QBManager

Matrix and Determinant

79424 If $f(x)=\left|\begin{array}{ccc}1 & x & x+1 \\ 2 x & x(x-1) & (x+1) x \\ 3 x(x-1) & x(x-1)(x-2) & (x+1) x(x-1)\end{array}\right|$

1 0

2 1

3 100

4 -100

AMU-2012

Matrix and Determinant

79425 If $b \quad c \quad b \alpha-c=0$ and $\alpha \neq \frac{1}{2}$, then $a, b, c$ are in $\begin{array}{lll} 2 & 1 & 0 \end{array}$

1 A.P.

2 G.P.

3 H.P.

4 none of these

Explanation:

(B) : According to given summation,
If $| \begin{array}{ccc} a & b & α a - b \\ b & c & α b - c \\ 2 & 1 & 0 \end{array} | = 0$
Operating $C_{3 \to} C_{3} - α C_{1} - C_{2}$ , we get
$| \begin{array}{ccc} a & b & 0 \\ b & c & 0 \\ 2 & 1 & - 2 α + 1 \end{array} | = 0$
$\Rightarrow a [c (- 2 α + 1) - b [b (- 2 α + 1] - 0$
$\Rightarrow (- 2 α + 1) (ac - b^{2}) = 0 \Rightarrow (- 2 α + 1) = 0 or (ac - b^{2}) = 0$
$But a \neq \frac{1}{2} and a c = b^{2}$

SEMJEEE-2012

Matrix and Determinant

79426 The solution for $| \begin{array}{ccc} 4 x & 6 x + 2 & 8 x + 1 \\ 6 x + 2 & 9 x + 3 & 12 x \\ 8 x + 1 & 12 x & 16 x + 2 \end{array} | = 0$ is
is given by

1

x = \frac{- 13}{71}

2

x = \frac{- 11}{97}

3

x = \frac{- 7}{43}

4

x = \frac{- 1}{11}

Explanation:

(B) : Given,
$| \begin{array}{ccc} 4 x & 6 x + 2 & 8 x + 1 \\ 6 x + 2 & 9 x + 3 & 12 x \\ 8 x + 1 & 12 x & 16 x + 2 \end{array} | = 0$
Operating $C_{3 \to} C_{3} - 2 C_{1}$
$| \begin{array}{ccc} 4 x & 6 x + 2 & 1 \\ 6 x + 2 & 9 x + 3 & - 4 \\ 8 x + 1 & 12 x & 0 \end{array} | = 0$
Now, taking 2 common from $C_{1}$ , we get
$| \begin{array}{ccc} 2 x & 6 x + 2 & 1 \\ 3 x + 1 & 9 x + 3 & - 4 \\ 4 x + \frac{1}{2} & 12 x & 0 \\ ng C_{2} \to_{\to}^{} C_{2} - 3 C_{1} \end{array} | = 0$
Operating $C_{2} \to C_{2} - 3 C$
$| \begin{array}{ccc} 2 x & 2 & 1 \\ 3 x + 1 & 0 & - 4 \\ 4 x + \frac{1}{2} & \frac{- 3}{2} & 0 \end{array} | = 0$
Multiply by 2 in column 2 , we get
$| \begin{array}{ccc} 2 x & 4 & 1 \\ 3 x + 1 & 0 & - 4 \\ 4 x + \frac{1}{2} & - 3 & 0 \end{array} | = 0$
Operating $R_{2} \overset{2}{\to} R_{2} + 4 R_{1}$
$| \begin{array}{ccc} 2 x & 4 & 1 \\ 11 x + 1 & 16 & 0 \\ 4 x + \frac{1}{2} & - 3 & 0 \end{array} | = 0$
$1 (- 33 x - 3) - 64 x - 8$
$- 33 x - 3 - 64 x - 8 = 0$
$- 97 x - 11 = 0$
$x = \frac{- 11}{97}$
$Now, 1 (- 33 x - 3) - 64 x - 8 = 0$

