Determinants 3 - Class 12 Maths MCQ

This set of Class 12 Maths Chapter 4 Multiple Choice Questions & Answers (MCQs) focuses on “Determinant – 3”.

1. What will be the value of \(\begin{vmatrix}0 & p-q & a – b\\q – p & 0 & x – y\\b – a & y – x & 0 \end {vmatrix}\)?
a) 0
b) a + b
c) x + y
d) p + q
View Answer

Answer: a
Explanation: The above matrix is a skew symmetric matrix and its order is odd
And we know that for any skew symmetric matrix with odd order has determinant = 0
Therefore, the value of the given determinant = 0.

2. What will be the value of f(x) if \(\begin{vmatrix}x & b & c\\a & y & c\\a & b & z \end {vmatrix}\)?
a) (x – a)(y – b)(z – c)(\(\frac{x}{x-a} + \frac{b}{y – b} – \frac{c}{z-c}\) – 2)
b) (x – a)(y – b)(z – c)(\(\frac{x}{x-a} – \frac{b}{y – b} – \frac{c}{z-c}\) – 2)
c) (x – a)(y – b)(z – c)(\(\frac{x}{x-a} + \frac{b}{y – b} + \frac{c}{z-c}\) – 2)
d) (x – a)(y – b)(z – c)(\(\frac{x}{x-a} + \frac{b}{y – b} + \frac{c}{z-c}\) + 2)
View Answer

Answer: c
Explanation: Given, \(\begin{vmatrix}x & b & c\\a & y & c\\a & b & z \end {vmatrix}\) = \(\begin{vmatrix}x & b & c\\a – x & y – b & 0\\0 & b – y & z – c \end {vmatrix}\)

Applying the operation R₂ = R₂ – R₁ and R₃ = R₃ – R₂
= (x – a)(y – b)(z – c)\(\begin{vmatrix}x/(x – a) & b/(y – b) & c/(z – c)\\-1 & y 1 & 0\\0 & -1 & 1 \end {vmatrix}\)
Now, expanding the determinant we get,
= (x – a)(y – b)(z – c)(\(\frac{x}{x-a} + \frac{b}{y – b} + \frac{c}{z-c}\))
= (x – a)(y – b)(z – c)(\(\frac{x}{x-a} + \frac{b}{y – b} + \frac{c}{z-c}\) – 2)
This is because,
\(\frac{b}{y – b} + \frac{c}{z-c} = \frac{y-(y-b)}{y-b} + \frac{z-(z-c)}{z-c} = \frac{y}{y-b}\) – 1 + \(\frac{z}{z-c}\) – 1 = \(\frac{y}{y-b} + \frac{z}{z-c}\) – 2

3. What will be the value of \(\begin{vmatrix}cos^2⁡ θ & cosθ \, sinθ & -sinθ \\cosθ\, sinθ & sin^2⁡θ & cosθ \\sinθ & -cosθ & 0 \end {vmatrix}\)?
a) -1
b) 0
c) 1
d) 2
View Answer

Answer: c
Explanation: The given matrix is, \(\begin{vmatrix}cos^2 θ & cosθ\, sinθ & -sinθ \\cosθ\, sinθ & sin^2⁡ θ & cosθ \\sinθ & -cosθ & 0 \end {vmatrix}\)
Now, performing the row operations R₁ = R₁ + sinθR₃ and R₂ = R₂ – cosθR₃
=\(\begin{vmatrix}cos^2⁡ θ + sin^2⁡ θ & cosθ\, sinθ – cosθ sinθ & -sinθ \\cosθ\, sinθ – cosθ sinθ & cos^2⁡ θ + sin^2 ⁡θ & cosθ \\sinθ & -cosθ & 0 \end {vmatrix}\)
Solving further,
= \(\begin{vmatrix}1 & 0 & -sinθ \\0 & 1 & cosθ \\sinθ & -cosθ & 0 \end {vmatrix}\)
Breaking the determinant, we get,
= 1(0 + cos²θ) – sinθ(0 – sinθ)
= 1

4. What is the value of \(\begin{vmatrix}sin^2 ⁡a & sina\, cosa & cos^2 ⁡a \\sin^2 ⁡b & sinb\, cosb & cos^2 ⁡b \\sin^2⁡ c & sinc\, cosc & cos^2⁡ c \end {vmatrix}\)?
a) -sin(a – b) sin(b – c) sin(c – a)
b) sin(a – b) sin(b – c) sin(c – a)
c) -sin(a + b) sin(b + c) sin(c + a)
d) sin(a + b) sin(b + c) sin(c + a)
View Answer

