QBManager

Matrix and Determinant

79147 If $a > 0, b > 0, c > 0$ are respectively the pth, qth, rth terms of G.P., then the value of the
$determinant | \begin{array}{lll} \log a & p & 1 \\ \log b & q & 1 \end{array} | is$
$\begin{array}{lll} \log c & r & 1 \end{array}$

1 0

2 1

3 -1

4 None of these

Explanation:

(A) : According to given summation,
Let $A$ be the $1^{st}$ term and $R$ the common ratio of G.P. Then, $a = T_{p} = {AR}^{P - 1}$
$∴ \log a = \log A + (p - 1) \log R$
Similarly, $\log b = \log A + (q - 1) \log R$
and $\log c = \log A + (r - 1) \log R$
$∴ Δ = | \begin{array}{lll} \log A + (p - 1) \log R & p & 1 \\ \log A + (q - 1) \log R & q & 1 \\ \log A + (r - 1) \log R & r & 1 \end{array} |$
Split into two determinants and in the first take $\log A$ common and in the second take $\log R$ common.
$Δ = \log A | \begin{array}{lll} 1 & p & 1 \\ 1 & q & 1 \\ 1 & r & 1 \end{array} | + \log R | \begin{array}{ccc} p - 1 & p & 1 \\ q - 1 & q & 1 \\ r - 1 & r & 1 \end{array} |$
Apply $C_{1} \to C_{1} - C_{2} + C_{3}$ in the second part of the above determinant-
$Δ = 0 + \log R | \begin{array}{ccc} 0 & p & 1 \\ 0 & q & 1 \\ 0 & r & 1 \end{array} | \Rightarrow Δ = 0$
$1 (x - 3) (x - 3)^{2}$

BITSAT-2009

Matrix and Determinant

79148 The determinant $1 (x - 4) (x - 4)^{2}$ vanishes
$1 (x - 5) (x - 5)^{2}$
for

1 3 values of

x

2 2 values of

x

3 1 values of

x

4 No value of

x

Explanation:

(D) : The given determinant vanishes, i.e.
$| \begin{array}{ccc} 1 & x - 3 & (x - 3)^{2} \\ 1 & x - 4 & (x - 4)^{2} \\ 1 & x - 5 & (x - 5)^{2} \end{array} | = 0$
Operating,
$R_{1} \to R_{1} - R_{2}$
$R_{2} \to R_{2} - R_{3}$
$| \begin{array}{ccc} 0 & 1 & (x - 3)^{2} - (x - 4)^{2} \\ 0 & 1 & (x - 4)^{2} - (x - 5)^{2} \\ 1 & x - 5 & (x - 5)^{2} \end{array} | = | \begin{array}{ccc} 0 & 1 & 2 x - 7 \\ 0 & 1 & 2 x - 9 \\ 1 & x - 5 & (x - 5)^{2} \end{array} |$
$=$ Expanding along ' $C_{1}$ ', we get -
$1 [2 x - 9 - (2 x - 7)] = 0$
$2 x - 9 - 2 x + 7 \Rightarrow - 2 \neq 0$
Thus the given determinant does not vanish for any value.
So, there is not any value of ' $x$ '

BITSAT-2013

Matrix and Determinant

79149 If [ ] denotes the greatest integer less than or equal to the real number under consideration and $- 1 \leq x < 0; 0 \leq y < 1; 1 \leq z < 2$ , then the value of the determinant
$| \begin{array}{ccc} [x] + 1 & [y] & [z] \\ [x] & [y] + 1 & [z] \\ [x] & [y] & [z] + 1 \end{array} |$ is

1

[z]

2

[y]

3

[x]

4 None of these

Explanation:

(A) : Here, It is given that
$\begin{array}{lll} - 1 \leq x < 0, & 0 \leq y < 1, & 1 \leq z < 2 \\ \Rightarrow [x] = - 1 & \Rightarrow [y] = 0 & \Rightarrow [z] = 1 \end{array}$
Now,
$| \begin{array}{ccc} [x] + 1 & [y] & [z] \\ [x] & [y] + 1 & [z] \\ [x] & [y] & [z] + 1 \end{array} |$
$= | \begin{array}{ccc} 0 & 0 & 1 \\ - 1 & 1 & 1 \\ - 1 & 0 & 2 \end{array} | = 1 = [z]$

BITSAT-2016

Matrix and Determinant

79150 Let $ω = \frac{- 1}{2} + i \frac{\sqrt{3}}{2}$ . Then, the value of the determinant $| \begin{array}{ccc} 1 & 1 & 1 \\ 1 & - 1 - ω^{2} & ω^{2} \\ 1 & ω^{2} & ω^{4} \end{array} |$ is

1

3 ω

2

ω (ω - 1)

