# GSEB Solutions Class 12 Maths Chapter 4 Determinants Miscellaneous Exercise

Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 4 Determinants Miscellaneous Exercise Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 12 Maths Chapter 4 Determinants Miscellaneous Exercise

Question 1.
Prove that the determinant $$\left|\begin{array}{ccc} x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x \end{array}\right|$$ independent of θ.
Solution:
Let ∆ = $$\left|\begin{array}{ccc} x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x \end{array}\right|$$
Expanding with the help of elements of I row, we get
∆ = x$$\left|\begin{array}{cc} -x & 1 \\ 1 & x \end{array}\right|$$ – sin θ $$\left|\begin{array}{cc} -\sin \theta & 1 \\ \cos \theta & x \end{array}\right|$$ + cos θ$$\left|\begin{array}{cc} -\sin \theta & -x \\ \cos \theta & 1 \end{array}\right|$$
= x (- x² – 1) – sin θ (- x sin θ – cos θ) + cos θ (- sin θ + x cos θ)
= – x³ – x + x sin²θ + sin θ cos θ – sin θ cos θ + x cos² θ
= – x³ – x + x (sin² θ + cos² θ) = – x³ – x + x = – x³,
which is independent of θ.

Question 2.
Without expanding the determinant, prove that
$$\left|\begin{array}{lll} a & a^{2} & b c \\ b & b^{2} & c a \\ c & c^{2} & a b \end{array}\right|$$ = $$\left|\begin{array}{lll} 1 & a^{2} & a^{3} \\ 1 & b^{2} & b^{3} \\ 1 & c^{2} & c^{3} \end{array}\right|$$
Solution:
Let ∆ = $$\left|\begin{array}{lll} a & a^{2} & b c \\ b & b^{2} & c a \\ c & c^{2} & a b \end{array}\right|$$.
Multiply R1, R2 and R3 by a, b and c respectively and then dividing the determined by abc, we get

Question 3.
Evaluate
$$\left|\begin{array}{ccc} \cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\ -\sin \beta & \cos \beta & 0 \\ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha \end{array}\right|$$
Solution:
∆ = $$\left|\begin{array}{ccc} \cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\ -\sin \beta & \cos \beta & 0 \\ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha \end{array}\right|$$
Taking out cos a common from I row and sin α common from III row, we get

Question 4.
If a, b and c are reals and
∆ = $$\left|\begin{array}{lll} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{array}\right|$$ = 0,
show that either a + b + c = 0 or a = b = c
Solution:
∆ = $$\left|\begin{array}{lll} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{array}\right|$$.
Operating C1 → C1 + C2 + C3, we get
∆ = $$\left|\begin{array}{ccc} 2(a+b+c) & c+a & a+b \\ 2(a+b+c) & a+b & b+c \\ 2(a+b+c) & b+c & c+a \end{array}\right|$$ = 2$$\left|\begin{array}{ccc} a+b+c & c+a & a+b \\ a+b+c & a+b & b+c \\ a+b+c & b+c & c+a \end{array}\right|$$
Operating C2 → C2 – C1, C3 → C3 – C1, we get
∆ = 2 $$\left|\begin{array}{lll} a+b+c & -b & -c \\ a+b+c & -c & -a \\ a+b+c & -a & -b \end{array}\right|$$
Taking out (a + b + c) common from C1(- 1) from C2 and C3 we get
∆ = 2 (a + b + c) $$\left|\begin{array}{lll} 1 & b & c \\ 1 & c & a \\ 1 & a & b \end{array}\right|$$.
Expanding with the help of elements of I column, we get
∆ = 2 (a + b + c) [(c b- a²) – (b² – ac) + (ab – c²)]
= 2 (a + b + c)[-(a²- bc) – (b² – ca) – (c² – ab)]
= – (a+ b + c) [2a² + 2b² + 2c² – 2 bc – 2 ca – 2 ab]
= – (a + b + c) [(a – b)² + (b – c)² + (c – a)²]
Now ∆ = 0, when either (a + b + c) = 0
or (a – b)² + (b – c)² + (c – a)² = 0
⇒ a = b = c.
Hence ∆ = 0, when either a + b + c = 0 or a = b = c.