Answer: a
Explanation: We have, \(\begin{vmatrix}sin^2 a & sina \,cosa & cos^2 a\\sin^2 b & sinb\, cosb & cos^2 b\\sin^2 c & sinc\, cosc & cos^2 c \end {vmatrix}\)
Now, multiplying by 2 in both numerator and denominator of column 2 and C₁ = C₁ + C₃ we get,
1/2 \(\begin{vmatrix}sin^2 a + cos^2 a & 2sina\, cosa & cos^2 a\\sin^2 b + cos^2 b & 2sinb\, cosb & cos^2 b\\sin^2 c + cos^2 c & 2sinc\, cosc & cos^2 c \end {vmatrix}\)
= 1/2 \(\begin{vmatrix}sin^2 a + cos^2 a & sin2a & cos^2 a\\sin^2 b + cos^2 b & sin2b & cos^2 b\\sin^2 c + cos^2 c & sin2c & cos^2 c \end {vmatrix}\)
= 1/2 \(\begin{vmatrix}1 & sin2a & cos^2 a\\1 & sin2b & cos^2 b\\1 & sin2c & cos^2 c \end {vmatrix}\)
Solving further,
= 1/2 \(\begin{vmatrix}1 & sin2a & cos^2⁡a \\0 & sin2b-sin2a & cos^2⁡ b-cos^2 ⁡a \\0 & sin2c-sin2a & cos^2 ⁡c-cos^2 ⁡a \end {vmatrix}\)
= 1/2 [(sin2b – sin2a)(cos²⁡c – cos²⁡a) – (cos² b – cos²a)(sin2c – sin2a)]
Now, since, [cos² ⁡A + cos² B = sin(A + B) * sin(B – A)]
So, 1/2 [2 cos(a + b) sin(b – a) * sin(c + a)sin(a – c) – sin(a + b)sin(a – b) * 2 cos(a + c)sin(c – a)]
= sin(a – b)sin(c – a)[sin(c + a)cos(a + b) – cos(c + a) sin(a + b)]
= sin(a – b) sin(c – a) sin(c + a – a – b)
= -sin(a – b) sin(b – c) sin(c – a)

5. What will be the value of f(x) if \(\begin{vmatrix}2ab & a^2 & b^2 \\a^2 & b^2 & 2ab \\b^2 & 2ab & a^2 \end {vmatrix}\)?
a) a² + b²
b) -(a² + b²)
c) -(a² + b²)³
d) -(a³ + b³)²
View Answer

Answer: d
Explanation: Given,\(\begin{vmatrix}2ab & a^2 & b^2 \\a^2 & b^2 & 2ab \\b^2 & 2ab & a^2 \end {vmatrix}\)
Using C₁ = C₁ + C₂ + C₃
= \(\begin{vmatrix}a^2 + b^2 + 2ab & a^2 & b^2 \\a^2 + b^2 + 2ab & b^2 & 2ab \\a^2 + b^2 + 2ab & 2ab & a^2 \end {vmatrix}\)
= (a + b)²\(\begin{vmatrix}1 & a^2 & b^2 \\1 & b^2 & 2ab \\1 & 2ab & a^2 \end {vmatrix}\)
= (a + b)²\(\begin{vmatrix}1 & a^2 & b^2 \\1 & b^2 – a^2 & 2ab – b^2 \\0 & 2ab – a^2 & a^2 – b^2 \end {vmatrix}\)
= (a + b)²[(b² – a²)(a² – b²) – (2ab – b²)( 2ab – a²)]
= -(a + b)²[(a² – b²)² + 4a²b² – 2ab(a² + b²) + a² b²)]
= –(a + b)²[(a²+b²)² – 2(a²+b²) (ab)+(ab)²]
= –(a + b)²(a² + b² – ab)²
= –[(a + b)²(a² + b² – ab)²]²
= –(a³ + b³)²

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6. What will be the value of f(x) if \(\begin{vmatrix}1 & ab & (\frac{1}{a} + \frac{1}{b}) \\1 & bc & (\frac{1}{b} + \frac{1}{c}) \\1 & ca & (\frac{1}{c} + \frac{1}{a})\end {vmatrix}\)?
a) -1
b) 0
c) 1
d) Can’t be predicted
View Answer

