QBManager

Matrix and Determinant

79419 A non-trivial solution of the system of equations $x+\lambda y+2 z=0,2 x+\lambda z=0$
$2 \lambda x-2 y+3 z=0$ is given by $x: y: z=$

1 $1: 2:-2$

2 $1:-2: 2$

3 $2: 1: 2$

4 $2: 1:-2$

COMEDK-2016

Matrix and Determinant

79420 If $\mathbf{a} \neq \mathbf{p}, \mathbf{b} \neq \mathbf{q}, \mathbf{c} \neq \mathbf{r}$ and $\left|\begin{array}{lll}p & b & c \\ \mathbf{a} & \mathbf{q} & \mathbf{c} \\ \mathbf{a} & \boldsymbol{b} & \mathbf{r}\end{array}\right|=0$, then the value of $\frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c}$ is

1 1

2 -1

3 0

4 2

Explanation:

(D) Given, $| \begin{array}{lll} p & b & c \\ a & q & c \\ a & b & r \end{array} | = 0$
Operating $R_{2} \to R_{2} - R_{1}, R_{3} \to R_{3} - R_{1}$ we get,
$| \begin{array}{ccc} p & b & c \\ a - p & q - b & 0 \\ a - p & 0 & r - c \end{array} | = 0$
Operating $C_{1} \to \frac{C_{1}}{p - a}, C_{2} \to \frac{C_{2}}{q - b}, C_{3} \to \frac{C_{3}}{r - c}$
$| \begin{array}{ccc} \frac{p}{p - a} & \frac{b}{q - b} & \frac{c}{r - c} \\ - 1 & 1 & 0 \\ - 1 & 0 & 1 \end{array} | = 0$
Expanding along $R_{1}$ , we get
$\frac{p}{p - a} + \frac{b}{q - b} + \frac{c}{r - c} = 0$
$\frac{p}{p - a} + \frac{q}{q - b} - 1 + \frac{r}{r - c} - 1 = 0$
$\frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c} = 2$

COMEDK-2018

Matrix and Determinant

79421 If the points $(a_{1}, b_{1}), (a_{2}, b_{2})$ and $(a_{1} + a_{2}, b_{1} + b_{2})$ are collinear, then

1

a_{1} b_{2} = a_{2} b_{1}

2

a_{1} + a_{2} = b_{1} + b_{2}

3

a_{2} b_{2} = a_{1} b

4

a_{1} + b_{1} = a_{2} + b_{2}

Explanation:

(A) : According to given summation,
The given points are collinear,
$∴ \frac{1}{2} | \begin{array}{ccc} a_{1} & b_{1} & 1 \\ a_{2} & b_{2} & 1 \\ a_{1} + a_{2} & b_{1} + b_{2} & 1 \end{array} | = 0$
Operating $R_{2} \to R_{2} - R_{1}, R_{3} \to R_{3} - R_{1}$ , we get
$| \begin{array}{ccc} a_{1} & b_{1} & 1 \\ a_{2} - a_{1} & b_{2} - b_{1} & 0 \\ a_{2} & b_{2} & 0 \end{array} | = 0$
Expanding along $C_{3}$ , we get
$b_{2} (a_{2} - a_{1}) - a_{2} (b_{2} - b_{1}) = 0$
$- a_{1} b_{2} + a_{2} b_{1} = 0$
$a_{1} b_{2} = a_{2} b_{1}$ .

COMEDK-2019

Matrix and Determinant

79422 For how many values of $x$ in the closed interval $[- 4, - 1]$ , the matrix
$[\begin{array}{ccc} 3 & - 1 + x & 2 \\ 3 & - 1 & x + 2 \\ x + 3 & - 1 & 2 \end{array}]$ is singular?

