Evaluate the determinants in Exercises 1 and 2
Question 1: \left| \begin{array}{cc} 2 & 4 \\ -5 & -1 \end{array} \right|
Solution: \left| \begin{array}{cc} 2 & 4 \\ -5 & -1 \end{array} \right| = (2)(-1) - (4)(-5)
\left| \begin{array}{cc} 2 & 4 \\ -5 & -1 \end{array} \right| = -2 + 20
\left| \begin{array}{cc} 2 & 4 \\ -5 & -1 \end{array} \right| = 18
Question 2
(i) \left| \begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array} \right|
Solution: \left| \begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array} \right| = \cos \theta \cdot \cos \theta - (-\sin \theta)(\sin \theta)
\left| \begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array} \right| = \cos^2 \theta + \sin^2
\left| \begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array} \right| = 1
(ii) \left| \begin{array}{cc} x^2 - x + 1 & x - 1 \\ x + 1 & x + 1 \end{array} \right|
Solution: \left| \begin{array}{cc} x^2 - x + 1 & x - 1 \\ x + 1 & x + 1 \end{array} \right| = (x^2 - x + 1)(x + 1) - (x - 1)(x + 1)
\left| \begin{array}{cc} x^2 - x + 1 & x - 1 \\ x + 1 & x + 1 \end{array} \right| = x^3 + x^2 + x + 1 - (x^2 - 1)
\left| \begin{array}{cc} x^2 - x + 1 & x - 1 \\ x + 1 & x + 1 \end{array} \right| = x^3 + x^2 + x + 1 - x^2 + 1
\left| \begin{array}{cc} x^2 - x + 1 & x - 1 \\ x + 1 & x + 1 \end{array} \right| = x^3 + x + 2
Question 3: If A = \left[ \begin{array}{cc} 1 & 2 \\ 4 & 2 \end{array} \right], Then show that |2A| = 4|A|
Solution: 2A = \left[ \begin{array}{cc} 2 & 4 \\ 8 & 4 \end{array} \right]
LHS = |2A| = (2)(4) - (4)(8) = 8 - 32 = -24
LHS = -24
Now, |A| = (1)(2) - (4)(2) = 2 - 8 = -6
RHS = 4|A| = 4(-6) = -24
RHS = -24
\therefore LHS =RHS , Hence Proved
Question 4: If A = \left[ \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{array} \right], then show that |3A| = 27 |A|
Solution: 3A = \left[ \begin{array}{ccc} 3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12 \end{array} \right]
determinant of triangular matrix is product of diagonal elements:
LHS = |3A| = 3 \cdot 3 \cdot 12 = 108
Now, A = \left[ \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{array} \right]
This is an upper triangular matrix, so:
|A| = 1 \cdot 1 \cdot 4 = 4
RHS = 27|A| = 27 \cdot 4 = 108
\therefore LHS =RHS , Hence Proved
Question 5: Evaluate the determinants:
(i) \left| \begin{array}{ccc} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{array} \right|
Solution: Expanding along second row (maximum zeros):
\left| \begin{array}{ccc} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{array} \right| =(-1) \times \left| \begin{array}{cc} 3 & -1 \\ 3 & -5 \end{array} \right|
\left| \begin{array}{ccc} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{array} \right| = (-1) \times (3 \times -5 - (-1) \times 3)
\left| \begin{array}{ccc} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{array} \right|= (-1) \times (-15 + 3)
\boxed{\left| \begin{array}{ccc} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{array} \right|= 12}
(ii) \left| \begin{array}{ccc} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{array} \right|
Solution: Expanding along first row:
\left| \begin{array}{ccc} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{array} \right| = 3 \times \left| \begin{array}{cc} 1 & -2 \\ 3 & 1 \end{array} \right| - (-4) \times \left| \begin{array}{cc} 1 & -2 \\ 2 & 1 \end{array} \right| + 5 \times \left| \begin{array}{cc} 1 & 1 \\ 2 & 3 \end{array} \right|
\left| \begin{array}{ccc} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{array} \right|= 3(1 \times 1 - (-2) \times 3) + 4(1 \times 1 - (-2) \times 2) + 5(1 \times 3 - 1 \times 2)
\left| \begin{array}{ccc} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{array} \right|= 3(1+6) + 4(1+4) + 5(3-2)
