ML Aggarwal Class 12 Maths Solutions Section A Chapter 4 Determinants MCQs

The availability of Understanding ISC Mathematics Class 12 Solutions Chapter 4 Determinants MCQs encourages students to tackle difficult exercises.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 4 Determinants MCQs

Choose the correct answer from the given four options in questions (1 to 37):

Question 1.
If A = [a_ij] is a square matrix of order 3 and A_ij denote cofactor of the element a_ij in[A], then the value of |A| is given by
(a) a₁₁A₁₁ + a₁₂A₁₂ + a₁₃A₁₃
(b) a₁₁A₁₁ + a₁₂A₂₁ + a₁₃A₃₁
(c) a₁₁A₂₁ + a₁₂A₂₂ + a₁₃A₂₃
(d) a₁₁A₁₃ + a₂₁A₂₃ + a₃₁A₃₃
Solution:
(a) a₁₁A₁₁ + a₁₂A₁₂ + a₁₃A₁₃

Given A = \(\left|\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right|\)
= \(a_{11}\left|\begin{array}{ll}
a_{22} & a_{23} \\
a_{32} & a_{33}
\end{array}\right|-a_{12}\left|\begin{array}{ll}
a_{21} & a_{23} \\
a_{31} & a_{33}
\end{array}\right|+a_{13}\left|\begin{array}{ll}
a_{21} & a_{22} \\
a_{31} & a_{32}
\end{array}\right|\)
= a₁₁A₁₁ + a₁₂A₁₂ + a₁₃A₁₃

Question 2.
The value of a determinant is unaltered ¡f
(a) two rows are interchanged
(b) two columns are interchanged
(c) every element in a row (or column) is multiplied by the same number
(d) to each element of a row (or a column) is added equi-multiples of the corresponding elements of another row or column
Solution:
(d) to each element of a row (or a column) is added equi-multiples of the corresponding elements of another row or column

For option (a) ; value of determinant is multiplied by – 1.
For option (b) ; value of det is multiplied by – 1.
For option (c) ; value of det is divided by same number.

Question 3.
If A is square matrix of order 3, then which of the following is not true?
(a) | A’ | = | A |
(b) | kA | = k3 | A |
(c) minor of an element of | A | can never be equal cofactor of the same element
(d) order of minors and of cofactors of elements of | A | is same
Solution:
(c) minor of an element of | A | can never be equal cofactor of the same element

[Cofactors of a₁₁ = Minor of a₁₁]

Question 4.
The value of \(\left|\begin{array}{ccc}
a+p d & a+q d & a+r d \\
p & q & r \\
d & d & d
\end{array}\right|\) is
(a) – 1
(b) 0
(c) 1
(d) p + q + r
Solution:
(b) 0

Let ∆ = \(\left|\begin{array}{ccc}
a+p d & a+q d & a+r d \\
p & q & r \\
d & d & d
\end{array}\right|\) ;
operate R₁ → R₁ – dR₂
= \(\left|\begin{array}{lll}
a & a & a \\
p & q & r \\
d & d & d
\end{array}\right|\) ;
Taking a common from R₁ and d common from R₃
= ad \(\left|\begin{array}{lll}
1 & 1 & 1 \\
p & q & r \\
1 & 1 & 1
\end{array}\right|\)
= ad × 0
= 0 [∵ R₁ and R₃ are identical]

Question 5.
The value of \(\left|\begin{array}{lll}
2^2 & 2^3 & 2^4 \\
2^3 & 2^4 & 2^5 \\
2^4 & 2^5 & 2^6
\end{array}\right|\) is
(a) 2⁹
(b) 2⁶
(c) 2¹³
(d) 0
Solution:
(d) 0

Let Δ = \(\left|\begin{array}{lll}
2^2 & 2^3 & 2^4 \\
2^3 & 2^4 & 2^5 \\
2^4 & 2^5 & 2^6
\end{array}\right|\) ;
Taking 2², 2³ and 2⁴ common from R₁, R₂ and R₃
= 2⁹ \(\left|\begin{array}{lll}
1 & 2 & 2^2 \\
1 & 2 & 2^2 \\
1 & 2 & 2^2
\end{array}\right|\)
= 2⁹ × 0
= 0 [∵ R₁, R₂ and R₃ are all identical]

