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 = [aij] is a square matrix of order 3 and Aij denote cofactor of the element aij 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.