SRM JEE-2013

Matrix and Determinant

79427 The value of $| \begin{array}{ccc} 1 + ω & ω^{2} & - ω \\ 1 + ω^{2} & ω & - ω^{2} \\ ω^{2} + ω & ω & - ω^{2} \end{array} |$ is equal to
( $ω$ being an imaginary cube root of unity)

1 0

2

2 ω

3

2 ω^{2}

4

- 3 ω^{2}

Explanation:

(D) : It is given that,
$Δ = | \begin{array}{ccc} 1 + ω & ω^{2} & - ω \\ 1 + ω^{2} & ω & - ω^{2} \\ ω^{2} + ω & ω & - ω^{2} \end{array} |$
We know that:-
$[1 + ω + ω^{2} = 0]$
$Δ = | \begin{array}{ccc} - ω^{2} & ω^{2} & - ω \\ - ω & ω & - ω^{2} \\ - 1 & ω & - ω^{2} \end{array} |$
Operating $C_{1} \to C_{1} + C_{2}$ , we get
$Δ = | \begin{array}{ccc} 0 & ω^{2} & - ω \\ 0 & ω & - ω^{2} \\ - 1 + ω & ω & - ω^{2} \end{array} |$
$Δ = (- 1 + ω) (- ω^{4} + ω^{2})$
$= (- 1 + ω) (- ω + ω^{2})$
$= ω - ω^{2} - ω^{2} + ω^{3}$
$= 1 + ω - 2 ω^{2} = - ω^{2} - 2 ω^{2}$
$Δ = - 3 ω^{2}$

SRM JEE-2013

Matrix and Determinant

79424 If $f (x) = | \begin{array}{ccc} 1 & x & x + 1 \\ 2 x & x (x - 1) & (x + 1) x \\ 3 x (x - 1) & x (x - 1) (x - 2) & (x + 1) x (x - 1) \end{array} |$

1 0

2 1

3 100

4 -100

Explanation:

(A) : Given,
$f (x) = | \begin{array}{ccc} 1 & x & x + 1 \\ 2 x & x (x - 1) & (x + 1) x \\ 3 x (x - 1) & x (x - 1) (x - 2) & (x - 1) x (x - 1) \end{array} |$
Operating $C_{3 \to} C_{3} - C_{1} - C_{2}$ , we get
$f (x) = | \begin{array}{ccc} 1 & x & 0 \\ 2 x & x (x - 1) & 0 \\ 3 x (x - 1) & x (x - 1) (x - 2) & 0 \end{array} |, ∵ C_{3} = 0)$
a $b \boldsymbol α a - b$

AMU-2012

Matrix and Determinant

79425 If $b c b α - c = 0$ and $α \neq \frac{1}{2}$ , then $a, b, c$ are in $\begin{array}{lll} 2 & 1 & 0 \end{array}$

1 A.P.

2 G.P.

3 H.P.

4 none of these

Explanation:

(B) : According to given summation,
If $| \begin{array}{ccc} a & b & α a - b \\ b & c & α b - c \\ 2 & 1 & 0 \end{array} | = 0$
Operating $C_{3 \to} C_{3} - α C_{1} - C_{2}$ , we get
$| \begin{array}{ccc} a & b & 0 \\ b & c & 0 \\ 2 & 1 & - 2 α + 1 \end{array} | = 0$
$\Rightarrow a [c (- 2 α + 1) - b [b (- 2 α + 1] - 0$
$\Rightarrow (- 2 α + 1) (ac - b^{2}) = 0 \Rightarrow (- 2 α + 1) = 0 or (ac - b^{2}) = 0$
$But a \neq \frac{1}{2} and a c = b^{2}$

SEMJEEE-2012

Matrix and Determinant

79426 The solution for $| \begin{array}{ccc} 4 x & 6 x + 2 & 8 x + 1 \\ 6 x + 2 & 9 x + 3 & 12 x \\ 8 x + 1 & 12 x & 16 x + 2 \end{array} | = 0$ is
is given by

1

x = \frac{- 13}{71}

2

x = \frac{- 11}{97}

3

x = \frac{- 7}{43}

4

x = \frac{- 1}{11}

Explanation:

(B) : Given,
$| \begin{array}{ccc} 4 x & 6 x + 2 & 8 x + 1 \\ 6 x + 2 & 9 x + 3 & 12 x \\ 8 x + 1 & 12 x & 16 x + 2 \end{array} | = 0$
Operating $C_{3 \to} C_{3} - 2 C_{1}$
$| \begin{array}{ccc} 4 x & 6 x + 2 & 1 \\ 6 x + 2 & 9 x + 3 & - 4 \\ 8 x + 1 & 12 x & 0 \end{array} | = 0$
Now, taking 2 common from $C_{1}$ , we get
$| \begin{array}{ccc} 2 x & 6 x + 2 & 1 \\ 3 x + 1 & 9 x + 3 & - 4 \\ 4 x + \frac{1}{2} & 12 x & 0 \\ ng C_{2} \to_{\to}^{} C_{2} - 3 C_{1} \end{array} | = 0$
Operating $C_{2} \to C_{2} - 3 C$
$| \begin{array}{ccc} 2 x & 2 & 1 \\ 3 x + 1 & 0 & - 4 \\ 4 x + \frac{1}{2} & \frac{- 3}{2} & 0 \end{array} | = 0$
Multiply by 2 in column 2 , we get
$| \begin{array}{ccc} 2 x & 4 & 1 \\ 3 x + 1 & 0 & - 4 \\ 4 x + \frac{1}{2} & - 3 & 0 \end{array} | = 0$
Operating $R_{2} \overset{2}{\to} R_{2} + 4 R_{1}$
$| \begin{array}{ccc} 2 x & 4 & 1 \\ 11 x + 1 & 16 & 0 \\ 4 x + \frac{1}{2} & - 3 & 0 \end{array} | = 0$
$1 (- 33 x - 3) - 64 x - 8$
$- 33 x - 3 - 64 x - 8 = 0$
$- 97 x - 11 = 0$
$x = \frac{- 11}{97}$
$Now, 1 (- 33 x - 3) - 64 x - 8 = 0$

SRM JEE-2013

Matrix and Determinant

79427 The value of $| \begin{array}{ccc} 1 + ω & ω^{2} & - ω \\ 1 + ω^{2} & ω & - ω^{2} \\ ω^{2} + ω & ω & - ω^{2} \end{array} |$ is equal to
( $ω$ being an imaginary cube root of unity)

1 0

2

2 ω

3

2 ω^{2}

4

- 3 ω^{2}

Explanation:

(D) : It is given that,
$Δ = | \begin{array}{ccc} 1 + ω & ω^{2} & - ω \\ 1 + ω^{2} & ω & - ω^{2} \\ ω^{2} + ω & ω & - ω^{2} \end{array} |$
We know that:-
$[1 + ω + ω^{2} = 0]$
$Δ = | \begin{array}{ccc} - ω^{2} & ω^{2} & - ω \\ - ω & ω & - ω^{2} \\ - 1 & ω & - ω^{2} \end{array} |$
Operating $C_{1} \to C_{1} + C_{2}$ , we get
$Δ = | \begin{array}{ccc} 0 & ω^{2} & - ω \\ 0 & ω & - ω^{2} \\ - 1 + ω & ω & - ω^{2} \end{array} |$
$Δ = (- 1 + ω) (- ω^{4} + ω^{2})$
$= (- 1 + ω) (- ω + ω^{2})$
$= ω - ω^{2} - ω^{2} + ω^{3}$
$= 1 + ω - 2 ω^{2} = - ω^{2} - 2 ω^{2}$
$Δ = - 3 ω^{2}$

SRM JEE-2013

Matrix and Determinant

79424 If $f (x) = | \begin{array}{ccc} 1 & x & x + 1 \\ 2 x & x (x - 1) & (x + 1) x \\ 3 x (x - 1) & x (x - 1) (x - 2) & (x + 1) x (x - 1) \end{array} |$

1 0

2 1

3 100

4 -100

Explanation:

(A) : Given,
$f (x) = | \begin{array}{ccc} 1 & x & x + 1 \\ 2 x & x (x - 1) & (x + 1) x \\ 3 x (x - 1) & x (x - 1) (x - 2) & (x - 1) x (x - 1) \end{array} |$
Operating $C_{3 \to} C_{3} - C_{1} - C_{2}$ , we get
$f (x) = | \begin{array}{ccc} 1 & x & 0 \\ 2 x & x (x - 1) & 0 \\ 3 x (x - 1) & x (x - 1) (x - 2) & 0 \end{array} |, ∵ C_{3} = 0)$
a $b \boldsymbol α a - b$

AMU-2012

Matrix and Determinant

79425 If $b c b α - c = 0$ and $α \neq \frac{1}{2}$ , then $a, b, c$ are in $\begin{array}{lll} 2 & 1 & 0 \end{array}$

1 A.P.

2 G.P.

3 H.P.

4 none of these

Explanation:

(B) : According to given summation,
If $| \begin{array}{ccc} a & b & α a - b \\ b & c & α b - c \\ 2 & 1 & 0 \end{array} | = 0$
Operating $C_{3 \to} C_{3} - α C_{1} - C_{2}$ , we get
$| \begin{array}{ccc} a & b & 0 \\ b & c & 0 \\ 2 & 1 & - 2 α + 1 \end{array} | = 0$
$\Rightarrow a [c (- 2 α + 1) - b [b (- 2 α + 1] - 0$
$\Rightarrow (- 2 α + 1) (ac - b^{2}) = 0 \Rightarrow (- 2 α + 1) = 0 or (ac - b^{2}) = 0$
$But a \neq \frac{1}{2} and a c = b^{2}$

SEMJEEE-2012

Matrix and Determinant

79426 The solution for $| \begin{array}{ccc} 4 x & 6 x + 2 & 8 x + 1 \\ 6 x + 2 & 9 x + 3 & 12 x \\ 8 x + 1 & 12 x & 16 x + 2 \end{array} | = 0$ is
is given by

1

x = \frac{- 13}{71}

2

x = \frac{- 11}{97}

3

x = \frac{- 7}{43}

4

x = \frac{- 1}{11}

Explanation:

(B) : Given,
$| \begin{array}{ccc} 4 x & 6 x + 2 & 8 x + 1 \\ 6 x + 2 & 9 x + 3 & 12 x \\ 8 x + 1 & 12 x & 16 x + 2 \end{array} | = 0$
Operating $C_{3 \to} C_{3} - 2 C_{1}$
$| \begin{array}{ccc} 4 x & 6 x + 2 & 1 \\ 6 x + 2 & 9 x + 3 & - 4 \\ 8 x + 1 & 12 x & 0 \end{array} | = 0$
Now, taking 2 common from $C_{1}$ , we get
$| \begin{array}{ccc} 2 x & 6 x + 2 & 1 \\ 3 x + 1 & 9 x + 3 & - 4 \\ 4 x + \frac{1}{2} & 12 x & 0 \\ ng C_{2} \to_{\to}^{} C_{2} - 3 C_{1} \end{array} | = 0$
Operating $C_{2} \to C_{2} - 3 C$
$| \begin{array}{ccc} 2 x & 2 & 1 \\ 3 x + 1 & 0 & - 4 \\ 4 x + \frac{1}{2} & \frac{- 3}{2} & 0 \end{array} | = 0$
Multiply by 2 in column 2 , we get
$| \begin{array}{ccc} 2 x & 4 & 1 \\ 3 x + 1 & 0 & - 4 \\ 4 x + \frac{1}{2} & - 3 & 0 \end{array} | = 0$
Operating $R_{2} \overset{2}{\to} R_{2} + 4 R_{1}$
$| \begin{array}{ccc} 2 x & 4 & 1 \\ 11 x + 1 & 16 & 0 \\ 4 x + \frac{1}{2} & - 3 & 0 \end{array} | = 0$
$1 (- 33 x - 3) - 64 x - 8$
$- 33 x - 3 - 64 x - 8 = 0$
$- 97 x - 11 = 0$
$x = \frac{- 11}{97}$
$Now, 1 (- 33 x - 3) - 64 x - 8 = 0$