Question 5.
Solve the equation $$\left|\begin{array}{ccc} x+a & x & x \\ x & x+a & x \\ x & x & x+a \end{array}\right|$$ = 0, a ≠ 0.
Solution:
∆ = $$\left|\begin{array}{ccc} x+a & x & x \\ x & x+a & x \\ x & x & x+a \end{array}\right|$$
Operating C1 → C1 + C2 + C3, we get
∆ = $$\left|\begin{array}{ccc} 3 x+a & x & x \\ 3 x+a & x+a & x \\ 3 x+a & x & x+a \end{array}\right|$$
= (3x + a) $$\left|\begin{array}{ccc} 1 & x & x \\ 1 & x+a & x \\ 1 & x & x+a \end{array}\right|$$
Operating R1 → R1 – R2, we get
∆ = (3x + a)$$\left|\begin{array}{ccc} 0 & -a & 0 \\ 1 & x+a & x \\ 1 & x & x+a \end{array}\right|$$.
Expanding with the help of elements of first row, we get
∆ = (3x + a) .a$$\left|\begin{array}{cc} 1 & x \\ 1 & x+a \end{array}\right|$$ = a (3x + a) (x + a – x)
= a³(3x + a).
But ∆ = 0 ∴ a²(3x + a).
Also, a ≠ 0 ∴ 3x + a = 0
or x = – $$\frac { a }{ 3 }$$.

Question 6.
Prove that $$\left|\begin{array}{ccc} a^{2} & b c & a c+c^{2} \\ a^{2}+a b & b^{2} & a c \\ a b & b^{2}+b c & c^{2} \end{array}\right|$$ = 4a² b² c².
Solution:
L.H.S = ∆ = $$\left|\begin{array}{ccc} a^{2} & b c & a c+c^{2} \\ a^{2}+a b & b^{2} & a c \\ a b & b^{2}+b c & c^{2} \end{array}\right|$$.
Taking out a common from C1, b from C2 and c from C3, we get

Taking out – a and – b common from C2 and C3, we get
∆ = 2a²b²c$$\left|\begin{array}{lll} a+c & 1 & 0 \\ a+b & 1 & 1 \\ b+c & 0 & 1 \end{array}\right|$$
Expanding with the help of elements of C3, we get
∆ = 2a² b²c [b + c + (a + c) – (a + b)]
= 2a²b²c.2c = 4a² b² c² = R.H.S.

Question 7.
If A-1 = $$\left[\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{array}\right]$$ and B = $$\left[\begin{array}{ccc} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{array}\right]$$, find (AB)-1.
Solution:

Question 8.
Let A = $$\left[\begin{array}{ccc} 1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5 \end{array}\right]$$. Verify that
Solution:

From (2) and (4), we have :

(ii) From part (i) A-1 = C (Say)

Hence, (A-1)-1 = A.

Question 9.
Evaluate $$\left|\begin{array}{ccc} x & y & x+y \\ y & x+y & x \\ x+y & x & y \end{array}\right|$$
Solution:
Let ∆ = $$\left|\begin{array}{ccc} x & y & x+y \\ y & x+y & x \\ x+y & x & y \end{array}\right|$$
Applying C1 → C1 + C2 + C3, we get
∆ = $$\left|\begin{array}{ccc} 2(x+y) & y & x+y \\ 2(x+y) & x+y & x \\ 2(x+y) & x & y \end{array}\right|$$
Taking out 2(x + y) common from C1, we get
∆ = 2 (x + y)
$$\left|\begin{array}{ccc} 1 & y & x+y \\ 1 & x+y & x \\ 1 & x & y \end{array}\right|$$
Applying R1 → R2 – R1 and R3 → R3 – R1, we get
∆ = 2 (x + y)$$\left|\begin{array}{ccc} 1 & y & x+y \\ 0 & x & -y \\ 0 & x-y & -x \end{array}\right|$$
Expanding by Cj, we get
∆ = 2 (x + y) [x (- x) – (- y)(x – y)]
= 2(x + y) [- x² + xy – y²] = – 2(x³ + y³).

