# Class 12 Maths MCQ – Determinants

This set of Class 12 Maths Chapter 4 Multiple Choice Questions & Answers (MCQs) focuses on “Determinants”. These MCQs are created based on the latest CBSE syllabus and the NCERT curriculum, offering valuable assistance for exam preparation.

1. Evaluate $$\begin{vmatrix}2&5\\-1&-1\end{vmatrix}$$.
a) 3
b) -7
c) 5
d) -2

Explanation: 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

Explanation: Evaluating along R1, we get
∆=5($$\sqrt{3}$$)-(-4)1=5$$\sqrt{3}$$+4.

3. Evaluate $$\begin{vmatrix}-sinθ&-1\\1&sin⁡θ\end{vmatrix}$$.
a) cos2⁡θ
b) -cos2⁡θ
c) cos⁡2θ
d) cos⁡θ

Explanation: Expanding along R1, we get
∆ = -sinθ(sinθ)-(-1)1=-sin2⁡θ+1=cos2⁡θ.

4. Evaluate $$\begin{vmatrix}i&-1\\-1&-i\end{vmatrix}$$.
a) 4
b) 3
c) 2
d) 0

Explanation: Expanding along R1, we get
∆=-i(i)-(-1)(-1)=-i2-1=-(-1)-1=0.

5. Evaluate $$\begin{vmatrix}1&1&-2\\3&4&5\\-1&2&1\end{vmatrix}$$.
a) -6
b) -34
c) 34
d) 22

Explanation: ∆=$$\begin{vmatrix}1&1&-2\\3&4&5\\-1&2&1\end{vmatrix}$$
Expanding along the first row, we get
∆=1$$\begin{vmatrix}4&5\\2&1\end{vmatrix}$$-1$$\begin{vmatrix}3&5\\-1&1\end{vmatrix}$$-2$$\begin{vmatrix}3&4\\-1&2\end{vmatrix}$$
=1(4-5(2))-1(3-5(-1))-2(6-4(-1))
=(4-10)-(3+5)-2(6+4)
=-6-8-20=-34.
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6. Evaluate $$\begin{vmatrix}5&4&3\\3&4&1\\5&6&1\end{vmatrix}$$.
a) 4
b) -24
c) -8
d) 8

Explanation: Expanding along the first row, we get
∆=5$$\begin{vmatrix}4&1\\6&1\end{vmatrix}$$-4$$\begin{vmatrix}3&1\\5&1\end{vmatrix}$$+3$$\begin{vmatrix}3&4\\5&6\end{vmatrix}$$
=5(4-6)-4(3-5)+3(18-20)
=5(-2)-4(-2)+3(-2)=-10+8-6=-8.

7. Evaluate $$\begin{vmatrix}8x+1&2x-2\\x^2-1&3x+5\end{vmatrix}$$.
a) -2x3-26x2+45x+3
b) -2x3+26x2+45x+3
c) -2x3+26x2+45x-3
d) -2x3-26x2-45x+3

Explanation: 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=$$\begin{bmatrix}2&5&9\\6&1&3\\4&8&2\end{bmatrix}$$, find |A|.
a) 352
b) 356
c) 325
d) 532

Explanation: Given that, A=$$\begin{bmatrix}2&5&9\\6&1&3\\4&8&2\end{bmatrix}$$
⇒|A|=$$\begin{vmatrix}2&5&9\\6&1&3\\4&8&2\end{vmatrix}$$
Evaluating along the first row, we get
∆=2$$\begin{vmatrix}1&3\\8&2\end{vmatrix}$$-5$$\begin{vmatrix}6&3\\4&2\end{vmatrix}$$+9$$\begin{vmatrix}6&1\\4&8\end{vmatrix}$$
∆=2(2-24)-5(12-12)+9(48-4)
∆=2(-22)-0+9(44)
∆=-44+9(44)=44(-1+9)=352

9. Evaluate $$\begin{vmatrix}\sqrt{3}&\sqrt{2}\\-1&2\sqrt{3}\end{vmatrix}$$.
a) 6-3$$\sqrt{2}$$
b) 6-$$\sqrt{2}$$
c) 6+3$$\sqrt{2}$$
d) 6+$$\sqrt{2}$$

Explanation: ∆=$$\begin{vmatrix}\sqrt{3}&\sqrt{2}\\-1&2\sqrt{3}\end{vmatrix}$$
∆=($$\sqrt{3}$$×2$$\sqrt{3}$$)+$$\sqrt{2}$$
∆=6+$$\sqrt{2}$$.

10. Find the value of x if $$\begin{vmatrix}3&x\\2&x^2 \end{vmatrix}$$=$$\begin{vmatrix}5&3\\3&2\end{vmatrix}$$.
a) x=1, –$$\frac{1}{3}$$
b) x=-1, –$$\frac{1}{3}$$
c) x=1, $$\frac{1}{3}$$
d) x=-1, $$\frac{1}{3}$$

Explanation: Given that $$\begin{vmatrix}3&x\\2&x^2 \end{vmatrix}$$=$$\begin{vmatrix}5&3\\3&2\end{vmatrix}$$
⇒3x2-2x=5(2)-3(3)
⇒3x2-2x=1
Solving for x, we get
x=1, –$$\frac{1}{3}$$.

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