -6 (D) 0
(A) 6 (B) ±6 (C) -6 (D) 0
Solve Equation
LHS: \( x^2 - 36 \).
RHS: \( 36 - 36 = 0 \).
\( x^2 - 36 = 0 \Rightarrow x^2 = 36 \Rightarrow x = \pm 6 \).
RHS: \( 36 - 36 = 0 \).
\( x^2 - 36 = 0 \Rightarrow x^2 = 36 \Rightarrow x = \pm 6 \).
Correct Option: (B) ±6