Question 6.
If \(\left|\begin{array}{cc}
3 x & 4 \\
5 & x
\end{array}\right|=\left|\begin{array}{cc}
4 & -3 \\
5 & -2
\end{array}\right|\), then x =
(a) 3 only
(b) – 3 only
(c) 3 or – 3
(d) 6 or – 6
Solution:
(c) 3 or – 3

Given \(\left|\begin{array}{cc}
3 x & 4 \\
5 & x
\end{array}\right|=\left|\begin{array}{cc}
4 & -3 \\
5 & -2
\end{array}\right|\)
⇒ 3x² – 20 = – 8 + 15
⇒ 3x² – 20 = 7
⇒ 3x² = 27
⇒ x² = 9
⇒ x = ± 3

Question 7.
The value of determinant \(\left|\begin{array}{lll}
a-b & b+c & a \\
b-c & c+a & b \\
c-a & a+b & c
\end{array}\right|\) is
(a) a³ + b³ + c³
(b) a³ + b³ + c³ – 3abc
(c) abc
(d) 3abc
Solution:
(b) a³ + b³ + c³ – 3abc

Let ∆ = \(\left|\begin{array}{lll}
a-b & b+c & a \\
b-c & c+a & b \\
c-a & a+b & c
\end{array}\right|\) ;
operate R₁ → R₁ + R₂ + R₃
= \(\left|\begin{array}{ccc}
0 & 2(a+b+c) & a+b+c \\
b-c & c+a & b \\
c-a & a+b & c
\end{array}\right|\) ;
Taking (a b + c) common from R₁
= (a + b + c) \(\left|\begin{array}{ccc}
0 & 2 & 1 \\
b-c & c+a & b \\
c-a & a+b & c
\end{array}\right|\) ;
operate C₂ → C₂ – 2C₃
= (a + b + c) \(\left|\begin{array}{ccc}
0 & 0 & 1 \\
b-c & c+a-2 b & b \\
c-a & a+b-2 c & c
\end{array}\right|\) ;
Expanding along R₁
= (a + b + c) [(b – c) (a + b – 2c) – (c – a) (c + a – 2b)]
= (a + b + c) [ab + b² – 2bc – ac – bc + 2c – c² + a² + 2bc – 2ab]
= (a + b + c) (a² + b² + c² – ab – bc – ca)
= a³ + b³ + c³ – 3abc.

Question 8.
The value of \(\left|\begin{array}{ccc}
a & a+2 b & a+4 b \\
a+2 b & a+4 b & a+6 b \\
a+4 b & a+6 b & a+8 b
\end{array}\right|\) is
(a) 3a + 6b
(b) 2b
(c) 0
(d) none of these
Solution:
(c) 0

Let Δ = \(\left|\begin{array}{ccc}
a & a+2 b & a+4 b \\
a+2 b & a+4 b & a+6 b \\
a+4 b & a+6 b & a+8 b
\end{array}\right|\) ;
operate R₂ → R₂ – R₁ ;
R₃ → R₃ – R₁
= \(\left|\begin{array}{ccc}
a & a+2 b & a+4 b \\
2 b & 2 b & 2 b \\
4 b & 4 b & 4 b
\end{array}\right|\) ;
Taking 2b and 4b common from R₂ and R₃
= 8b² \(\left|\begin{array}{ccc}
a & a+2 b & a+4 b \\
1 & 1 & 1 \\
1 & 1 & 1
\end{array}\right|\)
= 8b² × 0
= 0 [∵ R₂ and R₃ are identical]

Question 9.
If abc ≠ 0 and \(\left|\begin{array}{ccc}
1+a & 1 & 1 \\
1 & 1+b & 1 \\
1 & 1 & 1+c
\end{array}\right|\) = 0, then the value of a^-1 + b^-1 + c^-1 is
Solution:
Given \(\left|\begin{array}{ccc}
1+a & 1 & 1 \\
1 & 1+b & 1 \\
1 & 1 & 1+c
\end{array}\right|\) = 0
Taking a common from R₁, b common from R₂ and c common from R₃