\left| \begin{array}{ccc} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{array} \right| = 3(7) + 4(5) + 5(1)
\left| \begin{array}{ccc} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{array} \right|= 21 + 20 + 5 = 46
\boxed{\left| \begin{array}{ccc} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{array} \right| = 46}
(iii) \left| \begin{array}{ccc} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{array} \right|
Solution: Expanding along first row:
\left| \begin{array}{ccc} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{array} \right|= 0 \times \left| \begin{array}{cc} 0 & -3 \\ 3 & 0 \end{array} \right| - 1 \times \left| \begin{array}{cc} -1 & -3 \\ -2 & 0 \end{array} \right| + 2 \times \left| \begin{array}{cc} -1 & 0 \\ -2 & 3 \end{array} \right|
\left| \begin{array}{ccc} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{array} \right| = 0 - 1((-1)(0)-(-3)(-2)) + 2((-1)(3)-(0)(-2))
\left| \begin{array}{ccc} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{array} \right| = 0 - (0-6) + 2(-3-0)
\left| \begin{array}{ccc} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{array} \right| = 0 + 6 - 6
\boxed{\left| \begin{array}{ccc} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{array} \right| = 0}
(iv) \left| \begin{array}{ccc} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{array} \right|
Solution: Expanding along first row:
\left| \begin{array}{ccc} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{array} \right|= 2 \times \left| \begin{array}{cc} 2 & -1 \\ -5 & 0 \end{array} \right| - (-1) \times \left| \begin{array}{cc} 0 & -1 \\ 3 & 0 \end{array} \right| + (-2) \times \left| \begin{array}{cc} 0 & 2 \\ 3 & -5 \end{array} \right|
\left| \begin{array}{ccc} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{array} \right|= 2(2 \times 0 - (-1)(-5)) + 1(0 \times 0 - (-1)(3)) + (-2)(0 \times -5 - 2 \times 3)
\left| \begin{array}{ccc} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{array} \right|= 2(0-5) + 1(0+3) + (-2)(0-6)
\left| \begin{array}{ccc} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{array} \right|= 2(-5) + 3 + (-2)(-6)
\left| \begin{array}{ccc} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{array} \right|= -10 + 3 + 12
\left| \begin{array}{ccc} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{array} \right| =5
Question 6: Find |A| where:
A = \left[ \begin{array}{ccc} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{array} \right]Solution: Expanding along first row:
|A| = 1 \times \left| \begin{array}{cc} 1 & -3 \\ 4 & -9 \end{array} \right| - 1 \times \left| \begin{array}{cc} 2 & -3 \\ 5 & -9 \end{array} \right| + (-2) \times \left| \begin{array}{cc} 2 & 1 \\ 5 & 4 \end{array} \right|
|A| = 1(1 \times -9 - (-3)(4)) - 1(2 \times -9 - (-3)(5)) - 2(2 \times 4 - 1 \times 5)
|A| = 1(-9 + 12) - 1(-18 + 15) - 2(8 - 5)
|A| = 1(3) - 1(-3) - 2(3)
|A| = 3 + 3 - 6
\boxed {|A|= 0]}
Question 7: Find values of x:
(i) \left| \begin{array}{cc} 2 & 4 \\ 5 & 1 \end{array} \right| = \left| \begin{array}{cc} 2x & 4 \\ 6 & x \end{array} \right|
Solution:
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
\boxed{x = \pm \sqrt{3}}
(ii) \left| \begin{array}{cc} 2 & 3 \\ 4 & 5 \end{array} \right| = \left| \begin{array}{cc} x & 3 \\ 2x & 5 \end{array} \right|
Solution:
2 \times 5 - 3 \times 4 = x \times 5 - 3 \times 2x
10 - 12 = 5x - 6x
-2 = -x
\boxed{x = 2}
Question 8: If \left| \begin{array}{cc} x & 2 \\ 18 & x \end{array} \right| = \left| \begin{array}{cc} 6 & 2 \\ 18 & 6 \end{array} \right|,
then x is equal to
(A) 6
(B) \pm 6
(C) -6
(D) 0
Solution:
x^2 - (2)(18) = 6 \times 6 - (2)(18)
x^2 - 36 = 36 - 36
x^2 - 36 = 0
x^2 = 36
x = \pm 6
\boxed {Answer: (B)}