MATHEMATICS CLASS- 12
CHAPTER-4
(DETERMINANTS)
Exercise 4.1
Question 1
Evaluate the determinant
$$ \begin{vmatrix} 2&4\\ -5&-1 \end{vmatrix} $$Solution:
We know that for a matrix
$$ \begin{vmatrix} a&b\\ c&d \end{vmatrix}, $$the determinant is given by
$$ ad-bc. $$Therefore,
$$ \begin{aligned} \begin{vmatrix} 2&4\\ -5&-1 \end{vmatrix} &=2(-1)-4(-5)\\[2mm] &=-2+20\\[2mm] &=18. \end{aligned} $$Hence,
$$ \boxed{ \begin{vmatrix} 2&4\\ -5&-1 \end{vmatrix} =18 } $$Question 2 (i)
Evaluate the determinant
$$ \begin{vmatrix} \cos\theta&-\sin\theta\\ \sin\theta&\cos\theta \end{vmatrix} $$Solution:
Using the formula for the determinant of a matrix of order $2$,
$$ \begin{aligned} \begin{vmatrix} \cos\theta&-\sin\theta\\ \sin\theta&\cos\theta \end{vmatrix} &=\cos\theta(\cos\theta)-(-\sin\theta)(\sin\theta)\\[2mm] &=\cos^2\theta+\sin^2\theta. \end{aligned} $$Using the identity
$$ \sin^2\theta+\cos^2\theta=1, $$we get
$$ \begin{aligned} \begin{vmatrix} \cos\theta&-\sin\theta\\ \sin\theta&\cos\theta \end{vmatrix} &=1. \end{aligned} $$Hence,
$$ \boxed{ \begin{vmatrix} \cos\theta&-\sin\theta\\ \sin\theta&\cos\theta \end{vmatrix} =1 } $$Question 2 (ii)
Evaluate the determinant
$$ \begin{vmatrix} x^2-x+1&x-1\\ x+1&x+1 \end{vmatrix} $$Solution:
Using the formula for the determinant of a matrix of order $2$,
$$ \begin{aligned} \begin{vmatrix} x^2-x+1&x-1\\ x+1&x+1 \end{vmatrix} &=(x^2-x+1)(x+1)-(x-1)(x+1). \end{aligned} $$Expanding the first product,
$$ \begin{aligned} (x^2-x+1)(x+1) &=x^3+x^2-x^2-x+x+1\\[2mm] &=x^3+1. \end{aligned} $$Also,
$$ \begin{aligned} (x-1)(x+1) &=x^2-1. \end{aligned} $$Therefore,
$$ \begin{aligned} \begin{vmatrix} x^2-x+1&x-1\\ x+1&x+1 \end{vmatrix} &=(x^3+1)-(x^2-1)\\[2mm] &=x^3-x^2+2. \end{aligned} $$Hence,
$$ \boxed{ \begin{vmatrix} x^2-x+1&x-1\\ x+1&x+1 \end{vmatrix} =x^3-x^2+2 } $$Question 3
If
$$ A= \begin{bmatrix} 1&2\\ 4&2 \end{bmatrix}, $$then show that
$$ |2A|=4|A|. $$Solution:
First, find the determinant of matrix $A$.
$$ \begin{aligned} |A| &= \begin{vmatrix} 1&2\\ 4&2 \end{vmatrix} \\[2mm] &=1(2)-2(4) \\[2mm] &=2-8 \\[2mm] &=-6. \end{aligned} $$Now, multiply every element of matrix $A$ by $2$.
$$ 2A= \begin{bmatrix} 2&4\\ 8&4 \end{bmatrix} $$Now evaluate $|2A|$.
$$ \begin{aligned} |2A| &= \begin{vmatrix} 2&4\\ 8&4 \end{vmatrix} \\[2mm] &=2(4)-4(8) \\[2mm] &=8-32 \\[2mm] &=-24. \end{aligned} $$Also,
$$ 4|A|=4(-6)=-24. $$Since
$$ |2A|=-24=4|A|, $$therefore,
$$ \boxed{|2A|=4|A|.} $$Hence proved.