Question 10.
Evaluate $$\left|\begin{array}{ccc} 1 & x & y \\ 1 & x+y & y \\ 1 & x & x+y \end{array}\right|$$
Solution:
Let ∆ = $$\left|\begin{array}{ccc} 1 & x & y \\ 1 & x+y & y \\ 1 & x & x+y \end{array}\right|$$
Applying R2 → R2 – R1 and R3 → R3 – R1, we get
∆ = $$\left|\begin{array}{lll} 1 & x & y \\ 0 & y & 0 \\ 0 & 0 & x \end{array}\right|$$ = 1 x y x x = xy.

Question 11.
$$\left|\begin{array}{lll} \alpha & \alpha^{2} & \beta+\gamma \\ \beta & \beta^{2} & \gamma+\alpha \\ \gamma & \gamma^{2} & \alpha+\beta \end{array}\right|$$ = (α – γ) ( γ – α) (α – ß) (α + ß + γ)
Solution:

Question 12.
$$\left|\begin{array}{ccc} x & x^{2} & 1+p x^{3} \\ y & y^{2} & 1+p y^{3} \\ z & z^{2} & 1+p z^{3} \end{array}\right|$$ = (1 + pxyz)(x – y)(y – z)(z – x), where p is any scalar.
Solution:

Applying R1 → R1 – R2 and R2 → R2 – R3, we get
∆ = (1 + pxyz)$$\left|\begin{array}{ccc} 0 & x-y & x^{2}-y^{2} \\ 0 & y-z & y^{2}-z^{2} \\ 1 & z & z^{2} \end{array}\right|$$
Taking out common x-y from R x, y – z from R2 we get
∆ = (1 + pxyz) (x – y)(y – z)$$\left|\begin{array}{ccc} 0 & 1 & x+y \\ 0 & 1 & y+z \\ 1 & 2 & z^{2} \end{array}\right|$$
= (1 + pxyz) (y + z – x – y)
= (1 + pxyz) (x – y)(y – z)(z – x) = R.H.S.

Question 13.
$$\left|\begin{array}{ccc} 3 a & -a+b & -a+c \\ -b+a & 3 b & -b+c \\ -c+a & -c+b & 3 c \end{array}\right|$$ = 3 (a + b + c) (ab + bc + ca)
Solution:

Expanding by C1, we get
∆ = (a + b + c) [(2b + a) (2c + a) – (a – c) (a – b)]
= (a + b + c)(4bc + 2ab + 2ac + a² – a² + ab + ac – bc)
= (a + b + c) (3ab + 3be + 3ca)
= 3 (a + b + c) (ab + ba + ca) = R.H.S.

Question 14.
$$\left|\begin{array}{ccc} 1 & 1+p & 1+p+q \\ 2 & 3+2 p & 4+3 p+2 q \\ 3 & 6+3 p & 10+6 p+3 q \end{array}\right|$$ = 1
Solution:
Let ∆ = $$\left|\begin{array}{ccc} 1 & 1+p & 1+p+q \\ 2 & 3+2 p & 4+3 p+2 q \\ 3 & 6+3 p & 10+6 p+3 q \end{array}\right|$$
Applying R2 → R2 – 2R1 and R3 → R3 – 3R1, we get
∆ = $$\left|\begin{array}{ccc} 1 & 1+p & 1+p+q \\ 0 & 1 & 2+p \\ 0 & 3 & 7+3 p \end{array}\right|$$
= 1 (7 + 3p – 6 – 3p) = 1 (1) = 1. (Expanding by C1)
= R.H.S.

