Mathematics Questions and Answers – Determinant – 1

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This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Determinant – 1”.

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

Answer: a
Explanation: Expanding along R1, we get
∆=2(-1)-5(-1)=-2+5
=3.
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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
View Answer

Answer: c
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
View Answer

Answer: a
Explanation: Expanding along R1, we get
∆=-sinθ(sinθ)-(-1)1=-sin2⁡θ+1=cos2⁡θ.
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4. Evaluate \(\begin{vmatrix}i&-1\\-1&-i\end{vmatrix}\).
a) 4
b) 3
c) 2
d) 0
View Answer

Answer: d
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
View Answer

Answer: b
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
View Answer

Answer: c
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
View Answer

Answer: b
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.
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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
View Answer

Answer: a
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+\(\sqrt{2}\)
View Answer

Answer: d
Explanation: ∆=\(\begin{vmatrix}\sqrt{3}&\sqrt{2}\\-1&2\sqrt{3}\end{vmatrix}\)
∆=(\(\sqrt{3}\)×2\(\sqrt{3}\))+\(\sqrt{2}\)
∆=6+\(\sqrt{2}\).
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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, \(\frac{1}{3}\)
View Answer

Answer: a
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, –\(\frac{1}{3}\).

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