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
(i) [adj A]^-1 = adj(A^-1), (ii) (A^-1)^-1 = A.
Solution:


From (2) and (4), we have :
(adj A)^-1 = adj (A^-1),

(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.