SRM JEE-2013

Matrix and Determinant

79427 The value of $| \begin{array}{ccc} 1 + ω & ω^{2} & - ω \\ 1 + ω^{2} & ω & - ω^{2} \\ ω^{2} + ω & ω & - ω^{2} \end{array} |$ is equal to
( $ω$ being an imaginary cube root of unity)

1 0

2

2 ω

3

2 ω^{2}

4

- 3 ω^{2}

Explanation:

(D) : It is given that,
$Δ = | \begin{array}{ccc} 1 + ω & ω^{2} & - ω \\ 1 + ω^{2} & ω & - ω^{2} \\ ω^{2} + ω & ω & - ω^{2} \end{array} |$
We know that:-
$[1 + ω + ω^{2} = 0]$
$Δ = | \begin{array}{ccc} - ω^{2} & ω^{2} & - ω \\ - ω & ω & - ω^{2} \\ - 1 & ω & - ω^{2} \end{array} |$
Operating $C_{1} \to C_{1} + C_{2}$ , we get
$Δ = | \begin{array}{ccc} 0 & ω^{2} & - ω \\ 0 & ω & - ω^{2} \\ - 1 + ω & ω & - ω^{2} \end{array} |$
$Δ = (- 1 + ω) (- ω^{4} + ω^{2})$
$= (- 1 + ω) (- ω + ω^{2})$
$= ω - ω^{2} - ω^{2} + ω^{3}$
$= 1 + ω - 2 ω^{2} = - ω^{2} - 2 ω^{2}$
$Δ = - 3 ω^{2}$

SRM JEE-2013

Matrix and Determinant

79424 If $f (x) = | \begin{array}{ccc} 1 & x & x + 1 \\ 2 x & x (x - 1) & (x + 1) x \\ 3 x (x - 1) & x (x - 1) (x - 2) & (x + 1) x (x - 1) \end{array} |$

1 0

2 1

3 100

4 -100

Explanation:

(A) : Given,
$f (x) = | \begin{array}{ccc} 1 & x & x + 1 \\ 2 x & x (x - 1) & (x + 1) x \\ 3 x (x - 1) & x (x - 1) (x - 2) & (x - 1) x (x - 1) \end{array} |$
Operating $C_{3 \to} C_{3} - C_{1} - C_{2}$ , we get
$f (x) = | \begin{array}{ccc} 1 & x & 0 \\ 2 x & x (x - 1) & 0 \\ 3 x (x - 1) & x (x - 1) (x - 2) & 0 \end{array} |, ∵ C_{3} = 0)$
a $b \boldsymbol α a - b$

AMU-2012

Matrix and Determinant

79425 If $b c b α - c = 0$ and $α \neq \frac{1}{2}$ , then $a, b, c$ are in $\begin{array}{lll} 2 & 1 & 0 \end{array}$

1 A.P.

2 G.P.

3 H.P.

4 none of these

Explanation:

(B) : According to given summation,
If $| \begin{array}{ccc} a & b & α a - b \\ b & c & α b - c \\ 2 & 1 & 0 \end{array} | = 0$
Operating $C_{3 \to} C_{3} - α C_{1} - C_{2}$ , we get
$| \begin{array}{ccc} a & b & 0 \\ b & c & 0 \\ 2 & 1 & - 2 α + 1 \end{array} | = 0$
$\Rightarrow a [c (- 2 α + 1) - b [b (- 2 α + 1] - 0$
$\Rightarrow (- 2 α + 1) (ac - b^{2}) = 0 \Rightarrow (- 2 α + 1) = 0 or (ac - b^{2}) = 0$
$But a \neq \frac{1}{2} and a c = b^{2}$

