# Class 12 Maths MCQ – Determinant – 2

This set of Class 12 Maths Chapter 4 Multiple Choice Questions & Answers (MCQs) focuses on “Determinant – 2”.

1. Evaluate $$\begin{vmatrix}3&-1&3\\6&-5&4\\3&-2&3\end{vmatrix}$$
a) 100
b) 223
c) 240
d) 230

Explanation: Expanding along R1, we get
Δ=$$\begin{vmatrix}3&-1&3\\6&-5&4\\3&-2&3\end{vmatrix}$$
Δ=3$$\begin{vmatrix}-5&45\\-2&3\end{vmatrix}$$-(-1)$$\begin{vmatrix}6&4\\3&3\end{vmatrix}$$+3$$\begin{vmatrix}6&-5\\3&-2\end{vmatrix}$$
Δ=3(-15+90)+(18-12)+3(-12+15)
Δ=3(75)+6+9=240.

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

Explanation: Δ=$$\begin{vmatrix}1&0&1\\0&0&1\\1&0&1\end{vmatrix}$$
Δ=1$$\begin{vmatrix}0&1\\0&1\end{vmatrix}$$-0$$\begin{vmatrix}0&1\\1&1\end{vmatrix}$$+1$$\begin{vmatrix}0&0\\1&0\end{vmatrix}$$
Δ=1(0-0)-0(0-1)+1(0-0)
Δ=0-0+0=0.

3. Evaluate |A|2-5|A|+1, if A=$$\begin{bmatrix}7&4\\5&5\end{bmatrix}$$
a) 161
b) 251
c) 150
d) 151

Explanation: Given that, A=$$\begin{bmatrix}7&4\\5&5\end{bmatrix}$$
|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)

Explanation: 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 $$\begin{vmatrix}cos ⁡\,y&1\\sin \,⁡y&1\end{vmatrix}$$
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 $$\begin{vmatrix}2&5\\3&x\end{vmatrix}$$=$$\begin{vmatrix}x&-1\\5&3\end{vmatrix}$$.
a) 20
b) -20
c) 30
d) -30

Explanation: $$\begin{vmatrix}2&5\\3&x\end{vmatrix}$$=$$\begin{vmatrix}x&-1\\5&3\end{vmatrix}$$
⇒2x-15=3x+5
⇒x=-20
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6. Find the value of x, if $$\begin{vmatrix}1&-1\\3&-5\end{vmatrix}$$=$$\begin{vmatrix}x&x^2\\3&5\end{vmatrix}$$.
a) x=2, –$$\frac{1}{3}$$
b) x=-1, –$$\frac{1}{3}$$
c) x=-2, –$$\frac{1}{3}$$
d) x=0, –$$\frac{1}{3}$$

Explanation: Given that, $$\begin{vmatrix}1&-1\\3&-5\end{vmatrix}$$=$$\begin{vmatrix}x&x^2\\3&5\end{vmatrix}$$ -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) $$\begin{bmatrix}4&2\\5&4\end{bmatrix}$$
b) $$\begin{bmatrix}1&5&4\\3&6&2\\4&8&7\end{bmatrix}$$
c) $$\begin{bmatrix}5&8&9\\3&4&6\end{bmatrix}$$
d) $$\begin{bmatrix}1&2\\5&4\end{bmatrix}$$

Explanation: Determinant of the matrix A=$$\begin{bmatrix}5&8&9\\3&4&6\end{bmatrix}$$ 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=$$\begin{bmatrix}9&8\\7&6\end{bmatrix}$$
a) -1
b) 1
c) 2
d) -2

Explanation: Given that, A=$$\begin{bmatrix}9&8\\7&6\end{bmatrix}$$
⇒Δ=$$\begin{vmatrix}9&8\\7&6\end{vmatrix}$$=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

Explanation: 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 $$\begin{vmatrix}5&0&5\\1&4&3\\0&8&6\end{vmatrix}$$.
a) 20
b) 0
c) -40
d) 40

Explanation: Δ=$$\begin{vmatrix}5&0&5\\1&4&3\\0&8&6\end{vmatrix}$$
Expanding along R1, we get
Δ=5$$\begin{vmatrix}4&3\\8&6\end{vmatrix}$$-0$$\begin{vmatrix}1&3\\0&6\end{vmatrix}$$+5$$\begin{vmatrix}1&4\\0&8\end{vmatrix}$$
Δ=5(24-24)-0+5(8-0)
Δ=0-0+40=40.

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