QBManager

Matrix and Determinant

79143 If $| \begin{array}{lll} b & b^{2} & 1 + b^{3} \\ c & c^{2} & 1 + c^{3} \end{array} | = 0$ and the vectors
$(1, a, a^{2}), (1, b, b^{2}), (1, c, c^{2})$ are non-coplaner, then abc $=$

1 2

2 -1

3 1

4 0

Explanation:

(B) : According to given summation,
$| \begin{array}{lll} a & a^{2} & 1 + a^{3} \\ b & b^{2} & 1 + b^{3} \\ c & c^{2} & 1 + c^{3} \end{array} | = | \begin{array}{ccc} a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1 \end{array} | + | \begin{array}{lll} a & a^{2} & a^{3} \\ b & b^{2} & b^{3} \\ c & c^{2} & c^{3} \end{array} |$
$| \begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array} | + a b c | \begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array} | = 0$
$(1 + a b c) | \begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array} | = 0$
$1 + abc = 0$
$a b c = - 1$
$[∵$ the determinant $\neq 0]$

COMEDK-2018

Matrix and Determinant

79144 If $Δ_{1} = | \begin{array}{lll} x & b & b \\ a & x & b \\ a & a & x \end{array} |$ and $Δ_{2} = | \begin{array}{ll} x & b \\ a & x \end{array} |$ , then $\frac{d}{d x} (Δ_{1})$ is equal to

1

3 {(Δ_{2})}^{2}

2

3 {(Δ_{2})}^{1 / 2}

3

3 Δ_{2}

4

3 Δ_{2}^{2}

Explanation:

(C) : We have,,
$Δ_{1} = | \begin{array}{lll} x & b & b \\ a & x & b \\ a & a & x \end{array} |$ and $Δ_{2} = | \begin{array}{ll} x & b \\ a & x \end{array} |$
In the given matrix $Δ_{1}$ expand along $R_{1}$
$Δ_{1} = x (x^{2} - a b) - b (a x - a b) + b (a^{2} - a x)$
$Δ_{1} = x^{3} - a b x - a b x + a b^{2} + b a^{2} + b a^{2} - a b x$
$Δ_{1} = x^{3} - 3 a b x + a b (a + b)$
$\frac{d}{dx} (Δ_{1}) = \frac{d}{dx} (x^{3} - 3 abx + ab (a + b))$
$\frac{d}{dx} (Δ_{1}) = 3 x^{2} - 3 ab + 0$
$\frac{d}{dx} (Δ_{1}) = 3 (x^{2} - ab)$
Here, $Δ_{2} = | \begin{array}{ll} x & b \\ a & x \end{array} | = x^{2} - ab$
Substituting in equation (i),
$\frac{d}{dx} (Δ_{1}) = 3 (Δ_{2}) ∴ \frac{d}{dx} (Δ_{1}) = 3 Δ_{2}$

SRM JEE-2019

Matrix and Determinant

79145 If $1, ω, ω^{2}$ are the roots of unity, the
$Δ = | \begin{array}{ccc} 1 & ω^{n} & ω^{2 n} \\ ω^{n} & ω^{2 n} & 1 \end{array} | is equal to$

1

ω^{2}

2 0

3 1

4

ω

Explanation:

(B) : Given that,
$Δ = | \begin{array}{ccc} 1 & ω^{n} & ω^{2 n} \\ ω^{n} & ω^{2 n} & 1 \\ ω^{2 n} & 1 & ω^{n} \end{array} |$
$= 1 (ω^{3 n} - 1) - ω^{n} (ω^{2 n} - ω^{2 n}) + ω^{2 n} (ω^{n} - ω^{4 n})$
$= ω^{3 n} - 1 - ω^{3 n} + ω^{3 n} + ω^{3 n} - ω^{6 n}$
$= 1 - 1 - 1 + 1 + 1 - 1$
$[∵ ω^{3 n} = 1]$
$= 0$

BITSAT-2006

Matrix and Determinant

79146 If $a + b + c = 0$ then determinant
$| \begin{array}{ccc} a - b - c & 2 a & 2 a \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{array} | is equal to,$

1 0

2 1

3 2

4 3

Explanation:

(A) : It is given that,
$a + b + c = 0$
$And | \begin{array}{ccc} a - b - c & 2 a & 2 a \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{array} |$
Operating $R_{1} \to R_{1} + R_{2} + R_{3}$
$| \begin{array}{ccc} a + b + c & a + b + c & a + b + c \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{array} |$
Take $(a + b + c)$ common from $R_{1}$
$= (a + b + c) | \begin{array}{ccc} 1 & 1 & 1 \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{array} |$
$= 0$