Question 10.
The minimum value of \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & 1+\sin x & 1 \\
1 & 1 & 1+\cos x
\end{array}\right|\) is
(a) – 2
(b) – 1
(c) – \(\frac{1}{2}\)
(d) 0
Solution:
(c) – \(\frac{1}{2}\)

Δ = \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & 1+\sin x & 1 \\
1 & 1 & 1+\cos x
\end{array}\right|\) ;
operate R₂ → R₂ – R₁ ;
R₃ → R₃ – R₁
= \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
0 & \sin x & 0 \\
0 & 0 & \cos x
\end{array}\right|\) ;
expanding along C₁
= sin x cos x
= \(\frac{1}{2}\) sin 2x
since – 1 ≤ sin 2x ≤ 1
⇒ \(-\frac{1}{2} \leq \frac{1}{2} \sin 2 x \leq \frac{1}{2}\)
⇒ \(-\frac{1}{2} \leq \Delta \leq \frac{1}{2}\)
∴ Min. value of Δ = \(\frac{1}{2}\)

Question 11.
The value of \(\left|\begin{array}{ccc}
1+a & b & c \\
a & 1+b & c \\
a & b & 1+c
\end{array}\right|\) is
(a) abc
(b) a + b + c
(c) 1 + a + b + c
(d) 3 + abc
Solution:
(c) 1 + a + b + c

Let Δ = \(\left|\begin{array}{ccc}
1+a & b & c \\
a & 1+b & c \\
a & b & 1+c
\end{array}\right|\)
operate C₁ → C₁ + C₂ + C₃
= \(\left|\begin{array}{ccc}
1+a+b+c & b & c \\
1+a+b+c & 1+b & c \\
1+a+b+c & b & 1+c
\end{array}\right|\) ;
Taking (1 + a + b + c) common from C₁
= (1 + a + b + c) \(\left|\begin{array}{ccc}
1 & b & c \\
1 & 1+b & c \\
1 & b & 1+c
\end{array}\right|\) ;
operate R₂ → R₂ – R₁ ;
R₃ → R₃ – R₁
= (1 + a + b + c) \(\left|\begin{array}{lll}
1 & b & c \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right|\) ;
expanding along C₁
= (1 + a + b + c) \(\left|\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right|\)
= (1 + a + b + c)

Question 12.
If \(\left[\begin{array}{rrr}
1 & 3 & 9 \\
1 & x & x^2 \\
4 & 6 & 9
\end{array}\right]\) is singular matrix, then x =
(a) 3
(b) 3 or 6
(c) 3 or \(\frac{3}{2}\)
(d) – 3, \(\frac{3}{2}\)
Solution:
(c) 3 or \(\frac{3}{2}\)

Let A = \(\left[\begin{array}{rrr}
1 & 3 & 9 \\
1 & x & x^2 \\
4 & 6 & 9
\end{array}\right]\)
since A is singular matrix
∴ |A| = 0
⇒ \(\left|\begin{array}{rrr}
1 & 3 & 9 \\
1 & x & x^2 \\
4 & 6 & 9
\end{array}\right|\) = 0 ;
Expanding along R₁
⇒ 1 (9x – 6x²) – 3 (9 – 4x²) + 9 (6 – 4x) = 0
⇒ 6x² – 27x + 27 = 0
⇒ 2x² – 9x + 9 =0
⇒ (x – 3) (2x – 3) = 0
⇒ x = 3 or x = \(\frac{3}{2}\).

Question 13.
If x + y + z = π, then the value of \(\left|\begin{array}{ccc}
\sin (x+y+z) & \sin (x+z) & \cos y \\
-\sin y & 0 & \tan x \\
\cos (x+z) & \tan (y+z) & 0
\end{array}\right|\) is
(a) 0
(b) 1
(c) – 1
(d) none of these
Solution:
(a) 0

