250+ TOP MCQs on Determinant | Class 12 Maths

Mathematics Multiple Choice Questions on “Determinant – 1”.

1. Evaluate (begin{vmatrix}2&5\-1&-1end{vmatrix}).
a) 3
b) -7
c) 5
d) -2
Answer: a
Clarification: Expanding along R₁, we get
∆=2(-1)-5(-1)=-2+5
=3.

2. Evaluate (begin{vmatrix}5&-4\1&sqrt{3}end{vmatrix}).
a) 4(sqrt{3})+4
b) 4(sqrt{3})+5
c) 5(sqrt{3})+4
d) 5(sqrt{3})-4
Answer: c
Clarification: Evaluating along R₁, we get
∆=5((sqrt{3}))-(-4)1=5(sqrt{3})+4.

3. Evaluate (begin{vmatrix}-sinθ&-1\1&sin⁡θend{vmatrix}).
a) cos²⁡θ
b) -cos²⁡θ
c) cos⁡2θ
d) cos⁡θ
Answer: a
Clarification: Expanding along R₁, we get
∆=-sinθ(sinθ)-(-1)1=-sin²⁡θ+1=cos²⁡θ.

4. Evaluate (begin{vmatrix}i&-1\-1&-iend{vmatrix}).
a) 4
b) 3
c) 2
d) 0
Answer: d
Clarification: Expanding along R₁, we get
∆=-i(i)-(-1)(-1)=-i²-1=-(-1)-1=0.

5. Evaluate (begin{vmatrix}1&1&-2\3&4&5\-1&2&1end{vmatrix}).
a) -6
b) -34
c) 34
d) 22
Answer: b
Clarification: ∆=(begin{vmatrix}1&1&-2\3&4&5\-1&2&1end{vmatrix})
Expanding along the first row, we get
∆=1(begin{vmatrix}4&5\2&1end{vmatrix})-1(begin{vmatrix}3&5\-1&1end{vmatrix})-2(begin{vmatrix}3&4\-1&2end{vmatrix})
=1(4-5(2))-1(3-5(-1))-2(6-4(-1))
=(4-10)-(3+5)-2(6+4)
=-6-8-20=-34.

6. Evaluate (begin{vmatrix}5&4&3\3&4&1\5&6&1end{vmatrix}).
a) 4
b) -24
c) -8
d) 8
Answer: c
Clarification: Expanding along the first row, we get
∆=5(begin{vmatrix}4&1\6&1end{vmatrix})-4(begin{vmatrix}3&1\5&1end{vmatrix})+3(begin{vmatrix}3&4\5&6end{vmatrix})
=5(4-6)-4(3-5)+3(18-20)
=5(-2)-4(-2)+3(-2)=-10+8-6=-8.

7. Evaluate (begin{vmatrix}8x+1&2x-2\x^2-1&3x+5end{vmatrix}).
a) -2x³-26x²+45x+3
b) -2x³+26x²+45x+3
c) -2x³+26x²+45x-3
d) -2x³-26x²-45x+3
Answer: b
Clarification: Expanding along the first row, we get
∆=8x+1(3x+5)-(2x-2)(x²-1)
=(24x²+43x+5)-(2x³-2x²-2x+2)
=-2x³+26x²+45x+3.

8. If A=(begin{bmatrix}2&5&9\6&1&3\4&8&2end{bmatrix}), find |A|.
a) 352
b) 356
c) 325
d) 532
Answer: a
Clarification: Given that, A=(begin{bmatrix}2&5&9\6&1&3\4&8&2end{bmatrix})
⇒|A|=(begin{vmatrix}2&5&9\6&1&3\4&8&2end{vmatrix})
Evaluating along the first row, we get
∆=2(begin{vmatrix}1&3\8&2end{vmatrix})-5(begin{vmatrix}6&3\4&2end{vmatrix})+9(begin{vmatrix}6&1\4&8end{vmatrix})
∆=2(2-24)-5(12-12)+9(48-4)
∆=2(-22)-0+9(44)
∆=-44+9(44)=44(-1+9)=352

9. Evaluate (begin{vmatrix}sqrt{3}&sqrt{2}\-1&2sqrt{3}end{vmatrix}).
a) 6-3(sqrt{2})
b) 6-(sqrt{2})
c) 6+3(sqrt{2})
d) 6+(sqrt{2})
Answer: d
Clarification: ∆=(begin{vmatrix}sqrt{3}&sqrt{2}\-1&2sqrt{3}end{vmatrix})
∆=((sqrt{3})×2(sqrt{3}))+(sqrt{2})
∆=6+(sqrt{2}).

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