250+ TOP MCQs on Properties of Determinants | Class 12 Maths

Mathematics Multiple Choice Questions 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
Answer: a
Clarification: 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^2end{bmatrix}).
a) (x+1)
b) -2xy(x+1)
c) xy(x+1)
d) 2xy(x+1)
Answer: b
Clarification: Given that, A=(begin{bmatrix}1&x&y\1&x&-y\1&-x^2&y^2end{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&yend{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
Answer: a
Clarification: 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&1end{vmatrix}).
a) sin⁡θ+cos²⁡θ
b) -sin⁡θ-cos²⁡⁡θ
c) -sin⁡θ+cos²⁡⁡θ
d) sin⁡θ-cos²⁡⁡θ
Answer: d
Clarification: Δ=(begin{vmatrix}cos⁡θ&-cos⁡θ&1\sin^2⁡θ&cos^2⁡θ&1\sin⁡θ&-sin⁡θ&1end{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&1end{vmatrix})=(begin{vmatrix}0&-cos⁡θ&1\1&cos^2⁡θ&1\0&-sin⁡θ&1end{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-bend{vmatrix}).
a) 2abc
b) 2a{(b-c)(c-a+b)}
c) 2b{(a-c)(a+b+c)}
d) 2c{(b-c)(a-c+b)}
Answer: b
Clarification: Δ=(begin{vmatrix}b-c&b&c\a&c-a&c\a&b&a-bend{vmatrix})
Applying C₂→C₂-C₃
Δ=(begin{vmatrix}b-c&b-c&c\a&-a&c\a&-a&a-bend{vmatrix})
Applying C₁→C₁-C₂
Δ=(begin{vmatrix}0&b-c&c\2a&-a&c\2a&-a&a-bend{vmatrix})
Applying R₂→R₂-R₃
Δ=(begin{vmatrix}0&b-c&c\0&0&c-a+b\2a&-a&a-bend{vmatrix})
Expanding along C₁, we get
Δ=2a{(b-c)(c-a+b)}

6. If A=(begin{bmatrix}1&3\2&1end{bmatrix}), then ________
a) |2A|=4|A|
b) |2A|=2|A|
c) |A|=2|A|
d) |A|=|4A|
Answer: a
Clarification: Given that, A=(begin{bmatrix}1&3\2&1end{bmatrix})
2A=2(begin{bmatrix}1&3\2&1end{bmatrix})=(begin{bmatrix}2&6\4&2end{bmatrix})
|2A|=(begin{vmatrix}2&6\4&2end{vmatrix})=(4-24)=-20
4|A|=4(begin{vmatrix}1&3\2&1end{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&zend{vmatrix}).
a) 0
b) abc
c) 2abc
d) -1
Answer: a
Clarification: Δ=(begin{vmatrix}-a&b&c\-2a+4x&2b-4y&2c+4z\x&-y&zend{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&zend{vmatrix})+(begin{vmatrix}-a&b&c\4x&-4y&4z\x&-y&zend{vmatrix})
Δ=2(begin{vmatrix}-a&b&c\-a&b&c\x&-y&zend{vmatrix})+4(begin{vmatrix}-a&b&c\x&-y&z\x&-y&zend{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^2end{bmatrix})
a) abc(a³+b³+c³+abc)
b) abc(a³+b³+c³-abc)
c) abc(a³+b³+c³+abc)
d) (a³-b³+c³-abc)
Answer: b
Clarification: Given that, A=(begin{bmatrix}c^2&cb&ca\ab&a^2&-ac\ab&bc&-b^2end{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&-bend{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+qend{vmatrix}).
a) -1(1+m+n+q)
b) 1+m+n+q
c) 1+2q
d) 1+q
Answer: a
Clarification: Given that, Δ=(begin{vmatrix}1+m&n&q\m&1+n&q\n&m&1+qend{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+qend{vmatrix})=(1+m+n+q)(begin{vmatrix}1&n&q\1&1+n&q\1&m&1+qend{vmatrix})
Applying R₁→R₂-R₁
Δ=(1+m+n+q)(begin{vmatrix}0&1&0\1&1+n&q\1&m&1+qend{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&21end{vmatrix}).
a) 168
b) -1
c) -168
d) 0
Answer: d
Clarification: Δ=(begin{vmatrix}4&8&12\6&12&18\7&14&21end{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&3end{vmatrix})
Since the elements of all rows are identical, the determinant is zero.