QBManager

Matrix and Determinant

79252 If $3 x+2 y+z=0, x+4 y+z=0,2 x+y+4 z=0$
be a system of homogeneous equation, then

1 it is inconsistent.

2 only trivial solution exists.

3 it can be reduced to a single equation, so that no solution exists

4 determinant of the coefficient matrix is zero.

SRM JEE-2016

Matrix and Determinant

79253 The system of equations $2 x+y-5=0$, $x-2 y+1=9,2 x-14 y-a=0$, is consistent. Then, $a$ is equal to

1 1

2 2

3 5

4 None of these

VITEEE-2011

Matrix and Determinant

79254 The value of $x$ obtained from the equation
$\left|\begin{array}{ccc} \mathbf{x}+\boldsymbol{\alpha} & \boldsymbol{\beta} & \boldsymbol{\gamma} \\ \boldsymbol{\gamma} & \mathbf{x}+\boldsymbol{\beta} & \boldsymbol{\alpha} \\ \boldsymbol{\alpha} & \boldsymbol{\beta} & \mathbf{x}+\boldsymbol{\gamma} \end{array}\right|=\mathbf{0} \text { will be }$

1 0 and $-(\alpha+\beta+\gamma)$

2 0 and $\alpha+\beta+\gamma$

3 1 and $(\alpha-\beta-\gamma)$

4 0 and $\alpha^{2}+\beta^{2}+\gamma^{2}$

VITEEE-2017

Matrix and Determinant

79255 Find the value of $k$ for which the simultaneous equations $x+y+z=3 ; x+2 y+3 z=4$ and $x+$ $4 y + k z = 6$ will not have a unique solution.

1 0

2 5

3 6

4 7

Explanation:

(D) : It is given that
$x + y + z = 3, x + 2 y + 3 z = 4, x + 4 y + k z = 6$
The given system of equation will be consistent with unique solution, when
$| \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & k \end{array} | \neq 0$
$1 (2 k - 12) + 1 (3 - k) + 1 (4 - 2) \neq 0$
$2 k - 12 - k + 3 + 2 \neq 0$
$k - 7 \neq 0$
$k \neq 7$

VITEEE-2015

Matrix and Determinant

79252 If $3 x + 2 y + z = 0, x + 4 y + z = 0, 2 x + y + 4 z = 0$
be a system of homogeneous equation, then

1 it is inconsistent.

2 only trivial solution exists.

3 it can be reduced to a single equation, so that no solution exists

4 determinant of the coefficient matrix is zero.

Explanation:

(B)Here, we have
$3 x + 2 y + z = 0$
$x + 4 y + z = 0$
$2 x + y + 4 z = 0$
$[\begin{array}{lll} 3 & 2 & 1 \\ 1 & 4 & 1 \\ 2 & 1 & 4 \end{array}] [\begin{array}{l} x \\ y \\ Ax = 0 \end{array}] = [\begin{array}{l} 0 \\ 0 \\ 0 \end{array}]$
$∴ Coefficient matrix A = [\begin{array}{lll} 3 & 2 & 1 \\ 1 & 4 & 1 \\ 2 & 1 & 4 \end{array}]$
$| A | = 3 (16 - 1) - 2 (4 - 2) + 1 (1 - 8)$
$= 45 - 4 - 7 = 34$
$| A | \neq 0$
So, the given system of equation has only trivial solution.