BITSAT-2008

Matrix and Determinant

79143 If $| \begin{array}{lll} b & b^{2} & 1 + b^{3} \\ c & c^{2} & 1 + c^{3} \end{array} | = 0$ and the vectors
$(1, a, a^{2}), (1, b, b^{2}), (1, c, c^{2})$ are non-coplaner, then abc $=$

1 2

2 -1

3 1

4 0

Explanation:

(B) : According to given summation,
$| \begin{array}{lll} a & a^{2} & 1 + a^{3} \\ b & b^{2} & 1 + b^{3} \\ c & c^{2} & 1 + c^{3} \end{array} | = | \begin{array}{ccc} a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1 \end{array} | + | \begin{array}{lll} a & a^{2} & a^{3} \\ b & b^{2} & b^{3} \\ c & c^{2} & c^{3} \end{array} |$
$| \begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array} | + a b c | \begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array} | = 0$
$(1 + a b c) | \begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array} | = 0$
$1 + abc = 0$
$a b c = - 1$
$[∵$ the determinant $\neq 0]$

COMEDK-2018

Matrix and Determinant

79144 If $Δ_{1} = | \begin{array}{lll} x & b & b \\ a & x & b \\ a & a & x \end{array} |$ and $Δ_{2} = | \begin{array}{ll} x & b \\ a & x \end{array} |$ , then $\frac{d}{d x} (Δ_{1})$ is equal to

1

3 {(Δ_{2})}^{2}

2

3 {(Δ_{2})}^{1 / 2}

3

3 Δ_{2}

4

3 Δ_{2}^{2}

Explanation:

(C) : We have,,
$Δ_{1} = | \begin{array}{lll} x & b & b \\ a & x & b \\ a & a & x \end{array} |$ and $Δ_{2} = | \begin{array}{ll} x & b \\ a & x \end{array} |$
In the given matrix $Δ_{1}$ expand along $R_{1}$
$Δ_{1} = x (x^{2} - a b) - b (a x - a b) + b (a^{2} - a x)$
$Δ_{1} = x^{3} - a b x - a b x + a b^{2} + b a^{2} + b a^{2} - a b x$
$Δ_{1} = x^{3} - 3 a b x + a b (a + b)$
$\frac{d}{dx} (Δ_{1}) = \frac{d}{dx} (x^{3} - 3 abx + ab (a + b))$
$\frac{d}{dx} (Δ_{1}) = 3 x^{2} - 3 ab + 0$
$\frac{d}{dx} (Δ_{1}) = 3 (x^{2} - ab)$
Here, $Δ_{2} = | \begin{array}{ll} x & b \\ a & x \end{array} | = x^{2} - ab$
Substituting in equation (i),
$\frac{d}{dx} (Δ_{1}) = 3 (Δ_{2}) ∴ \frac{d}{dx} (Δ_{1}) = 3 Δ_{2}$

SRM JEE-2019

Matrix and Determinant

79145 If $1, ω, ω^{2}$ are the roots of unity, the
$Δ = | \begin{array}{ccc} 1 & ω^{n} & ω^{2 n} \\ ω^{n} & ω^{2 n} & 1 \end{array} | is equal to$

1

ω^{2}

2 0

3 1

4

ω

Explanation:

(B) : Given that,
$Δ = | \begin{array}{ccc} 1 & ω^{n} & ω^{2 n} \\ ω^{n} & ω^{2 n} & 1 \\ ω^{2 n} & 1 & ω^{n} \end{array} |$
$= 1 (ω^{3 n} - 1) - ω^{n} (ω^{2 n} - ω^{2 n}) + ω^{2 n} (ω^{n} - ω^{4 n})$
$= ω^{3 n} - 1 - ω^{3 n} + ω^{3 n} + ω^{3 n} - ω^{6 n}$
$= 1 - 1 - 1 + 1 + 1 - 1$
$[∵ ω^{3 n} = 1]$
$= 0$

BITSAT-2006

Matrix and Determinant

79146 If $a + b + c = 0$ then determinant
$| \begin{array}{ccc} a - b - c & 2 a & 2 a \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{array} | is equal to,$

1 0

2 1

3 2

4 3

Explanation:

(A) : It is given that,
$a + b + c = 0$
$And | \begin{array}{ccc} a - b - c & 2 a & 2 a \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{array} |$
Operating $R_{1} \to R_{1} + R_{2} + R_{3}$
$| \begin{array}{ccc} a + b + c & a + b + c & a + b + c \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{array} |$
Take $(a + b + c)$ common from $R_{1}$
$= (a + b + c) | \begin{array}{ccc} 1 & 1 & 1 \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{array} |$
$= 0$