Let Δ = \(\left|\begin{array}{ccc}
\sin (x+y+z) & \sin (x+z) & \cos y \\
-\sin y & 0 & \tan x \\
\cos (x+z) & \tan (y+z) & 0
\end{array}\right|\)
since x + y + z = π
= \(\left|\begin{array}{ccc}
0 & \sin (\pi-y) & \cos y \\
-\sin y & 0 & \tan x \\
\cos (\pi-x) & \tan (\pi-x) & 0
\end{array}\right|\)
= \(\left|\begin{array}{ccc}
0 & \sin y & \cos y \\
-\sin y & 0 & \tan x \\
-\cos y & -\tan x & 0
\end{array}\right|\)
= 0
[∵ Δ is skew symmetric matrix
∵ |Δ| = 0]

Question 14.
The value of \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
b+c & c+a & a+b \\
b+c-a & c+a-b & a+b-c
\end{array}\right|\) is
(a) a + b + c
(b) 2 (a + b + c)
(c) 1
(d) 0
Solution:
(d) 0

Let Δ = \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
b+c & c+a & a+b \\
b+c-a & c+a-b & a+b-c
\end{array}\right|\) ;
operate R₃ → R₃ – R₂ ;
= \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
b+c & c+a & a+b \\
-a & -b & -c
\end{array}\right|\) ;
operate R₂ → R₂ – R₃ ;
= \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
b+c+a & c+a+b & a+b+c \\
-a & -b & -c
\end{array}\right|\) ;
Taking (a + b + c) common from R₂
= (a + b + c) \(\left|\begin{array}{rrr}
1 & 1 & 1 \\
1 & 1 & 1 \\
-a & -b & -c
\end{array}\right|\)
= (a + b + c) × 0
= 0 [∵ R₁ and R₂ are identical rows]

Question 15.
If Δ₁ = \(\left|\begin{array}{lll}
\mathrm{A} x & x^2 & 1 \\
\mathrm{~B} y & y^2 & 1 \\
\mathrm{C} z & z^2 & 1
\end{array}\right|\) and Δ₂ = \(\left|\begin{array}{ccc}
\mathbf{A} & \mathbf{B} & \mathbf{C} \\
\boldsymbol{x} & \boldsymbol{y} & z \\
y z & z x & x y
\end{array}\right|\), then
(a) Δ₁ + Δ₂ = 0
(b) Δ₁ – Δ₂ = 0
(c) Δ₁ ≠ Δ₂
(d) none of these
Solution:
(b) Δ₁ – Δ₂ = 0

Δ₂ = \(\left|\begin{array}{ccc}
\mathbf{A} & \mathbf{B} & \mathbf{C} \\
\boldsymbol{x} & \boldsymbol{y} & z \\
y z & z x & x y
\end{array}\right|\) ;
Multiply C₁ by x ;
C₂ by y
and C₃ by z
= \(\frac{1}{x y z}\left|\begin{array}{ccc}
\mathrm{A} x & \mathrm{~B} y & \mathrm{C} z \\
x^2 & y^2 & z^2 \\
x y z & x y z & x y z
\end{array}\right|\) ;
Taking xyz common from R₃
= \(\frac{x y z}{x y z}\left|\begin{array}{ccc}
\mathrm{Ax} & \mathrm{B} y & \mathrm{C} z \\
x^2 & y^2 & z^2 \\
1 & 1 & 1
\end{array}\right|\)
= \(\left|\begin{array}{lll}
\mathrm{A} x & x^2 & 1 \\
\mathrm{~B} y & y^2 & 1 \\
\mathrm{C} z & z^2 & 1
\end{array}\right|\)
= Δ₁
⇒ Δ₁ – Δ₂ = 0

Question 16.
If \(\left|\begin{array}{ccc}
a & b & a \alpha-b \\
b & c & b \alpha-c \\
2 & 1 & 0
\end{array}\right|\) = 0, then
(a) a, b, c are in A.P.
(b) a, b, c are in G.P.
(c) α = 2
(d) none of these
Solution:
(b) a, b, c are in G.P.

Given \(\left|\begin{array}{ccc}
a & b & a \alpha-b \\
b & c & b \alpha-c \\
2 & 1 & 0
\end{array}\right|\) = 0
operate C₃ → C₃ – α C₁ + C₂
\(\left|\begin{array}{ccc}
a & b & 0 \\
b & c & 0 \\
2 & 1 & -2 \alpha+1
\end{array}\right|\) = 0;
expanding along C₃
(1 – 2α) (ac – b²) = 0
α = \(\frac{1}{2}\) or b² = ac
i.e. a, b, c are in G.P.