1 0

2 2

3 1

4 3

Explanation:

(C) : According to given summation,
If $| A | = 0$ , then $A$ is singular matrix
$= [\begin{array}{ccc} 3 & - 1 + x & 2 \\ 3 & - 1 & x + 2 \\ x + 3 & - 1 & 2 \end{array}]$
Operating $R_{1} \to R_{1} - R_{2}$
$= | \begin{array}{ccc} 3 - 3 & - 1 + x - (- 1) & 2 - (x + 2) \\ 3 & - 1 & x + 2 \\ x + 3 & - 1 & 2 \end{array} |$
$= | \begin{array}{ccc} 0 & x & - x \\ 3 & - 1 & x + 2 \\ x + 3 & - 1 & 2 \end{array} |$
Operating $R_{3} - R_{2}$
$= | \begin{array}{ccc} 0 & x & - x \\ 3 & - 1 & x + 2 \end{array}$
$= 0 \begin{array}{ccc} x - x & 0 - x \\ 0 - 3 x - {x (x + 2)} - x (0 - (- x)] \end{array}$
$= - x (- 3 x - x^{2} - 2 x) - x (x)$
$= - x [- x^{2} - 5 x + x] = - x (- x^{2} - 4 x)$
$x = 0 or x = - 4$
$\notin [- 4, - 1]$
So, number of values of $x$ is 1 i.e, $x = - 4$ .

SRM JEE-2008

Matrix and Determinant

79423 If $a, b, c$ are in A.P., then the value of
$| \begin{array}{lll} x + 1 & x + 2 & x + a \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{array} | is$

1 3

2 -3

3 0

4 2

Explanation:

(C) : According to given summation,
If $a, b, c$ are in A.P then, $2 b = a + c$
Now,
Let, $Δ = | \begin{array}{lll} x + 1 & x + 2 & x + a \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{array} |$
Operating $R_{1} \to R_{1} + R_{2} + R_{3}$
$Δ = | \begin{array}{ccc} 3 x + 6 & 3 x + 9 & 3 x + a + b + c \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{array} |$
$Δ = | \begin{array}{ccc} 3 x + 6 & 3 x + 9 & 3 x + b + 2 b \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{array} |$ from equaiton (i)
$Δ = | \begin{array}{ccc} 3 (x + 2) & 3 (x + 3) & 3 (x + b) \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{array} |$

SRM JEE-2011

Matrix and Determinant

79419 A non-trivial solution of the system of equations $x + λ y + 2 z = 0, 2 x + λ z = 0$
$2 λ x - 2 y + 3 z = 0$ is given by $x : y : z =$

1

1 : 2 : - 2

2

1 : - 2 : 2

3

2 : 1 : 2

4

2 : 1 : - 2

Explanation:

(D) : According to given summation,
$| \begin{array}{lll} 1 & λ & 2 \end{array} |$
$\begin{array}{lll} 2 & 0 & λ = 0 \end{array}$
$\begin{array}{lll} 2 λ & - 2 & 3 \end{array}$
Taking $λ = 2$ , the system becomes
$x + 2 y + 2 z = 0$
$2 x + 0 y + 2 z = 0$
$4 x - 2 y + 3 z = 0$
$∴ \frac{x}{4} = \frac{y}{2} = \frac{z}{- 4}$
Therefore, $x : y : z = 2 : 1 : - 2$ .

COMEDK-2016

Matrix and Determinant

79420 If $a \neq p, b \neq q, c \neq r$ and $| \begin{array}{lll} p & b & c \\ a & q & c \\ a & \boldsymbol b & r \end{array} | = 0$ , then the value of $\frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c}$ is

1 1

2 -1

3 0

4 2

Explanation:

(D) Given, $| \begin{array}{lll} p & b & c \\ a & q & c \\ a & b & r \end{array} | = 0$
Operating $R_{2} \to R_{2} - R_{1}, R_{3} \to R_{3} - R_{1}$ we get,
$| \begin{array}{ccc} p & b & c \\ a - p & q - b & 0 \\ a - p & 0 & r - c \end{array} | = 0$
Operating $C_{1} \to \frac{C_{1}}{p - a}, C_{2} \to \frac{C_{2}}{q - b}, C_{3} \to \frac{C_{3}}{r - c}$
$| \begin{array}{ccc} \frac{p}{p - a} & \frac{b}{q - b} & \frac{c}{r - c} \\ - 1 & 1 & 0 \\ - 1 & 0 & 1 \end{array} | = 0$
Expanding along $R_{1}$ , we get
$\frac{p}{p - a} + \frac{b}{q - b} + \frac{c}{r - c} = 0$
$\frac{p}{p - a} + \frac{q}{q - b} - 1 + \frac{r}{r - c} - 1 = 0$
$\frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c} = 2$