Question 4
If
$$ A= \begin{bmatrix} 1&0&1\\ 0&1&2\\ 0&0&4 \end{bmatrix}, $$then show that
$$ |3A|=27|A|. $$Solution:
First, find the determinant of matrix $A$.
Since $A$ is an upper triangular matrix, its determinant is the product of its diagonal elements.
$$ \begin{aligned} |A| &=1\times1\times4 \\[2mm] &=4. \end{aligned} $$Now, multiply every element of matrix $A$ by $3$.
$$ 3A= \begin{bmatrix} 3&0&3\\ 0&3&6\\ 0&0&12 \end{bmatrix} $$Again, $3A$ is an upper triangular matrix. Therefore,
$$ \begin{aligned} |3A| &=3\times3\times12 \\[2mm] &=108. \end{aligned} $$Also,
$$ 27|A| =27\times4 =108. $$Hence,
$$ |3A|=27|A|. $$Therefore,
$$ \boxed{|3A|=27|A|.} $$Hence proved.
Question 5 (i)
Evaluate the determinant
$$ \begin{vmatrix} 3&-1&-2\\ 0&0&-1\\ 3&-5&0 \end{vmatrix} $$Solution:
Expand the determinant along the second row, since it contains two zeros.
$$ \begin{aligned} \Delta &= \begin{vmatrix} 3&-1&-2\\ 0&0&-1\\ 3&-5&0 \end{vmatrix} \\[2mm] &= 0C_{21}+0C_{22}+(-1)C_{23}. \end{aligned} $$Now,
$$ C_{23}=(-1)^{2+3} \begin{vmatrix} 3&-1\\ 3&-5 \end{vmatrix}. $$Therefore,
$$ \begin{aligned} \Delta &=(-1)\left[(-1)^5 \begin{vmatrix} 3&-1\\ 3&-5 \end{vmatrix} \right] \\[2mm] &= \begin{vmatrix} 3&-1\\ 3&-5 \end{vmatrix}. \end{aligned} $$Now evaluate the determinant of order $2$.
$$ \begin{aligned} \begin{vmatrix} 3&-1\\ 3&-5 \end{vmatrix} &=3(-5)-(-1)(3) \\[2mm] &=-15+3 \\[2mm] &=-12. \end{aligned} $$Hence,
$$ \boxed{ \begin{vmatrix} 3&-1&-2\\ 0&0&-1\\ 3&-5&0 \end{vmatrix} =-12 } $$Question 5 (ii)
Evaluate the determinant
$$ \begin{vmatrix} 3&-4&5\\ 1&1&-2\\ 2&3&1 \end{vmatrix} $$Solution:
Expand the determinant along the first row.
$$ \begin{aligned} \Delta &= 3 \begin{vmatrix} 1&-2\\ 3&1 \end{vmatrix} -(-4) \begin{vmatrix} 1&-2\\ 2&1 \end{vmatrix} +5 \begin{vmatrix} 1&1\\ 2&3 \end{vmatrix}. \end{aligned} $$Now evaluate each minor.
$$ \begin{aligned} \begin{vmatrix} 1&-2\\ 3&1 \end{vmatrix} &=1(1)-(-2)(3) \\[2mm] &=1+6 \\[2mm] &=7. \end{aligned} $$ $$ \begin{aligned} \begin{vmatrix} 1&-2\\ 2&1 \end{vmatrix} &=1(1)-(-2)(2) \\[2mm] &=1+4 \\[2mm] &=5. \end{aligned} $$ $$ \begin{aligned} \begin{vmatrix} 1&1\\ 2&3 \end{vmatrix} &=1(3)-1(2) \\[2mm] &=3-2 \\[2mm] &=1. \end{aligned} $$Substitute these values in the expansion.