Question 17.
If a, b, c are distinct real numbers and \(\left|\begin{array}{lll}
a & a^2 & a^3-1 \\
b & b^2 & b^3-1 \\
c & c^2 & c^3-1
\end{array}\right|\) = 0, then
(a) a + b + c = 0
(b) abc = 1
(c) a + b + c = 1
(d) ab + bc + ca = 0
Solution:
(b) abc = 1

Given \(\left|\begin{array}{lll}
a & a^2 & a^3-1 \\
b & b^2 & b^3-1 \\
c & c^2 & c^3-1
\end{array}\right|\) = 0
⇒ \(\left|\begin{array}{lll}
a & a^2 & a^3 \\
b & b^2 & b^3 \\
c & c^2 & c^3
\end{array}\right|-\left|\begin{array}{ccc}
a & a^2 & 1 \\
b & b^2 & 1 \\
c & c^2 & 1
\end{array}\right|\) = 0
Taking a common from R₁, b from R₂ and c from R₃

expanding along C₁
(abc – 1) (b – a) (c – a) (c + a – b – a) = 0 …………..(1)
⇒ (abc – 1) (b – a) (c – a) (c – b) = 0
since a, b, c are distinct real numbers
∴ a ≠ b ≠ c
⇒ a – b ≠ 0 ;
b – c ≠ 0 and
c – a ≠ 0
⇒ (a – b) (b – c) (c – a) ≠ 0
∴ from (1) ;
abc – 1 = 0
⇒ abc = 1.

Question 18.
If a, b, c are distinct real numbers and \(\left|\begin{array}{lll}
a & a^2 & a^4-1 \\
b & b^2 & b^4-1 \\
c & c^2 & c^4-1
\end{array}\right|\) = 0, then
(a) abc = 1
(b) abc (a + b + c) = 1
(c) abc (a + b + c) = ab + bc + ca
(d) none of these
Solution:
(b) abc (a + b + c) = 1

Given \(\left|\begin{array}{lll}
a & a^2 & a^4-1 \\
b & b^2 & b^4-1 \\
c & c^2 & c^4-1
\end{array}\right|\) = 0
⇒ \(\left|\begin{array}{lll}
a & a^2 & a^4 \\
b & b^2 & b^4 \\
c & c^2 & c^4
\end{array}\right|-\left|\begin{array}{ccc}
a & a^2 & 1 \\
b & b^2 & 1 \\
c & c^2 & 1
\end{array}\right|\) = 0
Taking a common from R₁, b from R₂ and c from R₃ in first det and pass C₃ over first two columns in 2nd det.

(abc – 1) (b – a) (c – a) (c + a – b – a) = 0
⇒ (abc – 1) (b – a) (c – a) (c – b) = 0
since a, b, c are distinct real numbers = (a – b) (b – c) (c – a)
and Δ₁ = \(\left|\begin{array}{lll}
1 & a & a^3 \\
1 & b & b^3 \\
1 & c & c^3
\end{array}\right|\) ;
operate R₂ → R₂ – R₁ ;
R₃ → R₃ – R₁
= \(\left|\begin{array}{ccc}
1 & a & a^3 \\
0 & b-a & b^3-a^3 \\
0 & c-a & c^3-a^2
\end{array}\right|\) ;
Taking (b – a) common from R₂ and (c – a) from R₃
= (b – a) (c – a) \(\left|\begin{array}{ccc}
1 & a & a^3 \\
0 & 1 & b^2+a b+a^2 \\
0 & 1 & c^2+a c+a^2
\end{array}\right|\) ;
Expanding along C₁
= (b – a) (c – a) [c² + ac + a² – b² – ab – a²]
= (b – a) (c – a) [c² – b² + a (c – b)]
= (a – b) (b – c) (c – a) (a + b + c)
∴ from (1) ; we have
(a + b + c) abc (a – b) (b – c) (c – a) – (a – b) (b – c) (c – a) = 0
⇒ (a – b) (b – c) (c – a) [abc (a + b + c) – 1] = 0
Since a ≠ b ≠ c
∴ (a – b) (b – c) (c – a) ≠ 0
⇒ abc (a + b + c) = 1.