COMEDK-2018

Matrix and Determinant

79421 If the points $(a_{1}, b_{1}), (a_{2}, b_{2})$ and $(a_{1} + a_{2}, b_{1} + b_{2})$ are collinear, then

1

a_{1} b_{2} = a_{2} b_{1}

2

a_{1} + a_{2} = b_{1} + b_{2}

3

a_{2} b_{2} = a_{1} b

4

a_{1} + b_{1} = a_{2} + b_{2}

Explanation:

(A) : According to given summation,
The given points are collinear,
$∴ \frac{1}{2} | \begin{array}{ccc} a_{1} & b_{1} & 1 \\ a_{2} & b_{2} & 1 \\ a_{1} + a_{2} & b_{1} + b_{2} & 1 \end{array} | = 0$
Operating $R_{2} \to R_{2} - R_{1}, R_{3} \to R_{3} - R_{1}$ , we get
$| \begin{array}{ccc} a_{1} & b_{1} & 1 \\ a_{2} - a_{1} & b_{2} - b_{1} & 0 \\ a_{2} & b_{2} & 0 \end{array} | = 0$
Expanding along $C_{3}$ , we get
$b_{2} (a_{2} - a_{1}) - a_{2} (b_{2} - b_{1}) = 0$
$- a_{1} b_{2} + a_{2} b_{1} = 0$
$a_{1} b_{2} = a_{2} b_{1}$ .

COMEDK-2019

Matrix and Determinant

79422 For how many values of $x$ in the closed interval $[- 4, - 1]$ , the matrix
$[\begin{array}{ccc} 3 & - 1 + x & 2 \\ 3 & - 1 & x + 2 \\ x + 3 & - 1 & 2 \end{array}]$ is singular?

1 0

2 2

3 1

4 3

Explanation:

(C) : According to given summation,
If $| A | = 0$ , then $A$ is singular matrix
$= [\begin{array}{ccc} 3 & - 1 + x & 2 \\ 3 & - 1 & x + 2 \\ x + 3 & - 1 & 2 \end{array}]$
Operating $R_{1} \to R_{1} - R_{2}$
$= | \begin{array}{ccc} 3 - 3 & - 1 + x - (- 1) & 2 - (x + 2) \\ 3 & - 1 & x + 2 \\ x + 3 & - 1 & 2 \end{array} |$
$= | \begin{array}{ccc} 0 & x & - x \\ 3 & - 1 & x + 2 \\ x + 3 & - 1 & 2 \end{array} |$
Operating $R_{3} - R_{2}$
$= | \begin{array}{ccc} 0 & x & - x \\ 3 & - 1 & x + 2 \end{array}$
$= 0 \begin{array}{ccc} x - x & 0 - x \\ 0 - 3 x - {x (x + 2)} - x (0 - (- x)] \end{array}$
$= - x (- 3 x - x^{2} - 2 x) - x (x)$
$= - x [- x^{2} - 5 x + x] = - x (- x^{2} - 4 x)$
$x = 0 or x = - 4$
$\notin [- 4, - 1]$
So, number of values of $x$ is 1 i.e, $x = - 4$ .