Answer: c
Explanation: We have,\(\begin{vmatrix}1 & ab & (\frac{1}{a} + \frac{1}{b}) \\1 & bc & (\frac{1}{b} + \frac{1}{c}) \\1 & ca & (\frac{1}{c} + \frac{1}{a})\end {vmatrix}\)
= (1/abc)\(\begin{vmatrix}1 & ab & \frac{b + a}{ab} * abc \\1 & bc & \frac{b + c}{bc} * abc \\1 & ca & \frac{c + a}{ac} * abc \end {vmatrix}\)
= (1/abc)\(\begin{vmatrix}1 & ab & bc + ac \\1 & bc & ac + ab \\1 & ca & ab + bc \end {vmatrix}\)
Operating, C₃ = C₃ + C₂
= (1/abc)\(\begin{vmatrix}1 & ab & ab + bc + ac \\1 & bc & ab + bc + ac \\1 & ca & ab + bc + ac \end {vmatrix}\)
= ((ab + bc + ac)/abc)\(\begin{vmatrix}1 & ab & 1 \\1 & bc & 1 \\1 & ca & 1 \end {vmatrix}\)
= 0

7. What is the value of \(\begin{vmatrix}1 & cosx-sinx & cosx + sinx \\1 & cosy-siny & cosy + siny \\1 & cosz-sinz & cosz + sinz \end {vmatrix}\)?
a) 3\(\begin{vmatrix}1 & cosx & sinx \\1 & cosy & siny \\1 & cosz & sinz \end {vmatrix}\)
b) \(\begin{vmatrix}1 & cosx & sinx \\1 & cosy & siny \\1 & cosz & sinz \end {vmatrix}\)
c) 2\(\begin{vmatrix}1 & cosx & sinx \\1 & cosy & siny \\1 & cosz & sinz \end {vmatrix}\)
d) 4\(\begin{vmatrix}1 & cosx & sinx \\1 & cosy & siny \\1 & cosz & sinz \end {vmatrix}\)
View Answer

Answer: c
Explanation: Let, a = cosx, b = cosy, c = cosz, p =sinx, q = siny and r = sinz
So, \(\begin{vmatrix}1 & a – p & a + p \\1 & b – q & b + q \\1 & c – r & c + r \end {vmatrix}\)
Making C₃ = C₃ + C₂
= \(\begin{vmatrix}1 & a – p & 2a \\1 & b – q & 2b \\1 & c – r & 2c \end {vmatrix}\)

= 2\(\begin{vmatrix}1 & a – p & a \\1 & b – q & b \\1 & c – r & c \end {vmatrix}\)
Making C₂ = C₂ – C₃
= -2\(\begin{vmatrix}1 & p & a \\1 & q & b \\1 & r & c \end {vmatrix}\)
Interchanging 2^nd and 3^rd column, we get,
2\(\begin{vmatrix}1 & a & p \\1 & b & q \\1 & c & r \end {vmatrix}\)
= 2\(\begin{vmatrix}1 & cosx & sinx \\1 & cosy & siny \\1 & cosz & sinz \end {vmatrix}\)

8. What will be the value of \(\begin{vmatrix}a & b & c \\b & c & a \\c & a & b \end {vmatrix}\)?
a) (a³ + b³ + c³ + 3abc)
b) –(a³ + b³ + c³ + 3abc)
c) (a³ + b³ + c³ – 3abc)
d) –(a³ + b³ + c³ – 3abc)
View Answer

Answer: d
Explanation: Given, \(\begin{vmatrix}a & b & c \\b & c & a \\c & a & b \end {vmatrix}\)
Replacing R₁ = R₁ + R₂ + R₃
\(\begin{vmatrix}a + b + c & a + b + c & a + b + c \\b & c & a \\c & a & b \end {vmatrix}\)
= (a + b + c)\(\begin{vmatrix}1 & 1 & 1 \\b & c & a \\c & a & b \end {vmatrix}\)
Replacing 2^nd column by C₂ – C₁ and 3^rd column by C₃ – C₁
= (a + b + c)\(\begin{vmatrix}1 & 0 & 0 \\b & c-b & a-b \\c & a-c & b-c \end {vmatrix}\)
= (a + b + c)[(c – b)(b – c) – (a – b)(a – c)]
= (a + b + c)(bc – b² – c² + bc + a² + ac + ab – bc)
= -(a + b + c)(a² + b² + c² – ab – bc – ac)
= -(a³ + b³ + c³ – 3abc)