Question 19.
If α, β, γ are roots of the equation x³ + px + q = 0, then the value of \(\left|\begin{array}{lll}
\alpha & \beta & \gamma \\
\beta & \gamma & \alpha \\
\gamma & \alpha & \beta
\end{array}\right|\) is
(a) p
(b) q
(c) p² – 4q
(d) 0
Solution:
(d) 0

Given α, β, γ are roots of the equation x³ + px + q = 0
∴ α + β + γ = 0
= – \(\frac{\text { coeff. of } x^2}{\text { coeff. of } x^3}\)
∴ ∆ = \(\left|\begin{array}{lll}
\alpha & \beta & \gamma \\
\beta & \gamma & \alpha \\
\gamma & \alpha & \beta
\end{array}\right|\) ;
operate C₁ → C₁ + C₂ + C₃
= \(\left|\begin{array}{ccc}
\alpha+\beta+\gamma & \beta & \gamma \\
\alpha+\beta+\gamma & \gamma & \alpha \\
\alpha+\beta+\gamma & \alpha & \beta
\end{array}\right|\)
= \(\left|\begin{array}{lll}
0 & \beta & \gamma \\
0 & \gamma & \alpha \\
0 & \alpha & \beta
\end{array}\right|\)
= 0 [∵ C₁ is a zero column]

Question 20.
If A is a non-singular matrix, then
(a) |A| ≠ |A’|
(b) |A^-1| ≠ |A|^-1
(c) |AA’| ≠ |A²|
(d) |A| + |A’| ≠ 0
Solution:
(d) |A| + |A’| ≠ 0

|A| = |A’|
∴ option (a) is wrong
|A^-1| = |A|^-1
∴ option (b) is false
|AA’| = |A| |A’| = |A|²
∴ option (c) is false
|A| + |A’| = 2 |A| ≠ 0
[∵ A is non-singular matrix]
|A| + |A’| = 2 |A| ≠ 0

Question 21.
If A is a non-singular matrix of order 3, then which of the following is not true?
(a) |adj A| = |A²|
(b) (A^-1)^-1 = A
(c) If AB = AC, then B ≠ C where B and C are square matrices of order 3
(d) (AB)^-1 = B^-1A^-1, where B is a non-singular matrix of order 3
Solution:
(c) If AB = AC, then B ≠ C where B and C are square matrices of order 3

We know that, if A be a matrix of order n
Then |adj A| = |A|^{n – 1}
given A be a matrix of order 3
∴ |adj A| = |A|^{3 – 1}
= |A|²
∴ option (a) is correct.
option (b) is correct.
[∵ AA^-1 = I = A^-1A
⇒ (A^-1)^-1 = A]
Since (AB) (B^-1A^-1) = A (BB^-1) A^-1
= AIA^-1
= AA^-1 = I
∴ (AB)^-1 = B^-1A^-1
∴ option (d) is correct.
Since A is non-singular
∴ |A| ≠ 0
⇒ A^-1 exists
Given AB = AC
⇒ A^-1 (AB) = A^-1 (AC)
⇒ (A^-1 A)B = (A^-1 A) C
⇒ IB = IC
⇒ B = C

Question 22.
If A, B are two non-singular matrices of same order, then
(a) AB is non-singular
(b) AB is singular
(c) (AB)^-1 = A^-1B^-1
(d) AB is not invertible
Solution:
(a) AB is non-singular

Since A and B are non-singular matrices.
∴ |A| ≠ 0 ; |B| ≠ 0
Here |AB| = |A| |B| ≠ 0
⇒ AB is non-singular.