BITSAT-2008

Matrix and Determinant

79143 If $| \begin{array}{lll} b & b^{2} & 1 + b^{3} \\ c & c^{2} & 1 + c^{3} \end{array} | = 0$ and the vectors
$(1, a, a^{2}), (1, b, b^{2}), (1, c, c^{2})$ are non-coplaner, then abc $=$

1 2

2 -1

3 1

4 0

Explanation:

(B) : According to given summation,
$| \begin{array}{lll} a & a^{2} & 1 + a^{3} \\ b & b^{2} & 1 + b^{3} \\ c & c^{2} & 1 + c^{3} \end{array} | = | \begin{array}{ccc} a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1 \end{array} | + | \begin{array}{lll} a & a^{2} & a^{3} \\ b & b^{2} & b^{3} \\ c & c^{2} & c^{3} \end{array} |$
$| \begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array} | + a b c | \begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array} | = 0$
$(1 + a b c) | \begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array} | = 0$
$1 + abc = 0$
$a b c = - 1$
$[∵$ the determinant $\neq 0]$

COMEDK-2018

Matrix and Determinant

79144 If $Δ_{1} = | \begin{array}{lll} x & b & b \\ a & x & b \\ a & a & x \end{array} |$ and $Δ_{2} = | \begin{array}{ll} x & b \\ a & x \end{array} |$ , then $\frac{d}{d x} (Δ_{1})$ is equal to

1

3 {(Δ_{2})}^{2}

2

3 {(Δ_{2})}^{1 / 2}

3

3 Δ_{2}

4

3 Δ_{2}^{2}

Explanation:

(C) : We have,,
$Δ_{1} = | \begin{array}{lll} x & b & b \\ a & x & b \\ a & a & x \end{array} |$ and $Δ_{2} = | \begin{array}{ll} x & b \\ a & x \end{array} |$
In the given matrix $Δ_{1}$ expand along $R_{1}$
$Δ_{1} = x (x^{2} - a b) - b (a x - a b) + b (a^{2} - a x)$
$Δ_{1} = x^{3} - a b x - a b x + a b^{2} + b a^{2} + b a^{2} - a b x$
$Δ_{1} = x^{3} - 3 a b x + a b (a + b)$
$\frac{d}{dx} (Δ_{1}) = \frac{d}{dx} (x^{3} - 3 abx + ab (a + b))$
$\frac{d}{dx} (Δ_{1}) = 3 x^{2} - 3 ab + 0$
$\frac{d}{dx} (Δ_{1}) = 3 (x^{2} - ab)$
Here, $Δ_{2} = | \begin{array}{ll} x & b \\ a & x \end{array} | = x^{2} - ab$
Substituting in equation (i),
$\frac{d}{dx} (Δ_{1}) = 3 (Δ_{2}) ∴ \frac{d}{dx} (Δ_{1}) = 3 Δ_{2}$

SRM JEE-2019

Matrix and Determinant

79145 If $1, ω, ω^{2}$ are the roots of unity, the
$Δ = | \begin{array}{ccc} 1 & ω^{n} & ω^{2 n} \\ ω^{n} & ω^{2 n} & 1 \end{array} | is equal to$

1

ω^{2}

2 0

3 1

4

ω

Explanation:

(B) : Given that,
$Δ = | \begin{array}{ccc} 1 & ω^{n} & ω^{2 n} \\ ω^{n} & ω^{2 n} & 1 \\ ω^{2 n} & 1 & ω^{n} \end{array} |$
$= 1 (ω^{3 n} - 1) - ω^{n} (ω^{2 n} - ω^{2 n}) + ω^{2 n} (ω^{n} - ω^{4 n})$
$= ω^{3 n} - 1 - ω^{3 n} + ω^{3 n} + ω^{3 n} - ω^{6 n}$
$= 1 - 1 - 1 + 1 + 1 - 1$
$[∵ ω^{3 n} = 1]$
$= 0$

BITSAT-2006

Matrix and Determinant

79146 If $a + b + c = 0$ then determinant
$| \begin{array}{ccc} a - b - c & 2 a & 2 a \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{array} | is equal to,$

1 0

2 1

3 2

4 3

Explanation:

(A) : It is given that,
$a + b + c = 0$
$And | \begin{array}{ccc} a - b - c & 2 a & 2 a \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{array} |$
Operating $R_{1} \to R_{1} + R_{2} + R_{3}$
$| \begin{array}{ccc} a + b + c & a + b + c & a + b + c \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{array} |$
Take $(a + b + c)$ common from $R_{1}$
$= (a + b + c) | \begin{array}{ccc} 1 & 1 & 1 \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{array} |$
$= 0$