9. What will be the value of \(\begin{vmatrix}2bc – a^2 & c^2 & b^2 \\c^2 & 2ac – b^2 & a^2 \\b^2 & a^2 & 2ab – c^2 \end {vmatrix}\) if given another determinant \(\begin{vmatrix}a & b & c \\b & c & a \\c & a & b \end {vmatrix}\)?
a) (a³ + b³ + c³ + 3abc)²
b) –(a³ + b³ + c³ + 3abc)²
c) (a³ + b³ + c³ – 3abc)²
d) –(a³ + b³ + c³ – 3abc)²
View Answer

Answer: c
Explanation: Now, \(\begin{vmatrix}a & b & c \\b & c & a \\c & a & b \end {vmatrix}\)
Interchanging 2^nd and 3^rd columns,
= –\(\begin{vmatrix}a & c & b \\b & a & c \\c & b & a \end {vmatrix}\)
= \(\begin{vmatrix}-a & c & b \\ -b & a & c \\-c & b & a \end {vmatrix}\)
So, \(\begin{vmatrix}a & b & c \\b & c & a \\c & a & b \end {vmatrix}^2\) = \(\begin{vmatrix}a & b & c \\b & c & a \\c & a & b \end {vmatrix}\)\(\begin{vmatrix} -a & c & b \\ -b & a & c \\-c & b & a \end {vmatrix}\)
= {–(a³ + b³ + c³ – 3abc)}² = \(\begin{vmatrix}-a^2 + bc + bc & -ab + ab + c^2 & -ac + b^2 + ca \\-ab + ab + c^2 & -ac – b^2 + ca & -a^2 + bc + bc \\-ac + b^2 + ca & -a^2 + bc + bc & -ab + ab – c^2 \end {vmatrix}\)
=> \(\begin{vmatrix}2bc – a^2 & c^2 & b^2 \\c^2 & 2ac – b^2 & a^2 \\b^2 & a^2 & 2ab – c^2 \end {vmatrix}\) = (a³ + b³ + c³ – 3abc)²

10. What will be the value of f(x) if \(\begin{vmatrix}1 & 1 & 1 \\x & y & z \\x^3 & y^3 & z^3 \end {vmatrix}\)?
a) -1
b) 0
c) 1
d) 2
View Answer

Answer: b
Explanation: Given, \(\begin{vmatrix}1 & 1 & 1 \\x & y & z \\x^3 & y^3 & z^3 \end {vmatrix}\)
Operating, C₁ = C₁ – C₂ and C₂ = C₂ – C₃
= \(\begin{vmatrix}1 & 1 & 1 \\x – y & y – z & y \\x^3 – y^3 & y^3 – z^3 & z^3 \end {vmatrix}\)
Expanding by the 1^st row,
= (x – y)(y³ – z³) – (y – z)(x³ – y³)
= (x – y)(y – z)[(y² + yz + z²) – (x² + xy + y²)]
= (x – y)(y – z)(z – x)(x + y + z)
As, x + y + z = 0
= 0

11. If, x³ = 1, then, what will be the value of\(\begin{vmatrix}a & b & c \\b & c & a \\c & a & b \end {vmatrix}\)?
a) -(a + bx + cx²)\(\begin{vmatrix}1 & b & c \\x^2 & c & a \\x & a & b \end {vmatrix}\)
b) (a + bx + cx²)\(\begin{vmatrix}1 & b & c \\x^2 & c & a \\x & a & b \end {vmatrix}\)
c) (a – bx – cx²)\(\begin{vmatrix}1 & b & c \\x^2 & c & a \\x & a & b \end {vmatrix}\)
d) (a + bx – cx²)\(\begin{vmatrix}1 & b & c \\x^2 & c & a \\x & a & b \end {vmatrix}\)
View Answer

Answer: b
Explanation: We have, \(\begin{vmatrix}a & b & c \\b & c & a \\c & a & b \end {vmatrix}\)
As, x³ = 1,
= \(\begin{vmatrix}a & bx & cx^2 \\b & cx & ax^2 \\c & ax & bx^2 \end {vmatrix}\)
Replacing the 1^st column by C₁ + C₂ + C₃ we get,
= \(\begin{vmatrix}a + bx + cx^2 & bx & cx^2 \\ b + cx + ax^2 & cx & ax^2 \\c + ax + bx^2 & ax & bx^2 \end {vmatrix}\)
As, x³ = 1 so, x⁴ = x³ * x = x
= \(\begin{vmatrix}a + bx + cx^2 & b & c \\x^2 (a + bx + cx^2) & c & a \\x(a + bx + cx^2) & a & b \end {vmatrix}\)
= (a + bx + cx²)\(\begin{vmatrix}1 & b & c \\x^2 & c & a \\x & a & b \end {vmatrix}\)

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