SRM JEE-2016

Matrix and Determinant

79253 The system of equations $2 x + y - 5 = 0$ , $x - 2 y + 1 = 9, 2 x - 14 y - a = 0$ , is consistent. Then, $a$ is equal to

1 1

2 2

3 5

4 None of these

Explanation:

(D) : Here, we have
$2 x + y - 5 = 0$
$x - 2 y + 1 = 0$
This system is consistent,
$∴ | \begin{array}{ccc} 2 & 1 & - 5 \\ 1 & - 2 & 1 \\ 2 & - 14 & - a \end{array} | = 0$
$2 (2 a + 14) - 1 (- a - 2) - 5 (- 14 + 4) = 0$
$4 a + 28 + a + 2 + 50 = 0$
$5 a = - 80 \Rightarrow a = - 16 .$

VITEEE-2011

Matrix and Determinant

79254 The value of $x$ obtained from the equation
$| \begin{array}{ccc} x + \boldsymbol α & \boldsymbol β & \boldsymbol γ \\ \boldsymbol γ & x + \boldsymbol β & \boldsymbol α \\ \boldsymbol α & \boldsymbol β & x + \boldsymbol γ \end{array} | = 0 will be$

1 0 and

- (α + β + γ)

2 0 and

α + β + γ

3 1 and

(α - β - γ)

4 0 and

α^{2} + β^{2} + γ^{2}

Explanation:

(A) : Here, we have
$| \begin{array}{ccc} x + α & β & γ \\ γ & x + β & α \\ α & β & x + γ \end{array} | = 0$
Applying: $C_{1} \to C_{1} + C_{2} + C_{3}$
$| \begin{array}{ccc} x + α + β + γ & β & γ \\ x + α + β + γ & x + β & α \\ x + α + β + γ & β & x + γ \end{array} | = 0$
$(x + α + β + γ) | \begin{array}{ccc} 1 & β & γ \\ 1 & x + β & α \\ 1 & β & x + γ \end{array} | = 0$
$x + α + β + γ = 0$
$x = - (α + β + γ)$
Again if
$| \begin{array}{ccc} 1 & β & γ \\ 1 & x + β & α \\ 1 & β & x + γ \end{array} | = 0$
Operating $R_{2} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{1}$
$| \begin{array}{lll} 1 & β & γ \end{array}$
$\begin{array}{lll} 0 & x & α - γ \end{array} = 0$
$\begin{array}{lll} 0 & 0 & γ \end{array}$
$∴$ Solutions of the equation are $x = 0, - (α + β + γ)$

VITEEE-2017

Matrix and Determinant

79255 Find the value of $k$ for which the simultaneous equations $x + y + z = 3; x + 2 y + 3 z = 4$ and $x +$ $4 y + k z = 6$ will not have a unique solution.

1 0

2 5

3 6

4 7

Explanation:

(D) : It is given that
$x + y + z = 3, x + 2 y + 3 z = 4, x + 4 y + k z = 6$
The given system of equation will be consistent with unique solution, when
$| \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & k \end{array} | \neq 0$
$1 (2 k - 12) + 1 (3 - k) + 1 (4 - 2) \neq 0$
$2 k - 12 - k + 3 + 2 \neq 0$
$k - 7 \neq 0$
$k \neq 7$

VITEEE-2015

Matrix and Determinant

79252 If $3 x + 2 y + z = 0, x + 4 y + z = 0, 2 x + y + 4 z = 0$
be a system of homogeneous equation, then

1 it is inconsistent.

2 only trivial solution exists.

3 it can be reduced to a single equation, so that no solution exists

4 determinant of the coefficient matrix is zero.

Explanation:

(B)Here, we have
$3 x + 2 y + z = 0$
$x + 4 y + z = 0$
$2 x + y + 4 z = 0$
$[\begin{array}{lll} 3 & 2 & 1 \\ 1 & 4 & 1 \\ 2 & 1 & 4 \end{array}] [\begin{array}{l} x \\ y \\ Ax = 0 \end{array}] = [\begin{array}{l} 0 \\ 0 \\ 0 \end{array}]$
$∴ Coefficient matrix A = [\begin{array}{lll} 3 & 2 & 1 \\ 1 & 4 & 1 \\ 2 & 1 & 4 \end{array}]$
$| A | = 3 (16 - 1) - 2 (4 - 2) + 1 (1 - 8)$
$= 45 - 4 - 7 = 34$
$| A | \neq 0$
So, the given system of equation has only trivial solution.