Question 23.
If for matrix A, A³ = I, then A^-1 =
(a) A
(b) A²
(c) A³
(d) none of these
Solution:
(b) A²

Given A³ = I
⇒ A^-1A³ = A^-1I
⇒ (A^-1 A) A² = A^-1
⇒ IA² = A^-1
⇒ A^-1 = A²

Question 24.
If A, B, C are non-singular matrices of same order, then (AB^-1C)^-1 =
(a) CBA^-1
(b) C^-1B^-1A^-1
(c) C^-1BA^-1
(d) C^-1BA
Solution:
(c) C^-1BA^-1

Since A, B and C are non-singular matrices.
∴ B^-1 , C^-1 and A^-1 exists.
(AB^-1C)^-1 = C^-1 (B^-1)^-1 A^-1
= C^-1 BA^-1

Question 25.
If A = \(\left[\begin{array}{cc}
2 x & 0 \\
x & x
\end{array}\right]\) and A^-1 = \(\left[\begin{array}{rr}
1 & 0 \\
-1 & 2
\end{array}\right]\), then the value of x is
(a) 2
(b) – \(\frac{1}{2}\)
(c) 1
(d) \(\frac{1}{2}\)
Solution:
(b) – \(\frac{1}{2}\)
and (d) \(\frac{1}{2}\)

Given A = \(\left[\begin{array}{cc}
2 x & 0 \\
x & x
\end{array}\right]\)
∴ |A| = 2x²
∴ A^-1 = \(\left[\begin{array}{rr}
1 & 0 \\
-1 & 2
\end{array}\right]\) = 2
Since |A^-1| = |A|^-1
⇒ 2 = (2x²)^-1
⇒ 2 = \(\frac{1}{2 x^2}\)
⇒ x² = \(\frac{1}{4}\)
⇒ x = ± \(\frac{1}{2}\)

Question 26.
If A is a square matrix of order 3 such that A (adj A) = \(\left[\begin{array}{rrr}
-3 & 0 & 0 \\
0 & -3 & 0 \\
0 & 0 & -3
\end{array}\right]\), then |A| is equal to
(a) 9
(b) – 3
(c) – 6
(d) 3
Solution:
(b) – 3

Given A (adj A) = \(\left[\begin{array}{rrr}
-3 & 0 & 0 \\
0 & -3 & 0 \\
0 & 0 & -3
\end{array}\right]\)
= – 3I
We know that A adj A = |A| I
∴ |A| = – 3

Question 27.
If A is square matrix of order 3 such that A (adj A) = \(\left[\begin{array}{rrr}
-2 & 0 & 0 \\
0 & -2 & 0 \\
0 & 0 & -2
\end{array}\right]\), then |adj A| is equal to
(a) – 2
(b) – 4
(c) 4
(d) – 8
Solution:
(c) 4

Given A (adj A) = \(\left[\begin{array}{rrr}
-2 & 0 & 0 \\
0 & -2 & 0 \\
0 & 0 & -2
\end{array}\right]\)
= – 2I
Also A adj A = |A| I
∴ |A| = – 2
We know that, if A be a matrix of order n
Then |adj A| = |A|^n-1
Here A be a matrix of order 3.
∴ |adj A| = |A|^3-1
= |A|²
= (- 2)² = 4.

Question 28.
If A = \(\left[\begin{array}{lll}
a & 0 & 0 \\
0 & a & 0 \\
0 & 0 & a
\end{array}\right]\), a ≠ 0, then |adj A| is equal to
(a) a³
(b) a⁹
(c) a⁶
(d) a²⁷
Solution:
(c) a⁶

Given |A| = \(\left[\begin{array}{lll}
a & 0 & 0 \\
0 & a & 0 \\
0 & 0 & a
\end{array}\right]\) = a³
Since A be a matrix of order 3.
∴ |adj A| = |A|^{3 – 1}
= |A|²
= (a³)²
= a⁶

Question 29.
If A² – A + I = O, then A^-1 is equal to
(a) A + I
(b) A – I
(c) A + 2I
(d) I – A
Solution:
Given A² = A + I = O
⇒ I = A – A²
⇒ A^-1 I = A^-1 A – A^-1 . A²
⇒ A^-1 = I – (A^-1 A) A
⇒ A^-1 = I – IA
[∵ A^-1 A = I]
⇒ A^-1 = I – A.

Question 30.
If A and B are Invertible matrices of same order, then which of the following statement is not true?
(a) |A^-1| = |A|^-1
(b) adj A = |A| A^-1
(c) (A + B)^-1 = B^-1 + A^-1
(d) (AB)^-1 = B^-1 A^-1
Solution:
(c) (A + B)^-1 = B^-1 + A^-1

Given A and B are invertible matrices.
∴ A^-1 and B^-1 both are exists.
|A^-1| = |A|^-1 which is true.
Since A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adjA
⇒ adj A = |A| A^-1
∴ option (b) is true.
By reversal law for inverse.
∴ option (d) is true.