SRM JEE-2008

Matrix and Determinant

79423 If $a, b, c$ are in A.P., then the value of
$| \begin{array}{lll} x + 1 & x + 2 & x + a \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{array} | is$

1 3

2 -3

3 0

4 2

Explanation:

(C) : According to given summation,
If $a, b, c$ are in A.P then, $2 b = a + c$
Now,
Let, $Δ = | \begin{array}{lll} x + 1 & x + 2 & x + a \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{array} |$
Operating $R_{1} \to R_{1} + R_{2} + R_{3}$
$Δ = | \begin{array}{ccc} 3 x + 6 & 3 x + 9 & 3 x + a + b + c \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{array} |$
$Δ = | \begin{array}{ccc} 3 x + 6 & 3 x + 9 & 3 x + b + 2 b \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{array} |$ from equaiton (i)
$Δ = | \begin{array}{ccc} 3 (x + 2) & 3 (x + 3) & 3 (x + b) \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{array} |$

SRM JEE-2011

Matrix and Determinant

79419 A non-trivial solution of the system of equations $x + λ y + 2 z = 0, 2 x + λ z = 0$
$2 λ x - 2 y + 3 z = 0$ is given by $x : y : z =$

1

1 : 2 : - 2

2

1 : - 2 : 2

3

2 : 1 : 2

4

2 : 1 : - 2

Explanation:

(D) : According to given summation,
$| \begin{array}{lll} 1 & λ & 2 \end{array} |$
$\begin{array}{lll} 2 & 0 & λ = 0 \end{array}$
$\begin{array}{lll} 2 λ & - 2 & 3 \end{array}$
Taking $λ = 2$ , the system becomes
$x + 2 y + 2 z = 0$
$2 x + 0 y + 2 z = 0$
$4 x - 2 y + 3 z = 0$
$∴ \frac{x}{4} = \frac{y}{2} = \frac{z}{- 4}$
Therefore, $x : y : z = 2 : 1 : - 2$ .

COMEDK-2016

Matrix and Determinant

79420 If $a \neq p, b \neq q, c \neq r$ and $| \begin{array}{lll} p & b & c \\ a & q & c \\ a & \boldsymbol b & r \end{array} | = 0$ , then the value of $\frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c}$ is

1 1

2 -1

3 0

4 2

Explanation:

(D) Given, $| \begin{array}{lll} p & b & c \\ a & q & c \\ a & b & r \end{array} | = 0$
Operating $R_{2} \to R_{2} - R_{1}, R_{3} \to R_{3} - R_{1}$ we get,
$| \begin{array}{ccc} p & b & c \\ a - p & q - b & 0 \\ a - p & 0 & r - c \end{array} | = 0$
Operating $C_{1} \to \frac{C_{1}}{p - a}, C_{2} \to \frac{C_{2}}{q - b}, C_{3} \to \frac{C_{3}}{r - c}$
$| \begin{array}{ccc} \frac{p}{p - a} & \frac{b}{q - b} & \frac{c}{r - c} \\ - 1 & 1 & 0 \\ - 1 & 0 & 1 \end{array} | = 0$
Expanding along $R_{1}$ , we get
$\frac{p}{p - a} + \frac{b}{q - b} + \frac{c}{r - c} = 0$
$\frac{p}{p - a} + \frac{q}{q - b} - 1 + \frac{r}{r - c} - 1 = 0$
$\frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c} = 2$

COMEDK-2018

Matrix and Determinant

79421 If the points $(a_{1}, b_{1}), (a_{2}, b_{2})$ and $(a_{1} + a_{2}, b_{1} + b_{2})$ are collinear, then

1

a_{1} b_{2} = a_{2} b_{1}

2

a_{1} + a_{2} = b_{1} + b_{2}

3

a_{2} b_{2} = a_{1} b

4

a_{1} + b_{1} = a_{2} + b_{2}

Explanation:

(A) : According to given summation,
The given points are collinear,
$∴ \frac{1}{2} | \begin{array}{ccc} a_{1} & b_{1} & 1 \\ a_{2} & b_{2} & 1 \\ a_{1} + a_{2} & b_{1} + b_{2} & 1 \end{array} | = 0$
Operating $R_{2} \to R_{2} - R_{1}, R_{3} \to R_{3} - R_{1}$ , we get
$| \begin{array}{ccc} a_{1} & b_{1} & 1 \\ a_{2} - a_{1} & b_{2} - b_{1} & 0 \\ a_{2} & b_{2} & 0 \end{array} | = 0$
Expanding along $C_{3}$ , we get
$b_{2} (a_{2} - a_{1}) - a_{2} (b_{2} - b_{1}) = 0$
$- a_{1} b_{2} + a_{2} b_{1} = 0$
$a_{1} b_{2} = a_{2} b_{1}$ .

