Mathematics Questions and Answers – Properties of Determinants

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This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Properties of Determinants”.

1. Which of the following is not a property of determinant?
a) The value of determinant changes if all of its rows and columns are interchanged
b) The value of determinant changes if any two rows or columns are interchanged
c) The value of determinant is zero if any two rows and columns are identical
d) The value of determinant gets multiplied by k, if each element of row or column is multiplied by k
View Answer

Answer: a
Explanation: The value of determinant remains unchanged if all of its rows and columns are interchanged i.e. |A|=|A’|, where A is a square matrix and A’ is the transpose of the matrix A.
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2. Find the determinant of the matrix A=\(\begin{bmatrix}1&x&y\\1&x&-y\\1&-x^2&y^2\end{bmatrix}\).
a) (x+1)
b) -2xy(x+1)
c) xy(x+1)
d) 2xy(x+1)
View Answer

Answer: b
Explanation: Given that, A=\(\begin{bmatrix}1&x&y\\1&x&-y\\1&-x^2&y^2\end{bmatrix}\)
Δ=\(\begin{vmatrix}1&x&y\\1&x&-y\\1 &-x^2&y^2 \end{vmatrix}\)
Taking x common C2 and y common from C3, we get
Δ=xy\(\begin{vmatrix}1&1&1\\1&1&-1\\1&-x&y\end{vmatrix}\)
Expanding along R1, we get
Δ=xy{1(y-x)-1(y+1)+1(-x-1)}
Δ=xy(y-x-y-1-x-1)
Δ=xy(-2x-2)=-2xy(x+1).

3. Evaluate \(\begin{vmatrix}x^2&x^3&x^4\\x&y&z\\x^2&x^3&x^4 \end{vmatrix}\).
a) 0
b) 1
c) xyz
d) x2 yz3
View Answer

Answer: a
Explanation: Δ=\(\begin{vmatrix}x^2&x^3&x^4\\x&y&z\\x^2&x^3&x^4 \end{vmatrix}\)
If the elements of any two rows or columns are identical, then the value of determinant is zero. Here, the elements of row 1 and row 3 are identical. Hence, its determinant is 0.
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4. Evaluate \(\begin{vmatrix}cos⁡θ&-cos⁡θ&1\\sin^2⁡θ&cos^2⁡θ&1\\sin⁡θ&-sin⁡θ&1\end{vmatrix}\).
a) sin⁡θ+cos2⁡θ
b) -sin⁡θ-cos2⁡⁡θ
c) -sin⁡θ+cos2⁡⁡θ
d) sin⁡θ-cos2⁡⁡θ
View Answer

Answer: d
Explanation: Δ=\(\begin{vmatrix}cos⁡θ&-cos⁡θ&1\\sin^2⁡θ&cos^2⁡θ&1\\sin⁡θ&-sin⁡θ&1\end{vmatrix}\)
Applying C1→C1+C2
Δ=\(\begin{vmatrix}cos⁡θ-cos⁡θ&-cos⁡θ&1\\sin^2⁡θ+cos^2⁡θ&cos^2⁡θ&1\\sinθ-sin⁡θ&-sin⁡θ&1\end{vmatrix}\)=\(\begin{vmatrix}0&-cos⁡θ&1\\1&cos^2⁡θ&1\\0&-sin⁡θ&1\end{vmatrix}\)
Expanding along C1, we get
0-1(cos2⁡⁡θ+sinθ)=sin⁡θ-cos2⁡⁡θ.

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

Answer: b
Explanation: Δ=\(\begin{vmatrix}b-c&b&c\\a&c-a&c\\a&b&a-b\end{vmatrix}\)
Applying C2→C2-C3
Δ=\(\begin{vmatrix}b-c&b-c&c\\a&-a&c\\a&-a&a-b\end{vmatrix}\)
Applying C1→C1-C2
Δ=\(\begin{vmatrix}0&b-c&c\\2a&-a&c\\2a&-a&a-b\end{vmatrix}\)
Applying R2→R2-R3
Δ=\(\begin{vmatrix}0&b-c&c\\0&0&c-a+b\\2a&-a&a-b\end{vmatrix}\)
Expanding along C1, we get
Δ=2a{(b-c)(c-a+b)}
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6. If A=\(\begin{bmatrix}1&3\\2&1\end{bmatrix}\), then ________
a) |2A|=4|A|
b) |2A|=2|A|
c) |A|=2|A|
d) |A|=|4A|
View Answer

Answer: a
Explanation: Given that, A=\(\begin{bmatrix}1&3\\2&1\end{bmatrix}\)
2A=2\(\begin{bmatrix}1&3\\2&1\end{bmatrix}\)=\(\begin{bmatrix}2&6\\4&2\end{bmatrix}\)
|2A|=\(\begin{vmatrix}2&6\\4&2\end{vmatrix}\)=(4-24)=-20
4|A|=4\(\begin{vmatrix}1&3\\2&1\end{vmatrix}\)=4(1-6)=4(-5)=-20
∴|2A|=4|A|.

7. Evaluate \(\begin{vmatrix}-a&b&c\\-2a+4x&2b-4y&2c+4z\\x&-y&z\end{vmatrix}\).
a) 0
b) abc
c) 2abc
d) -1
View Answer

Answer: a
Explanation: Δ=\(\begin{vmatrix}-a&b&c\\-2a+4x&2b-4y&2c+4z\\x&-y&z\end{vmatrix}\)
Using the properties of determinants, the given determinant can be expressed as a sum of two determinants.
Δ=\(\begin{vmatrix}-a&b&c\\-2a&2b&2c\\x&-y&z\end{vmatrix}\)+\(\begin{vmatrix}-a&b&c\\4x&-4y&4z\\x&-y&z\end{vmatrix}\)
Δ=2\(\begin{vmatrix}-a&b&c\\-a&b&c\\x&-y&z\end{vmatrix}\)+4\(\begin{vmatrix}-a&b&c\\x&-y&z\\x&-y&z\end{vmatrix}\)
Since two rows are similar in each of the determinants, the determinant is 0.
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8. Find the determinant of A=\(\begin{bmatrix}c^2&cb&ca\\ab&a^2&-ac\\ab&bc&-b^2\end{bmatrix}\)
a) abc(a3+b3+c3+abc)
b) abc(a3+b3+c3-abc)
c) abc(a3+b3+c3+abc)
d) (a3-b3+c3-abc)
View Answer

Answer: b
Explanation: Given that, A=\(\begin{bmatrix}c^2&cb&ca\\ab&a^2&-ac\\ab&bc&-b^2\end{bmatrix}\)
Taking c a, b common from R1, R2, R3 respectively, we get
Δ=abc\(\begin{bmatrix}c&b&a\\b&a&-c\\a&c&-b\end{bmatrix}\)
Δ=abc{(c(-ab+c2)-b(-b2+ac)+a(bc-a2)
Δ=abc(-abc+c3+b3-abc+abc-a3)
Δ=abc(a3+b3+c3-abc).

9. Evaluate \(\begin{vmatrix}1+m&n&q\\m&1+n&q\\n&m&1+q\end{vmatrix}\).
a) -1(1+m+n+q)
b) 1+m+n+q
c) 1+2q
d) 1+q
View Answer

Answer: a
Explanation: Given that, Δ=\(\begin{vmatrix}1+m&n&q\\m&1+n&q\\n&m&1+q\end{vmatrix}\)
Applying C1→C1+C2+C3
Δ=\(\begin{vmatrix}1+m+n+q&n&q\\1+m+n+q&1+n&q\\1+m+n+q&m&1+q\end{vmatrix}\)=(1+m+n+q)\(\begin{vmatrix}1&n&q\\1&1+n&q\\1&m&1+q\end{vmatrix}\)
Applying R1→R2-R1
Δ=(1+m+n+q)\(\begin{vmatrix}0&1&0\\1&1+n&q\\1&m&1+q\end{vmatrix}\)
Expanding along the first row, we get
Δ=(1+m+n+q)(0-1(1+q-q)+0)
Δ=-1(1+m+n+q).
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10. Evaluate \(\begin{vmatrix}4&8&12\\6&12&18\\7&14&21\end{vmatrix}\).
a) 168
b) -1
c) -168
d) 0
View Answer

Answer: d
Explanation: Δ=\(\begin{vmatrix}4&8&12\\6&12&18\\7&14&21\end{vmatrix}\)
Taking 4, 6 and 7 from R1, R2, R3 respectively
Δ=4×6×7\(\begin{vmatrix}1&2&3\\1&2&3\\1&2&3\end{vmatrix}\)
Since the elements of all rows are identical, the determinant is zero.

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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