Mathematics Questions and Answers – Determinant – 2

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This set of Mathematics Assessment Questions and Answers 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
View Answer

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

Answer: b
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
View Answer

Answer: d
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.
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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)
View Answer

Answer: a
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
View Answer

Answer: b
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, –\(\frac{1}{3}\)
View Answer

Answer: a
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) \(\begin{bmatrix}1&2\\5&4\end{bmatrix}\)
View Answer

Answer: c
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.
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8. Find the determinant of the matrix A=\(\begin{bmatrix}9&8\\7&6\end{bmatrix}\)
a) -1
b) 1
c) 2
d) -2
View Answer

Answer: d
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
View Answer

Answer: a
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.
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10. Evaluate \(\begin{vmatrix}5&0&5\\1&4&3\\0&8&6\end{vmatrix}\).
a) 20
b) 0
c) -40
d) 40
View Answer

Answer: b
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.

Sanfoundry Global Education & Learning Series – Mathematics – Class 12.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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