3

3 ω

4

3 ω (1 - ω)

Explanation:

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

BITSAT-2016

Matrix and Determinant

79151 The points represented by the complex numbers $1 + i, - 2 + 3 i, 5 / 3 i$ on the argand plane are

1 vertices of an equilateral triangle

2 vertices of an isosceles triangle

3 collinear

4 None of these

Explanation:

(C) : According to given summation,
let $z_{1} = 1 + i, z_{2} = - 2 + 3 i$ and $z_{3} = 0 + \frac{5}{3} i$
Then, $| \begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array} | = | \begin{array}{ccc} 1 & 1 & 1 \\ - 2 & 3 & 1 \\ 0 & 5 / 3 & 1 \end{array} |$
$= 1 [3 - \frac{5}{3}] + 1 (2) + 1 [\frac{- 10}{3}] = \frac{4}{3} + 2 - \frac{10}{3} = \frac{4 + 6 - 10}{3} = 0$
$∴$ The points are collinear.

BITSAT-2018

Matrix and Determinant

79147 If $a > 0, b > 0, c > 0$ are respectively the pth, qth, rth terms of G.P., then the value of the
$determinant | \begin{array}{lll} \log a & p & 1 \\ \log b & q & 1 \end{array} | is$
$\begin{array}{lll} \log c & r & 1 \end{array}$

1 0

2 1

3 -1

4 None of these

Explanation:

(A) : According to given summation,
Let $A$ be the $1^{st}$ term and $R$ the common ratio of G.P. Then, $a = T_{p} = {AR}^{P - 1}$
$∴ \log a = \log A + (p - 1) \log R$
Similarly, $\log b = \log A + (q - 1) \log R$
and $\log c = \log A + (r - 1) \log R$
$∴ Δ = | \begin{array}{lll} \log A + (p - 1) \log R & p & 1 \\ \log A + (q - 1) \log R & q & 1 \\ \log A + (r - 1) \log R & r & 1 \end{array} |$
Split into two determinants and in the first take $\log A$ common and in the second take $\log R$ common.
$Δ = \log A | \begin{array}{lll} 1 & p & 1 \\ 1 & q & 1 \\ 1 & r & 1 \end{array} | + \log R | \begin{array}{ccc} p - 1 & p & 1 \\ q - 1 & q & 1 \\ r - 1 & r & 1 \end{array} |$
Apply $C_{1} \to C_{1} - C_{2} + C_{3}$ in the second part of the above determinant-
$Δ = 0 + \log R | \begin{array}{ccc} 0 & p & 1 \\ 0 & q & 1 \\ 0 & r & 1 \end{array} | \Rightarrow Δ = 0$
$1 (x - 3) (x - 3)^{2}$

BITSAT-2009

Matrix and Determinant

79148 The determinant $1 (x - 4) (x - 4)^{2}$ vanishes
$1 (x - 5) (x - 5)^{2}$
for

1 3 values of

x

2 2 values of

x

3 1 values of

x

4 No value of

x

Explanation:

(D) : The given determinant vanishes, i.e.
$| \begin{array}{ccc} 1 & x - 3 & (x - 3)^{2} \\ 1 & x - 4 & (x - 4)^{2} \\ 1 & x - 5 & (x - 5)^{2} \end{array} | = 0$
Operating,
$R_{1} \to R_{1} - R_{2}$
$R_{2} \to R_{2} - R_{3}$
$| \begin{array}{ccc} 0 & 1 & (x - 3)^{2} - (x - 4)^{2} \\ 0 & 1 & (x - 4)^{2} - (x - 5)^{2} \\ 1 & x - 5 & (x - 5)^{2} \end{array} | = | \begin{array}{ccc} 0 & 1 & 2 x - 7 \\ 0 & 1 & 2 x - 9 \\ 1 & x - 5 & (x - 5)^{2} \end{array} |$
$=$ Expanding along ' $C_{1}$ ', we get -
$1 [2 x - 9 - (2 x - 7)] = 0$
$2 x - 9 - 2 x + 7 \Rightarrow - 2 \neq 0$
Thus the given determinant does not vanish for any value.
So, there is not any value of ' $x$ '

BITSAT-2013

Matrix and Determinant

79149 If [ ] denotes the greatest integer less than or equal to the real number under consideration and $- 1 \leq x < 0; 0 \leq y < 1; 1 \leq z < 2$ , then the value of the determinant
$| \begin{array}{ccc} [x] + 1 & [y] & [z] \\ [x] & [y] + 1 & [z] \\ [x] & [y] & [z] + 1 \end{array} |$ is

1

[z]

2

[y]

3

[x]

4 None of these

Explanation:

(A) : Here, It is given that
$\begin{array}{lll} - 1 \leq x < 0, & 0 \leq y < 1, & 1 \leq z < 2 \\ \Rightarrow [x] = - 1 & \Rightarrow [y] = 0 & \Rightarrow [z] = 1 \end{array}$
Now,
$| \begin{array}{ccc} [x] + 1 & [y] & [z] \\ [x] & [y] + 1 & [z] \\ [x] & [y] & [z] + 1 \end{array} |$
$= | \begin{array}{ccc} 0 & 0 & 1 \\ - 1 & 1 & 1 \\ - 1 & 0 & 2 \end{array} | = 1 = [z]$

BITSAT-2016

Matrix and Determinant

79150 Let $ω = \frac{- 1}{2} + i \frac{\sqrt{3}}{2}$ . Then, the value of the determinant $| \begin{array}{ccc} 1 & 1 & 1 \\ 1 & - 1 - ω^{2} & ω^{2} \\ 1 & ω^{2} & ω^{4} \end{array} |$ is

1

3 ω

2

ω (ω - 1)

3

3 ω

4

3 ω (1 - ω)

Explanation:

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

BITSAT-2016

Matrix and Determinant

79151 The points represented by the complex numbers $1 + i, - 2 + 3 i, 5 / 3 i$ on the argand plane are

1 vertices of an equilateral triangle

2 vertices of an isosceles triangle

3 collinear

4 None of these

Explanation:

(C) : According to given summation,
let $z_{1} = 1 + i, z_{2} = - 2 + 3 i$ and $z_{3} = 0 + \frac{5}{3} i$
Then, $| \begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array} | = | \begin{array}{ccc} 1 & 1 & 1 \\ - 2 & 3 & 1 \\ 0 & 5 / 3 & 1 \end{array} |$
$= 1 [3 - \frac{5}{3}] + 1 (2) + 1 [\frac{- 10}{3}] = \frac{4}{3} + 2 - \frac{10}{3} = \frac{4 + 6 - 10}{3} = 0$
$∴$ The points are collinear.

BITSAT-2018

Matrix and Determinant

79147 If $a > 0, b > 0, c > 0$ are respectively the pth, qth, rth terms of G.P., then the value of the
$determinant | \begin{array}{lll} \log a & p & 1 \\ \log b & q & 1 \end{array} | is$
$\begin{array}{lll} \log c & r & 1 \end{array}$

1 0

2 1

3 -1

4 None of these

Explanation:

(A) : According to given summation,
Let $A$ be the $1^{st}$ term and $R$ the common ratio of G.P. Then, $a = T_{p} = {AR}^{P - 1}$
$∴ \log a = \log A + (p - 1) \log R$
Similarly, $\log b = \log A + (q - 1) \log R$
and $\log c = \log A + (r - 1) \log R$
$∴ Δ = | \begin{array}{lll} \log A + (p - 1) \log R & p & 1 \\ \log A + (q - 1) \log R & q & 1 \\ \log A + (r - 1) \log R & r & 1 \end{array} |$
Split into two determinants and in the first take $\log A$ common and in the second take $\log R$ common.
$Δ = \log A | \begin{array}{lll} 1 & p & 1 \\ 1 & q & 1 \\ 1 & r & 1 \end{array} | + \log R | \begin{array}{ccc} p - 1 & p & 1 \\ q - 1 & q & 1 \\ r - 1 & r & 1 \end{array} |$
Apply $C_{1} \to C_{1} - C_{2} + C_{3}$ in the second part of the above determinant-
$Δ = 0 + \log R | \begin{array}{ccc} 0 & p & 1 \\ 0 & q & 1 \\ 0 & r & 1 \end{array} | \Rightarrow Δ = 0$
$1 (x - 3) (x - 3)^{2}$

BITSAT-2009

Matrix and Determinant

79148 The determinant $1 (x - 4) (x - 4)^{2}$ vanishes
$1 (x - 5) (x - 5)^{2}$
for

1 3 values of

x

2 2 values of

x

3 1 values of

x

4 No value of

x

Explanation:

(D) : The given determinant vanishes, i.e.
$| \begin{array}{ccc} 1 & x - 3 & (x - 3)^{2} \\ 1 & x - 4 & (x - 4)^{2} \\ 1 & x - 5 & (x - 5)^{2} \end{array} | = 0$
Operating,
$R_{1} \to R_{1} - R_{2}$
$R_{2} \to R_{2} - R_{3}$
$| \begin{array}{ccc} 0 & 1 & (x - 3)^{2} - (x - 4)^{2} \\ 0 & 1 & (x - 4)^{2} - (x - 5)^{2} \\ 1 & x - 5 & (x - 5)^{2} \end{array} | = | \begin{array}{ccc} 0 & 1 & 2 x - 7 \\ 0 & 1 & 2 x - 9 \\ 1 & x - 5 & (x - 5)^{2} \end{array} |$
$=$ Expanding along ' $C_{1}$ ', we get -
$1 [2 x - 9 - (2 x - 7)] = 0$
$2 x - 9 - 2 x + 7 \Rightarrow - 2 \neq 0$
Thus the given determinant does not vanish for any value.
So, there is not any value of ' $x$ '