COMEDK-2019

Matrix and Determinant

79422 For how many values of $x$ in the closed interval $[- 4, - 1]$ , the matrix
$[\begin{array}{ccc} 3 & - 1 + x & 2 \\ 3 & - 1 & x + 2 \\ x + 3 & - 1 & 2 \end{array}]$ is singular?

1 0

2 2

3 1

4 3

Explanation:

(C) : According to given summation,
If $| A | = 0$ , then $A$ is singular matrix
$= [\begin{array}{ccc} 3 & - 1 + x & 2 \\ 3 & - 1 & x + 2 \\ x + 3 & - 1 & 2 \end{array}]$
Operating $R_{1} \to R_{1} - R_{2}$
$= | \begin{array}{ccc} 3 - 3 & - 1 + x - (- 1) & 2 - (x + 2) \\ 3 & - 1 & x + 2 \\ x + 3 & - 1 & 2 \end{array} |$
$= | \begin{array}{ccc} 0 & x & - x \\ 3 & - 1 & x + 2 \\ x + 3 & - 1 & 2 \end{array} |$
Operating $R_{3} - R_{2}$
$= | \begin{array}{ccc} 0 & x & - x \\ 3 & - 1 & x + 2 \end{array}$
$= 0 \begin{array}{ccc} x - x & 0 - x \\ 0 - 3 x - {x (x + 2)} - x (0 - (- x)] \end{array}$
$= - x (- 3 x - x^{2} - 2 x) - x (x)$
$= - x [- x^{2} - 5 x + x] = - x (- x^{2} - 4 x)$
$x = 0 or x = - 4$
$\notin [- 4, - 1]$
So, number of values of $x$ is 1 i.e, $x = - 4$ .

SRM JEE-2008

Matrix and Determinant

79423 If $a, b, c$ are in A.P., then the value of
$| \begin{array}{lll} x + 1 & x + 2 & x + a \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{array} | is$

1 3

2 -3

3 0

4 2

Explanation:

(C) : According to given summation,
If $a, b, c$ are in A.P then, $2 b = a + c$
Now,
Let, $Δ = | \begin{array}{lll} x + 1 & x + 2 & x + a \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{array} |$
Operating $R_{1} \to R_{1} + R_{2} + R_{3}$
$Δ = | \begin{array}{ccc} 3 x + 6 & 3 x + 9 & 3 x + a + b + c \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{array} |$
$Δ = | \begin{array}{ccc} 3 x + 6 & 3 x + 9 & 3 x + b + 2 b \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{array} |$ from equaiton (i)
$Δ = | \begin{array}{ccc} 3 (x + 2) & 3 (x + 3) & 3 (x + b) \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{array} |$

SRM JEE-2011

Matrix and Determinant

79419 A non-trivial solution of the system of equations $x + λ y + 2 z = 0, 2 x + λ z = 0$
$2 λ x - 2 y + 3 z = 0$ is given by $x : y : z =$

1

1 : 2 : - 2

2

1 : - 2 : 2

3

2 : 1 : 2

4

2 : 1 : - 2

Explanation:

(D) : According to given summation,
$| \begin{array}{lll} 1 & λ & 2 \end{array} |$
$\begin{array}{lll} 2 & 0 & λ = 0 \end{array}$
$\begin{array}{lll} 2 λ & - 2 & 3 \end{array}$
Taking $λ = 2$ , the system becomes
$x + 2 y + 2 z = 0$
$2 x + 0 y + 2 z = 0$
$4 x - 2 y + 3 z = 0$
$∴ \frac{x}{4} = \frac{y}{2} = \frac{z}{- 4}$
Therefore, $x : y : z = 2 : 1 : - 2$ .