BITSAT-2008

Matrix and Determinant

79143 If $| \begin{array}{lll} b & b^{2} & 1 + b^{3} \\ c & c^{2} & 1 + c^{3} \end{array} | = 0$ and the vectors
$(1, a, a^{2}), (1, b, b^{2}), (1, c, c^{2})$ are non-coplaner, then abc $=$

1 2

2 -1

3 1

4 0

Explanation:

(B) : According to given summation,
$| \begin{array}{lll} a & a^{2} & 1 + a^{3} \\ b & b^{2} & 1 + b^{3} \\ c & c^{2} & 1 + c^{3} \end{array} | = | \begin{array}{ccc} a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1 \end{array} | + | \begin{array}{lll} a & a^{2} & a^{3} \\ b & b^{2} & b^{3} \\ c & c^{2} & c^{3} \end{array} |$
$| \begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array} | + a b c | \begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array} | = 0$
$(1 + a b c) | \begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array} | = 0$
$1 + abc = 0$
$a b c = - 1$
$[∵$ the determinant $\neq 0]$

COMEDK-2018

Matrix and Determinant

79144 If $Δ_{1} = | \begin{array}{lll} x & b & b \\ a & x & b \\ a & a & x \end{array} |$ and $Δ_{2} = | \begin{array}{ll} x & b \\ a & x \end{array} |$ , then $\frac{d}{d x} (Δ_{1})$ is equal to

1

3 {(Δ_{2})}^{2}

2

3 {(Δ_{2})}^{1 / 2}

3

3 Δ_{2}

4

3 Δ_{2}^{2}

Explanation:

(C) : We have,,
$Δ_{1} = | \begin{array}{lll} x & b & b \\ a & x & b \\ a & a & x \end{array} |$ and $Δ_{2} = | \begin{array}{ll} x & b \\ a & x \end{array} |$
In the given matrix $Δ_{1}$ expand along $R_{1}$
$Δ_{1} = x (x^{2} - a b) - b (a x - a b) + b (a^{2} - a x)$
$Δ_{1} = x^{3} - a b x - a b x + a b^{2} + b a^{2} + b a^{2} - a b x$
$Δ_{1} = x^{3} - 3 a b x + a b (a + b)$
$\frac{d}{dx} (Δ_{1}) = \frac{d}{dx} (x^{3} - 3 abx + ab (a + b))$
$\frac{d}{dx} (Δ_{1}) = 3 x^{2} - 3 ab + 0$
$\frac{d}{dx} (Δ_{1}) = 3 (x^{2} - ab)$
Here, $Δ_{2} = | \begin{array}{ll} x & b \\ a & x \end{array} | = x^{2} - ab$
Substituting in equation (i),
$\frac{d}{dx} (Δ_{1}) = 3 (Δ_{2}) ∴ \frac{d}{dx} (Δ_{1}) = 3 Δ_{2}$

SRM JEE-2019

Matrix and Determinant

79145 If $1, ω, ω^{2}$ are the roots of unity, the
$Δ = | \begin{array}{ccc} 1 & ω^{n} & ω^{2 n} \\ ω^{n} & ω^{2 n} & 1 \end{array} | is equal to$

1

ω^{2}

2 0

3 1

4

ω

Explanation:

(B) : Given that,
$Δ = | \begin{array}{ccc} 1 & ω^{n} & ω^{2 n} \\ ω^{n} & ω^{2 n} & 1 \\ ω^{2 n} & 1 & ω^{n} \end{array} |$
$= 1 (ω^{3 n} - 1) - ω^{n} (ω^{2 n} - ω^{2 n}) + ω^{2 n} (ω^{n} - ω^{4 n})$
$= ω^{3 n} - 1 - ω^{3 n} + ω^{3 n} + ω^{3 n} - ω^{6 n}$
$= 1 - 1 - 1 + 1 + 1 - 1$
$[∵ ω^{3 n} = 1]$
$= 0$

BITSAT-2006

Matrix and Determinant

79146 If $a + b + c = 0$ then determinant
$| \begin{array}{ccc} a - b - c & 2 a & 2 a \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{array} | is equal to,$

1 0

2 1

3 2

4 3

Explanation:

(A) : It is given that,
$a + b + c = 0$
$And | \begin{array}{ccc} a - b - c & 2 a & 2 a \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{array} |$
Operating $R_{1} \to R_{1} + R_{2} + R_{3}$
$| \begin{array}{ccc} a + b + c & a + b + c & a + b + c \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{array} |$
Take $(a + b + c)$ common from $R_{1}$
$= (a + b + c) | \begin{array}{ccc} 1 & 1 & 1 \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{array} |$
$= 0$

BITSAT-2008

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