250+ TOP MCQs on Determinant | Class 12 Maths

Mathematics Assessment Questions and Answers on “Determinant – 2”.

1. Evaluate (begin{vmatrix}3&-1&3\6&-5&4\3&-2&3end{vmatrix})
a) 100
b) 223
c) 240
d) 230
Answer: c
Clarification: Expanding along R₁, we get
Δ=(begin{vmatrix}3&-1&3\6&-5&4\3&-2&3end{vmatrix})
Δ=3(begin{vmatrix}-5&45\-2&3end{vmatrix})-(-1)(begin{vmatrix}6&4\3&3end{vmatrix})+3(begin{vmatrix}6&-5\3&-2end{vmatrix})
Δ=3(-15+90)+(18-12)+3(-12+15)
Δ=3(75)+6+9=240.

2. Evaluate (begin{vmatrix}1&0&1\0&0&1\1&0&1end{vmatrix}).
a) 2
b) 0
c) 1
d) -1
Answer: b
Clarification: Δ=(begin{vmatrix}1&0&1\0&0&1\1&0&1end{vmatrix})
Δ=1(begin{vmatrix}0&1\0&1end{vmatrix})-0(begin{vmatrix}0&1\1&1end{vmatrix})+1(begin{vmatrix}0&0\1&0end{vmatrix})
Δ=1(0-0)-0(0-1)+1(0-0)
Δ=0-0+0=0.

3. Evaluate |A|²-5|A|+1, if A=(begin{bmatrix}7&4\5&5end{bmatrix})
a) 161
b) 251
c) 150
d) 151
Answer: d
Clarification: Given that, A=(begin{bmatrix}7&4\5&5end{bmatrix})
|A|=(7(5)-5(4))=35-20=15
|A|²-5|A|+1=(15)²-5(15)+1=225-75+1=151.

4. Evaluate (begin{vmatrix}sin⁡ ,y&0&sin⁡ ,y\cos⁡ ,y&1&cos⁡ ,x\sin⁡ ,y&1&sin ⁡,y end{vmatrix})
a) sin⁡ y (cos⁡ y-cos⁡ x)
b) sin⁡ x (cos⁡ y-cos⁡ x)
c) sin ⁡x (cos⁡ x-cos⁡ y)
d) sin ⁡y (cos⁡ 2y-cos⁡ x)
Answer: a
Clarification: Δ=(begin{vmatrix}sin⁡ ,y&0&sin⁡ ,y\cos⁡ ,y&1&cos⁡ ,x\sin⁡ ,y&1&sin⁡ ,y end{vmatrix})
Δ=sin⁡ y (begin{vmatrix}1&cos⁡ ,x\1&sin ,⁡y end{vmatrix})-0(begin{vmatrix}cos ⁡,y&cos⁡ ,x \sin⁡ ,y&sin ,y end{vmatrix})+sin⁡ y (begin{vmatrix}cos ⁡,y&1\sin ,⁡y&1end{vmatrix})
Δ=sin ⁡y (sin⁡ y-cos⁡ x)-0+sin ⁡y (cos⁡ y-sin ⁡y)
Δ=sin²⁡y-sin ⁡y cos⁡ x+sin⁡ y cos ⁡y-sin²⁡y=sin ⁡y (cos⁡ y-cos⁡ x)

5. Find the value of x, if (begin{vmatrix}2&5\3&xend{vmatrix})=(begin{vmatrix}x&-1\5&3end{vmatrix}).
a) 20
b) -20
c) 30
d) -30
Answer: b
Clarification: (begin{vmatrix}2&5\3&xend{vmatrix})=(begin{vmatrix}x&-1\5&3end{vmatrix})
⇒2x-15=3x+5
⇒x=-20

6. Find the value of x, if (begin{vmatrix}1&-1\3&-5end{vmatrix})=(begin{vmatrix}x&x^2\3&5end{vmatrix}).
a) x=2, –(frac{1}{3})
b) x=-1, –(frac{1}{3})
c) x=-2, –(frac{1}{3})
d) x=0, –(frac{1}{3})
Answer: a
Clarification: Given that, (begin{vmatrix}1&-1\3&-5end{vmatrix})=(begin{vmatrix}x&x^2\3&5end{vmatrix})
-5—(-3)=5x-3x²
-2=5x-3x²
3x²-5x-2=0
Solving for x, we get
x=2, –(frac{1}{3}).

7. Which of the following matrices will not have a determinant?
a) (begin{bmatrix}4&2\5&4end{bmatrix})
b) (begin{bmatrix}1&5&4\3&6&2\4&8&7end{bmatrix})
c) (begin{bmatrix}5&8&9\3&4&6end{bmatrix})
d) (begin{bmatrix}1&2\5&4end{bmatrix})
Answer: c
Clarification: Determinant of the matrix A=(begin{bmatrix}5&8&9\3&4&6end{bmatrix}) is not possible as it is a rectangular matrix and not a square matrix. Determinants can be calculated only if the matrix is a square matrix.

8. Find the determinant of the matrix A=(begin{bmatrix}9&8\7&6end{bmatrix})
a) -1
b) 1
c) 2
d) -2
Answer: d
Clarification: Given that, A=(begin{bmatrix}9&8\7&6end{bmatrix})
⇒Δ=(begin{vmatrix}9&8\7&6end{vmatrix})=9(6)-7(8)=54-56=-2

9. Find the determinant of the matrix A=(begin{bmatrix}-cos⁡θ&-tan⁡θ\cot⁡θ &cos⁡θ end{bmatrix}).
a) sin²⁡θ
b) sin⁡θ
c) -sin⁡θ
d) -sin²⁡θ
Answer: a
Clarification: Given that, A=(begin{bmatrix}-cos⁡θ&-tan⁡θ\cot⁡ θ&cos⁡θ end{bmatrix})
|A|=(begin{vmatrix}-cos⁡θ&-tan⁡θ\cot⁡θ&cos⁡θ end{vmatrix})
|A|=-cos⁡θ (cos⁡θ )-cotθ(-tan⁡θ)
|A|=-cos²⁡θ+1=sin²⁡θ.

10. Evaluate (begin{vmatrix}5&0&5\1&4&3\0&8&6end{vmatrix}).
a) 20
b) 0
c) -40
d) 40
Answer: b
Clarification: Δ=(begin{vmatrix}5&0&5\1&4&3\0&8&6end{vmatrix})
Expanding along R₁, we get
Δ=5(begin{vmatrix}4&3\8&6end{vmatrix})-0(begin{vmatrix}1&3\0&6end{vmatrix})+5(begin{vmatrix}1&4\0&8end{vmatrix})
Δ=5(24-24)-0+5(8-0)
Δ=0-0+40=40.

Mathematics Assessment Questions,