COMEDK-2016

Matrix and Determinant

79420 If $a \neq p, b \neq q, c \neq r$ and $| \begin{array}{lll} p & b & c \\ a & q & c \\ a & \boldsymbol b & r \end{array} | = 0$ , then the value of $\frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c}$ is

1 1

2 -1

3 0

4 2

Explanation:

(D) Given, $| \begin{array}{lll} p & b & c \\ a & q & c \\ a & b & r \end{array} | = 0$
Operating $R_{2} \to R_{2} - R_{1}, R_{3} \to R_{3} - R_{1}$ we get,
$| \begin{array}{ccc} p & b & c \\ a - p & q - b & 0 \\ a - p & 0 & r - c \end{array} | = 0$
Operating $C_{1} \to \frac{C_{1}}{p - a}, C_{2} \to \frac{C_{2}}{q - b}, C_{3} \to \frac{C_{3}}{r - c}$
$| \begin{array}{ccc} \frac{p}{p - a} & \frac{b}{q - b} & \frac{c}{r - c} \\ - 1 & 1 & 0 \\ - 1 & 0 & 1 \end{array} | = 0$
Expanding along $R_{1}$ , we get
$\frac{p}{p - a} + \frac{b}{q - b} + \frac{c}{r - c} = 0$
$\frac{p}{p - a} + \frac{q}{q - b} - 1 + \frac{r}{r - c} - 1 = 0$
$\frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c} = 2$

COMEDK-2018

Matrix and Determinant

79421 If the points $(a_{1}, b_{1}), (a_{2}, b_{2})$ and $(a_{1} + a_{2}, b_{1} + b_{2})$ are collinear, then

1

a_{1} b_{2} = a_{2} b_{1}

2

a_{1} + a_{2} = b_{1} + b_{2}

3

a_{2} b_{2} = a_{1} b

4

a_{1} + b_{1} = a_{2} + b_{2}

Explanation:

(A) : According to given summation,
The given points are collinear,
$∴ \frac{1}{2} | \begin{array}{ccc} a_{1} & b_{1} & 1 \\ a_{2} & b_{2} & 1 \\ a_{1} + a_{2} & b_{1} + b_{2} & 1 \end{array} | = 0$
Operating $R_{2} \to R_{2} - R_{1}, R_{3} \to R_{3} - R_{1}$ , we get
$| \begin{array}{ccc} a_{1} & b_{1} & 1 \\ a_{2} - a_{1} & b_{2} - b_{1} & 0 \\ a_{2} & b_{2} & 0 \end{array} | = 0$
Expanding along $C_{3}$ , we get
$b_{2} (a_{2} - a_{1}) - a_{2} (b_{2} - b_{1}) = 0$
$- a_{1} b_{2} + a_{2} b_{1} = 0$
$a_{1} b_{2} = a_{2} b_{1}$ .

COMEDK-2019

Matrix and Determinant

79422 For how many values of $x$ in the closed interval $[- 4, - 1]$ , the matrix
$[\begin{array}{ccc} 3 & - 1 + x & 2 \\ 3 & - 1 & x + 2 \\ x + 3 & - 1 & 2 \end{array}]$ is singular?

1 0

2 2

3 1

4 3

Explanation:

(C) : According to given summation,
If $| A | = 0$ , then $A$ is singular matrix
$= [\begin{array}{ccc} 3 & - 1 + x & 2 \\ 3 & - 1 & x + 2 \\ x + 3 & - 1 & 2 \end{array}]$
Operating $R_{1} \to R_{1} - R_{2}$
$= | \begin{array}{ccc} 3 - 3 & - 1 + x - (- 1) & 2 - (x + 2) \\ 3 & - 1 & x + 2 \\ x + 3 & - 1 & 2 \end{array} |$
$= | \begin{array}{ccc} 0 & x & - x \\ 3 & - 1 & x + 2 \\ x + 3 & - 1 & 2 \end{array} |$
Operating $R_{3} - R_{2}$
$= | \begin{array}{ccc} 0 & x & - x \\ 3 & - 1 & x + 2 \end{array}$
$= 0 \begin{array}{ccc} x - x & 0 - x \\ 0 - 3 x - {x (x + 2)} - x (0 - (- x)] \end{array}$
$= - x (- 3 x - x^{2} - 2 x) - x (x)$
$= - x [- x^{2} - 5 x + x] = - x (- x^{2} - 4 x)$
$x = 0 or x = - 4$
$\notin [- 4, - 1]$
So, number of values of $x$ is 1 i.e, $x = - 4$ .