SRM JEE-2016

Matrix and Determinant

79253 The system of equations $2 x + y - 5 = 0$ , $x - 2 y + 1 = 9, 2 x - 14 y - a = 0$ , is consistent. Then, $a$ is equal to

1 1

2 2

3 5

4 None of these

Explanation:

(D) : Here, we have
$2 x + y - 5 = 0$
$x - 2 y + 1 = 0$
This system is consistent,
$∴ | \begin{array}{ccc} 2 & 1 & - 5 \\ 1 & - 2 & 1 \\ 2 & - 14 & - a \end{array} | = 0$
$2 (2 a + 14) - 1 (- a - 2) - 5 (- 14 + 4) = 0$
$4 a + 28 + a + 2 + 50 = 0$
$5 a = - 80 \Rightarrow a = - 16 .$

VITEEE-2011

Matrix and Determinant

79254 The value of $x$ obtained from the equation
$| \begin{array}{ccc} x + \boldsymbol α & \boldsymbol β & \boldsymbol γ \\ \boldsymbol γ & x + \boldsymbol β & \boldsymbol α \\ \boldsymbol α & \boldsymbol β & x + \boldsymbol γ \end{array} | = 0 will be$

1 0 and

- (α + β + γ)

2 0 and

α + β + γ

3 1 and

(α - β - γ)

4 0 and

α^{2} + β^{2} + γ^{2}

Explanation:

(A) : Here, we have
$| \begin{array}{ccc} x + α & β & γ \\ γ & x + β & α \\ α & β & x + γ \end{array} | = 0$
Applying: $C_{1} \to C_{1} + C_{2} + C_{3}$
$| \begin{array}{ccc} x + α + β + γ & β & γ \\ x + α + β + γ & x + β & α \\ x + α + β + γ & β & x + γ \end{array} | = 0$
$(x + α + β + γ) | \begin{array}{ccc} 1 & β & γ \\ 1 & x + β & α \\ 1 & β & x + γ \end{array} | = 0$
$x + α + β + γ = 0$
$x = - (α + β + γ)$
Again if
$| \begin{array}{ccc} 1 & β & γ \\ 1 & x + β & α \\ 1 & β & x + γ \end{array} | = 0$
Operating $R_{2} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{1}$
$| \begin{array}{lll} 1 & β & γ \end{array}$
$\begin{array}{lll} 0 & x & α - γ \end{array} = 0$
$\begin{array}{lll} 0 & 0 & γ \end{array}$
$∴$ Solutions of the equation are $x = 0, - (α + β + γ)$

VITEEE-2017

Matrix and Determinant

79255 Find the value of $k$ for which the simultaneous equations $x + y + z = 3; x + 2 y + 3 z = 4$ and $x +$ $4 y + k z = 6$ will not have a unique solution.

1 0

2 5

3 6

4 7

Explanation:

(D) : It is given that
$x + y + z = 3, x + 2 y + 3 z = 4, x + 4 y + k z = 6$
The given system of equation will be consistent with unique solution, when
$| \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & k \end{array} | \neq 0$
$1 (2 k - 12) + 1 (3 - k) + 1 (4 - 2) \neq 0$
$2 k - 12 - k + 3 + 2 \neq 0$
$k - 7 \neq 0$
$k \neq 7$

VITEEE-2015

Matrix and Determinant

79252 If $3 x + 2 y + z = 0, x + 4 y + z = 0, 2 x + y + 4 z = 0$
be a system of homogeneous equation, then

1 it is inconsistent.

2 only trivial solution exists.

3 it can be reduced to a single equation, so that no solution exists

4 determinant of the coefficient matrix is zero.