Question 31.
Which of the following statements is correct?
(a) Determinant is a number associated to a matrix.
(b) Determinant is a square matrix.
(c) Determinant is a number associated to a square matrix.
(d) None of the above
Solution:
Ans. (c)
[By def. of determinant]

Question 32.
For what vatue of k inverse does not exist for the matrix \(\left[\begin{array}{ll}
1 & 2 \\
k & 6
\end{array}\right]\) ?
(a) 0
(b) 3
(c) 6
(d) 2
Solution:
(b) 3

Let A = \(\left[\begin{array}{ll}
1 & 2 \\
k & 6
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{ll}
1 & 2 \\
k & 6
\end{array}\right|\)
= 6 – 2k
We know that,
A^-1 does not exists iff |A| = 0
⇒ 6 – 2k = 0
⇒ 2 k = 6
⇒ k = 3.

Question 33.
If \(\left|\begin{array}{lll}
2 & 3 & 2 \\
x & x & x \\
4 & 9 & 1
\end{array}\right|\) + 3 = 0, then the value of x is
(a) 3
(b) 0
(c) – 1
(d) 1
Solution:
(c) – 1

Given \(\left|\begin{array}{lll}
2 & 3 & 2 \\
x & x & x \\
4 & 9 & 1
\end{array}\right|\) + 3 = 0
Expanding along R₁ ;
2 (x – 9x) – 3 (x – 4x) + 2 (9x – 4x) + 3 = 0
⇒ – 16x + 9x + 10x + 3 = 0
⇒ 3x + 3 = 0
⇒ x = – 1.

Question 34.
The matrix \(\left[\begin{array}{rrr}
2 & -1 & 3 \\
\lambda & 0 & 7 \\
-1 & 1 & 4
\end{array}\right]\) is not invertible for
(a) λ = – 1
(b) λ = 1
(c) λ = 0
(d) λ ∈ R – {1}
Solution:
Let A = \(\left[\begin{array}{rrr}
2 & -1 & 3 \\
\lambda & 0 & 7 \\
-1 & 1 & 4
\end{array}\right]\)
Now A is not invertible iff |A| = 0
iff \(\left|\begin{array}{rrr}
2 & -1 & 3 \\
\lambda & 0 & 7 \\
-1 & 1 & 4
\end{array}\right|\) = 0 ;
Expanding along R₁
⇔ 2 (0 – 7) + 1 (4λ + 7) + 3 (λ) = 0
⇔ – 14 + 4 λ + 7 + 3λ = 0
⇔ – 7 + 7λ = 0
⇔ λ = 1.

Question 35.
If A is a square matrix of order 3 and |A| = 2, then the value of |- AA’| is
(a) 4
(b) 2
(c) – 2
(d) – 4
Solution:
(d) – 4

|- AA’| = (- 1)³ |A| |A’|
[∵ |AB| = |A| |B| and if A be a matrix of order n.
Then |kA| = kⁿ |A|]
= – |A| |A|
[∵ |A’| = |A|]
= – |A|²
= – 2²
= 4

Question 36.
If A is a square matrix of order 3 and |A| = 5, then the value of |2A’| is
(a) – 10
(b) 10
(c) – 40
(d) 40
Solution:
(d) 40

Given A be a square matrix of order 3
∴ |kA| = k³ |A|
∴ |2A’| = 2³ |A’|
= 8 |A|
= 8 × 5 = 40

Question 37.
If A ¡s non-singular square matrix of order 3 such that A² = 3A, then the value of | A | ¡s
(a) – 3
(b) 3
(c) 9
(d) 27
Solution:
(d) 27

Given A is a non-singular square matrix of order 3.
∴ |A| ≠ 0 and
A^-1 exists.
Given A² = 3A
⇒ (A^-1 A) A = 3A^-1 A
⇒ IA = 3I
⇒ A = 3I
⇒ |A| = |3I|
= 3³ |I|
= 27 × 1 = 27.