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 - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter