Properties of Determinants - Class 12 Maths MCQ

This set of Class 12 Maths Chapter 4 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.

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.

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.

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).

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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