Question 15.
$$\left|\begin{array}{lll} \sin \alpha & \cos \alpha & \cos (\alpha+\delta) \\ \sin \beta & \cos \beta & \cos (\beta+\delta) \\ \sin \gamma & \cos \gamma & \cos (\gamma+\delta) \end{array}\right|$$ = 0
Solution:

Question 16.
Solve the system of equations
$$\frac { 2 }{ x }$$ + $$\frac { 3 }{ y }$$ + $$\frac { 10 }{ z }$$ = 4
$$\frac { 4 }{ x }$$ – $$\frac { 6 }{ y }$$ + $$\frac { 5 }{ z }$$ = 1
$$\frac { 6 }{ x }$$ + $$\frac { 9 }{ y }$$ – $$\frac { 20 }{ z }$$ = 2
Solution:
Let $$\frac { 1 }{ x }$$ = u, $$\frac { 1 }{ y }$$ = v, $$\frac { 1 }{ z }$$ = w
∴ System of equations is
2u + 3v + 10w = 4
4u – 6v + 5w = 1
6u + 9v – 20w = 2,
which may be written as AX = B,
where A = $$\left[\begin{array}{ccc} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{array}\right]$$, X = $$\left[\begin{array}{l} u \\ v \\ w \end{array}\right]$$ + $$\left[\begin{array}{l} 4 \\ 1 \\ 2 \end{array}\right]$$
To find adj A, we find the cofactors of each of the elements:

Hence, x = 2, y = 3, z = 5

Question 17.
If a, b, c are in A. P., then the determinant $$\left|\begin{array}{ccc} x+2 & x+3 & x+2 a \\ x+3 & x+4 & x+2 b \\ x+4 & x+5 & x+2 c \end{array}\right|$$ is equal to
(A) 0
(B) 1
(C) x
(D) 2a
Solution:

Question 18.
If x, y, z are non-zero real numbers, then the inverse of the matrix A = $$\left(\begin{array}{lll} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{array}\right)$$ is
(A) $$\left(\begin{array}{ccc} x^{-1} & 0 & 0 \\ 0 & y^{-1} & 0 \\ 0 & 0 & z^{-1} \end{array}\right)$$
(B) xyz$$\left(\begin{array}{ccc} x^{-1} & 0 & 0 \\ 0 & y^{-1} & 0 \\ 0 & 0 & z^{-1} \end{array}\right)$$
(C) $$\frac { 1 }{ xyz }$$ $$\left(\begin{array}{lll} x & 0 & 0 \\ 0 & y & 0 \\ 1 & 0 & z \end{array}\right)$$
(D) $$\frac { 1 }{ xyz }$$$$\left(\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{array}\right)$$
Solution:
$$\left(\begin{array}{lll} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{array}\right)$$
The cofactors of the elements are

∴ Part (A) is the required answer.

Question 19.
Let A = $$\left[\begin{array}{ccc} 1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1 \end{array}\right]$$, where 0 ≤ θ ≤ 2π, then
(A) Det (A) = 0
(B) Det A ∈ (2, ∞)
(C) Det (A) ∈ (2, 4)
(D) Det (A) ∈ [2,4]
Solution:
A = $$\left[\begin{array}{ccc} 1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1 \end{array}\right]$$
So, Det A = $$\left|\begin{array}{ccc} 1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1 \end{array}\right|$$,
= 1 . (1 + sin² θ) – sin θ (- sin θ + sin θ) + 1 . (sin² θ + 1)
= 2 (1 + sin² θ)
For θ = 0, π, 2π, Det A = 2.
For θ = $$\frac { π }{ 2 }$$, $$\frac { 3π }{ 2 }$$, Det A = 2. (1 + 1) = 4.
⇒ Part (D) is the correct answer.