$$ \begin{aligned} \Delta &=3(7)-(-4)(5)+5(1) \\[2mm] &=21+20+5 \\[2mm] &=46. \end{aligned} $$Hence,
$$ \boxed{ \begin{vmatrix} 3&-4&5\\ 1&1&-2\\ 2&3&1 \end{vmatrix} =46 } $$Question 5 (iii)
Evaluate the determinant
$$ \begin{vmatrix} 0&1&2\\ -1&0&-3\\ -2&3&0 \end{vmatrix} $$Solution:
Expand the determinant along the first row.
$$ \begin{aligned} \Delta &= 0 \begin{vmatrix} 0&-3\\ 3&0 \end{vmatrix} - 1 \begin{vmatrix} -1&-3\\ -2&0 \end{vmatrix} + 2 \begin{vmatrix} -1&0\\ -2&3 \end{vmatrix}. \end{aligned} $$The first term is zero. Therefore,
$$ \begin{aligned} \Delta &= - \begin{vmatrix} -1&-3\\ -2&0 \end{vmatrix} + 2 \begin{vmatrix} -1&0\\ -2&3 \end{vmatrix}. \end{aligned} $$Now evaluate the minors.
$$ \begin{aligned} \begin{vmatrix} -1&-3\\ -2&0 \end{vmatrix} &=(-1)(0)-(-3)(-2) \\[2mm] &=0-6 \\[2mm] &=-6. \end{aligned} $$ $$ \begin{aligned} \begin{vmatrix} -1&0\\ -2&3 \end{vmatrix} &=(-1)(3)-0(-2) \\[2mm] &=-3. \end{aligned} $$Substituting these values,
$$ \begin{aligned} \Delta &=-(-6)+2(-3) \\[2mm] &=6-6 \\[2mm] &=0. \end{aligned} $$Hence,
$$ \boxed{ \begin{vmatrix} 0&1&2\\ -1&0&-3\\ -2&3&0 \end{vmatrix} =0 } $$Question 5 (iv)
Evaluate the determinant
$$ \begin{vmatrix} 2&-1&-2\\ 0&2&-1\\ 3&-5&0 \end{vmatrix} $$Solution:
Expand the determinant along the first row.
$$ \begin{aligned} \Delta &= 2 \begin{vmatrix} 2&-1\\ -5&0 \end{vmatrix} -(-1) \begin{vmatrix} 0&-1\\ 3&0 \end{vmatrix} +(-2) \begin{vmatrix} 0&2\\ 3&-5 \end{vmatrix}. \end{aligned} $$Now evaluate each minor.
$$ \begin{aligned} \begin{vmatrix} 2&-1\\ -5&0 \end{vmatrix} &=2(0)-(-1)(-5) \\[2mm] &=0-5 \\[2mm] &=-5. \end{aligned} $$ $$ \begin{aligned} \begin{vmatrix} 0&-1\\ 3&0 \end{vmatrix} &=0(0)-(-1)(3) \\[2mm] &=3. \end{aligned} $$ $$ \begin{aligned} \begin{vmatrix} 0&2\\ 3&-5 \end{vmatrix} &=0(-5)-2(3) \\[2mm] &=-6. \end{aligned} $$Substituting these values,
$$ \begin{aligned} \Delta &=2(-5)-(-1)(3)+(-2)(-6) \\[2mm] &=-10+3+12 \\[2mm] &=5. \end{aligned} $$Hence,
$$ \boxed{ \begin{vmatrix} 2&-1&-2\\ 0&2&-1\\ 3&-5&0 \end{vmatrix} =5 } $$Question 6
If
$$ A= \begin{bmatrix} 1&1&-2\\ 2&1&-3\\ 5&4&-9 \end{bmatrix}, $$find $|A|$.