SEMJEEE-2012

Matrix and Determinant

79426 The solution for $| \begin{array}{ccc} 4 x & 6 x + 2 & 8 x + 1 \\ 6 x + 2 & 9 x + 3 & 12 x \\ 8 x + 1 & 12 x & 16 x + 2 \end{array} | = 0$ is
is given by

1

x = \frac{- 13}{71}

2

x = \frac{- 11}{97}

3

x = \frac{- 7}{43}

4

x = \frac{- 1}{11}

Explanation:

(B) : Given,
$| \begin{array}{ccc} 4 x & 6 x + 2 & 8 x + 1 \\ 6 x + 2 & 9 x + 3 & 12 x \\ 8 x + 1 & 12 x & 16 x + 2 \end{array} | = 0$
Operating $C_{3 \to} C_{3} - 2 C_{1}$
$| \begin{array}{ccc} 4 x & 6 x + 2 & 1 \\ 6 x + 2 & 9 x + 3 & - 4 \\ 8 x + 1 & 12 x & 0 \end{array} | = 0$
Now, taking 2 common from $C_{1}$ , we get
$| \begin{array}{ccc} 2 x & 6 x + 2 & 1 \\ 3 x + 1 & 9 x + 3 & - 4 \\ 4 x + \frac{1}{2} & 12 x & 0 \\ ng C_{2} \to_{\to}^{} C_{2} - 3 C_{1} \end{array} | = 0$
Operating $C_{2} \to C_{2} - 3 C$
$| \begin{array}{ccc} 2 x & 2 & 1 \\ 3 x + 1 & 0 & - 4 \\ 4 x + \frac{1}{2} & \frac{- 3}{2} & 0 \end{array} | = 0$
Multiply by 2 in column 2 , we get
$| \begin{array}{ccc} 2 x & 4 & 1 \\ 3 x + 1 & 0 & - 4 \\ 4 x + \frac{1}{2} & - 3 & 0 \end{array} | = 0$
Operating $R_{2} \overset{2}{\to} R_{2} + 4 R_{1}$
$| \begin{array}{ccc} 2 x & 4 & 1 \\ 11 x + 1 & 16 & 0 \\ 4 x + \frac{1}{2} & - 3 & 0 \end{array} | = 0$
$1 (- 33 x - 3) - 64 x - 8$
$- 33 x - 3 - 64 x - 8 = 0$
$- 97 x - 11 = 0$
$x = \frac{- 11}{97}$
$Now, 1 (- 33 x - 3) - 64 x - 8 = 0$

SRM JEE-2013

Matrix and Determinant

79427 The value of $| \begin{array}{ccc} 1 + ω & ω^{2} & - ω \\ 1 + ω^{2} & ω & - ω^{2} \\ ω^{2} + ω & ω & - ω^{2} \end{array} |$ is equal to
( $ω$ being an imaginary cube root of unity)

1 0

2

2 ω

3

2 ω^{2}

4

- 3 ω^{2}

Explanation:

(D) : It is given that,
$Δ = | \begin{array}{ccc} 1 + ω & ω^{2} & - ω \\ 1 + ω^{2} & ω & - ω^{2} \\ ω^{2} + ω & ω & - ω^{2} \end{array} |$
We know that:-
$[1 + ω + ω^{2} = 0]$
$Δ = | \begin{array}{ccc} - ω^{2} & ω^{2} & - ω \\ - ω & ω & - ω^{2} \\ - 1 & ω & - ω^{2} \end{array} |$
Operating $C_{1} \to C_{1} + C_{2}$ , we get
$Δ = | \begin{array}{ccc} 0 & ω^{2} & - ω \\ 0 & ω & - ω^{2} \\ - 1 + ω & ω & - ω^{2} \end{array} |$
$Δ = (- 1 + ω) (- ω^{4} + ω^{2})$
$= (- 1 + ω) (- ω + ω^{2})$
$= ω - ω^{2} - ω^{2} + ω^{3}$
$= 1 + ω - 2 ω^{2} = - ω^{2} - 2 ω^{2}$
$Δ = - 3 ω^{2}$

SRM JEE-2013

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