# 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

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)

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

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

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)}

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)}

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|

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

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)

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

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

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.

Sanfoundry Global Education & Learning Series – Mathematics – Class 12.

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