Explanation:

(B)Here, we have
$3 x + 2 y + z = 0$
$x + 4 y + z = 0$
$2 x + y + 4 z = 0$
$[\begin{array}{lll} 3 & 2 & 1 \\ 1 & 4 & 1 \\ 2 & 1 & 4 \end{array}] [\begin{array}{l} x \\ y \\ Ax = 0 \end{array}] = [\begin{array}{l} 0 \\ 0 \\ 0 \end{array}]$
$∴ Coefficient matrix A = [\begin{array}{lll} 3 & 2 & 1 \\ 1 & 4 & 1 \\ 2 & 1 & 4 \end{array}]$
$| A | = 3 (16 - 1) - 2 (4 - 2) + 1 (1 - 8)$
$= 45 - 4 - 7 = 34$
$| A | \neq 0$
So, the given system of equation has only trivial solution.

SRM JEE-2016

Matrix and Determinant

79253 The system of equations $2 x + y - 5 = 0$ , $x - 2 y + 1 = 9, 2 x - 14 y - a = 0$ , is consistent. Then, $a$ is equal to

1 1

2 2

3 5

4 None of these

Explanation:

(D) : Here, we have
$2 x + y - 5 = 0$
$x - 2 y + 1 = 0$
This system is consistent,
$∴ | \begin{array}{ccc} 2 & 1 & - 5 \\ 1 & - 2 & 1 \\ 2 & - 14 & - a \end{array} | = 0$
$2 (2 a + 14) - 1 (- a - 2) - 5 (- 14 + 4) = 0$
$4 a + 28 + a + 2 + 50 = 0$
$5 a = - 80 \Rightarrow a = - 16 .$

VITEEE-2011

Matrix and Determinant

79254 The value of $x$ obtained from the equation
$| \begin{array}{ccc} x + \boldsymbol α & \boldsymbol β & \boldsymbol γ \\ \boldsymbol γ & x + \boldsymbol β & \boldsymbol α \\ \boldsymbol α & \boldsymbol β & x + \boldsymbol γ \end{array} | = 0 will be$

1 0 and

- (α + β + γ)

2 0 and

α + β + γ

3 1 and

(α - β - γ)

4 0 and

α^{2} + β^{2} + γ^{2}

Explanation:

(A) : Here, we have
$| \begin{array}{ccc} x + α & β & γ \\ γ & x + β & α \\ α & β & x + γ \end{array} | = 0$
Applying: $C_{1} \to C_{1} + C_{2} + C_{3}$
$| \begin{array}{ccc} x + α + β + γ & β & γ \\ x + α + β + γ & x + β & α \\ x + α + β + γ & β & x + γ \end{array} | = 0$
$(x + α + β + γ) | \begin{array}{ccc} 1 & β & γ \\ 1 & x + β & α \\ 1 & β & x + γ \end{array} | = 0$
$x + α + β + γ = 0$
$x = - (α + β + γ)$
Again if
$| \begin{array}{ccc} 1 & β & γ \\ 1 & x + β & α \\ 1 & β & x + γ \end{array} | = 0$
Operating $R_{2} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{1}$
$| \begin{array}{lll} 1 & β & γ \end{array}$
$\begin{array}{lll} 0 & x & α - γ \end{array} = 0$
$\begin{array}{lll} 0 & 0 & γ \end{array}$
$∴$ Solutions of the equation are $x = 0, - (α + β + γ)$

VITEEE-2017

Matrix and Determinant

79255 Find the value of $k$ for which the simultaneous equations $x + y + z = 3; x + 2 y + 3 z = 4$ and $x +$ $4 y + k z = 6$ will not have a unique solution.

1 0

2 5

3 6

4 7

Explanation:

(D) : It is given that
$x + y + z = 3, x + 2 y + 3 z = 4, x + 4 y + k z = 6$
The given system of equation will be consistent with unique solution, when
$| \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & k \end{array} | \neq 0$
$1 (2 k - 12) + 1 (3 - k) + 1 (4 - 2) \neq 0$
$2 k - 12 - k + 3 + 2 \neq 0$
$k - 7 \neq 0$
$k \neq 7$

VITEEE-2015