Solution:
Expand the determinant along the first row.
$$ \begin{aligned} |A| &= 1 \begin{vmatrix} 1&-3\\ 4&-9 \end{vmatrix} - 1 \begin{vmatrix} 2&-3\\ 5&-9 \end{vmatrix} +(-2) \begin{vmatrix} 2&1\\ 5&4 \end{vmatrix}. \end{aligned} $$Now evaluate each minor.
$$ \begin{aligned} \begin{vmatrix} 1&-3\\ 4&-9 \end{vmatrix} &=1(-9)-(-3)(4) \\[2mm] &=-9+12 \\[2mm] &=3. \end{aligned} $$ $$ \begin{aligned} \begin{vmatrix} 2&-3\\ 5&-9 \end{vmatrix} &=2(-9)-(-3)(5) \\[2mm] &=-18+15 \\[2mm] &=-3. \end{aligned} $$ $$ \begin{aligned} \begin{vmatrix} 2&1\\ 5&4 \end{vmatrix} &=2(4)-1(5) \\[2mm] &=8-5 \\[2mm] &=3. \end{aligned} $$Substituting these values,
$$ \begin{aligned} |A| &=1(3)-1(-3)+(-2)(3) \\[2mm] &=3+3-6 \\[2mm] &=0. \end{aligned} $$Hence,
$$ \boxed{|A|=0} $$Question 7 (i)
Find the value of $x$, if
$$ \begin{vmatrix} 2&4\\ 5&1 \end{vmatrix} = \begin{vmatrix} 2x&4\\ 6&x \end{vmatrix}. $$Solution:
First, evaluate the determinant on the left-hand side.
$$ \begin{aligned} \begin{vmatrix} 2&4\\ 5&1 \end{vmatrix} &=2(1)-4(5) \\[2mm] &=2-20 \\[2mm] &=-18. \end{aligned} $$Now evaluate the determinant on the right-hand side.
$$ \begin{aligned} \begin{vmatrix} 2x&4\\ 6&x \end{vmatrix} &=2x(x)-4(6) \\[2mm] &=2x^2-24. \end{aligned} $$Since the two determinants are equal,
$$ 2x^2-24=-18. $$Therefore,
$$ 2x^2=6 $$ $$ x^2=3 $$ $$ x=\pm\sqrt3. $$Hence,
$$ \boxed{x=\pm\sqrt3} $$Question 7 (ii)
Find the value of $x$, if
$$ \begin{vmatrix} 2&3\\ 4&5 \end{vmatrix} = \begin{vmatrix} x&3\\ 2x&5 \end{vmatrix}. $$Solution:
First, evaluate the determinant on the left-hand side.
$$ \begin{aligned} \begin{vmatrix} 2&3\\ 4&5 \end{vmatrix} &=2(5)-3(4) \\[2mm] &=10-12 \\[2mm] &=-2. \end{aligned} $$Now evaluate the determinant on the right-hand side.
$$ \begin{aligned} \begin{vmatrix} x&3\\ 2x&5 \end{vmatrix} &=x(5)-3(2x) \\[2mm] &=5x-6x \\[2mm] &=-x. \end{aligned} $$Since the two determinants are equal,
$$ -x=-2. $$Therefore,
$$ x=2. $$Hence,
$$ \boxed{x=2} $$Question 8
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) $\pm6$
(C) $-6$
(D) $0$
Solution:
First, evaluate the determinant on the left-hand side.
$$ \begin{aligned} \begin{vmatrix} x&2\\ 18&x \end{vmatrix} &=x(x)-2(18) \\[2mm] &=x^2-36. \end{aligned} $$Now, evaluate the determinant on the right-hand side.
$$ \begin{aligned} \begin{vmatrix} 6&2\\ 18&6 \end{vmatrix} &=6(6)-2(18) \\[2mm] &=36-36 \\[2mm] &=0. \end{aligned} $$Since the two determinants are equal,
$$ x^2-36=0. $$Therefore,
$$ x^2=36 $$ $$ x=\pm6. $$Hence, the correct answer is (B) $\pm6$.
$$ \boxed{x=\pm6} $$