SRM JEE-2008

Matrix and Determinant

79423 If $a, b, c$ are in A.P., then the value of
$| \begin{array}{lll} x + 1 & x + 2 & x + a \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{array} | is$

1 3

2 -3

3 0

4 2

Explanation:

(C) : According to given summation,
If $a, b, c$ are in A.P then, $2 b = a + c$
Now,
Let, $Δ = | \begin{array}{lll} x + 1 & x + 2 & x + a \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{array} |$
Operating $R_{1} \to R_{1} + R_{2} + R_{3}$
$Δ = | \begin{array}{ccc} 3 x + 6 & 3 x + 9 & 3 x + a + b + c \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{array} |$
$Δ = | \begin{array}{ccc} 3 x + 6 & 3 x + 9 & 3 x + b + 2 b \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{array} |$ from equaiton (i)
$Δ = | \begin{array}{ccc} 3 (x + 2) & 3 (x + 3) & 3 (x + b) \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{array} |$

SRM JEE-2011

Matrix and Determinant

79419 A non-trivial solution of the system of equations $x + λ y + 2 z = 0, 2 x + λ z = 0$
$2 λ x - 2 y + 3 z = 0$ is given by $x : y : z =$

1

1 : 2 : - 2

2

1 : - 2 : 2

3

2 : 1 : 2

4

2 : 1 : - 2

Explanation:

(D) : According to given summation,
$| \begin{array}{lll} 1 & λ & 2 \end{array} |$
$\begin{array}{lll} 2 & 0 & λ = 0 \end{array}$
$\begin{array}{lll} 2 λ & - 2 & 3 \end{array}$
Taking $λ = 2$ , the system becomes
$x + 2 y + 2 z = 0$
$2 x + 0 y + 2 z = 0$
$4 x - 2 y + 3 z = 0$
$∴ \frac{x}{4} = \frac{y}{2} = \frac{z}{- 4}$
Therefore, $x : y : z = 2 : 1 : - 2$ .

COMEDK-2016

Matrix and Determinant

79420 If $a \neq p, b \neq q, c \neq r$ and $| \begin{array}{lll} p & b & c \\ a & q & c \\ a & \boldsymbol b & r \end{array} | = 0$ , then the value of $\frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c}$ is

1 1

2 -1

3 0

4 2

Explanation:

(D) Given, $| \begin{array}{lll} p & b & c \\ a & q & c \\ a & b & r \end{array} | = 0$
Operating $R_{2} \to R_{2} - R_{1}, R_{3} \to R_{3} - R_{1}$ we get,
$| \begin{array}{ccc} p & b & c \\ a - p & q - b & 0 \\ a - p & 0 & r - c \end{array} | = 0$
Operating $C_{1} \to \frac{C_{1}}{p - a}, C_{2} \to \frac{C_{2}}{q - b}, C_{3} \to \frac{C_{3}}{r - c}$
$| \begin{array}{ccc} \frac{p}{p - a} & \frac{b}{q - b} & \frac{c}{r - c} \\ - 1 & 1 & 0 \\ - 1 & 0 & 1 \end{array} | = 0$
Expanding along $R_{1}$ , we get
$\frac{p}{p - a} + \frac{b}{q - b} + \frac{c}{r - c} = 0$
$\frac{p}{p - a} + \frac{q}{q - b} - 1 + \frac{r}{r - c} - 1 = 0$
$\frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c} = 2$

COMEDK-2018

Matrix and Determinant

79421 If the points $(a_{1}, b_{1}), (a_{2}, b_{2})$ and $(a_{1} + a_{2}, b_{1} + b_{2})$ are collinear, then

1

a_{1} b_{2} = a_{2} b_{1}

2

a_{1} + a_{2} = b_{1} + b_{2}

3

a_{2} b_{2} = a_{1} b

4

a_{1} + b_{1} = a_{2} + b_{2}

Explanation:

(A) : According to given summation,
The given points are collinear,
$∴ \frac{1}{2} | \begin{array}{ccc} a_{1} & b_{1} & 1 \\ a_{2} & b_{2} & 1 \\ a_{1} + a_{2} & b_{1} + b_{2} & 1 \end{array} | = 0$
Operating $R_{2} \to R_{2} - R_{1}, R_{3} \to R_{3} - R_{1}$ , we get
$| \begin{array}{ccc} a_{1} & b_{1} & 1 \\ a_{2} - a_{1} & b_{2} - b_{1} & 0 \\ a_{2} & b_{2} & 0 \end{array} | = 0$
Expanding along $C_{3}$ , we get
$b_{2} (a_{2} - a_{1}) - a_{2} (b_{2} - b_{1}) = 0$
$- a_{1} b_{2} + a_{2} b_{1} = 0$
$a_{1} b_{2} = a_{2} b_{1}$ .

COMEDK-2019

Matrix and Determinant

79422 For how many values of $x$ in the closed interval $[- 4, - 1]$ , the matrix
$[\begin{array}{ccc} 3 & - 1 + x & 2 \\ 3 & - 1 & x + 2 \\ x + 3 & - 1 & 2 \end{array}]$ is singular?

1 0

2 2

3 1

4 3

Explanation:

(C) : According to given summation,
If $| A | = 0$ , then $A$ is singular matrix
$= [\begin{array}{ccc} 3 & - 1 + x & 2 \\ 3 & - 1 & x + 2 \\ x + 3 & - 1 & 2 \end{array}]$
Operating $R_{1} \to R_{1} - R_{2}$
$= | \begin{array}{ccc} 3 - 3 & - 1 + x - (- 1) & 2 - (x + 2) \\ 3 & - 1 & x + 2 \\ x + 3 & - 1 & 2 \end{array} |$
$= | \begin{array}{ccc} 0 & x & - x \\ 3 & - 1 & x + 2 \\ x + 3 & - 1 & 2 \end{array} |$
Operating $R_{3} - R_{2}$
$= | \begin{array}{ccc} 0 & x & - x \\ 3 & - 1 & x + 2 \end{array}$
$= 0 \begin{array}{ccc} x - x & 0 - x \\ 0 - 3 x - {x (x + 2)} - x (0 - (- x)] \end{array}$
$= - x (- 3 x - x^{2} - 2 x) - x (x)$
$= - x [- x^{2} - 5 x + x] = - x (- x^{2} - 4 x)$
$x = 0 or x = - 4$
$\notin [- 4, - 1]$
So, number of values of $x$ is 1 i.e, $x = - 4$ .

SRM JEE-2008

Matrix and Determinant

79423 If $a, b, c$ are in A.P., then the value of
$| \begin{array}{lll} x + 1 & x + 2 & x + a \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{array} | is$

1 3

2 -3

3 0

4 2

Explanation:

(C) : According to given summation,
If $a, b, c$ are in A.P then, $2 b = a + c$
Now,
Let, $Δ = | \begin{array}{lll} x + 1 & x + 2 & x + a \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{array} |$
Operating $R_{1} \to R_{1} + R_{2} + R_{3}$
$Δ = | \begin{array}{ccc} 3 x + 6 & 3 x + 9 & 3 x + a + b + c \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{array} |$
$Δ = | \begin{array}{ccc} 3 x + 6 & 3 x + 9 & 3 x + b + 2 b \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{array} |$ from equaiton (i)
$Δ = | \begin{array}{ccc} 3 (x + 2) & 3 (x + 3) & 3 (x + b) \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{array} |$

SRM JEE-2011