BITSAT-2013

Matrix and Determinant

79149 If [ ] denotes the greatest integer less than or equal to the real number under consideration and $- 1 \leq x < 0; 0 \leq y < 1; 1 \leq z < 2$ , then the value of the determinant
$| \begin{array}{ccc} [x] + 1 & [y] & [z] \\ [x] & [y] + 1 & [z] \\ [x] & [y] & [z] + 1 \end{array} |$ is

1

[z]

2

[y]

3

[x]

4 None of these

Explanation:

(A) : Here, It is given that
$\begin{array}{lll} - 1 \leq x < 0, & 0 \leq y < 1, & 1 \leq z < 2 \\ \Rightarrow [x] = - 1 & \Rightarrow [y] = 0 & \Rightarrow [z] = 1 \end{array}$
Now,
$| \begin{array}{ccc} [x] + 1 & [y] & [z] \\ [x] & [y] + 1 & [z] \\ [x] & [y] & [z] + 1 \end{array} |$
$= | \begin{array}{ccc} 0 & 0 & 1 \\ - 1 & 1 & 1 \\ - 1 & 0 & 2 \end{array} | = 1 = [z]$

BITSAT-2016

Matrix and Determinant

79150 Let $ω = \frac{- 1}{2} + i \frac{\sqrt{3}}{2}$ . Then, the value of the determinant $| \begin{array}{ccc} 1 & 1 & 1 \\ 1 & - 1 - ω^{2} & ω^{2} \\ 1 & ω^{2} & ω^{4} \end{array} |$ is

1

3 ω

2

ω (ω - 1)

3

3 ω

4

3 ω (1 - ω)

Explanation:

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

BITSAT-2016

Matrix and Determinant

79151 The points represented by the complex numbers $1 + i, - 2 + 3 i, 5 / 3 i$ on the argand plane are

1 vertices of an equilateral triangle

2 vertices of an isosceles triangle

3 collinear

4 None of these

Explanation:

(C) : According to given summation,
let $z_{1} = 1 + i, z_{2} = - 2 + 3 i$ and $z_{3} = 0 + \frac{5}{3} i$
Then, $| \begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array} | = | \begin{array}{ccc} 1 & 1 & 1 \\ - 2 & 3 & 1 \\ 0 & 5 / 3 & 1 \end{array} |$
$= 1 [3 - \frac{5}{3}] + 1 (2) + 1 [\frac{- 10}{3}] = \frac{4}{3} + 2 - \frac{10}{3} = \frac{4 + 6 - 10}{3} = 0$
$∴$ The points are collinear.

BITSAT-2018

Matrix and Determinant

79147 If $a > 0, b > 0, c > 0$ are respectively the pth, qth, rth terms of G.P., then the value of the
$determinant | \begin{array}{lll} \log a & p & 1 \\ \log b & q & 1 \end{array} | is$
$\begin{array}{lll} \log c & r & 1 \end{array}$

1 0

2 1

3 -1

4 None of these

Explanation:

(A) : According to given summation,
Let $A$ be the $1^{st}$ term and $R$ the common ratio of G.P. Then, $a = T_{p} = {AR}^{P - 1}$
$∴ \log a = \log A + (p - 1) \log R$
Similarly, $\log b = \log A + (q - 1) \log R$
and $\log c = \log A + (r - 1) \log R$
$∴ Δ = | \begin{array}{lll} \log A + (p - 1) \log R & p & 1 \\ \log A + (q - 1) \log R & q & 1 \\ \log A + (r - 1) \log R & r & 1 \end{array} |$
Split into two determinants and in the first take $\log A$ common and in the second take $\log R$ common.
$Δ = \log A | \begin{array}{lll} 1 & p & 1 \\ 1 & q & 1 \\ 1 & r & 1 \end{array} | + \log R | \begin{array}{ccc} p - 1 & p & 1 \\ q - 1 & q & 1 \\ r - 1 & r & 1 \end{array} |$
Apply $C_{1} \to C_{1} - C_{2} + C_{3}$ in the second part of the above determinant-
$Δ = 0 + \log R | \begin{array}{ccc} 0 & p & 1 \\ 0 & q & 1 \\ 0 & r & 1 \end{array} | \Rightarrow Δ = 0$
$1 (x - 3) (x - 3)^{2}$

BITSAT-2009

Matrix and Determinant

79148 The determinant $1 (x - 4) (x - 4)^{2}$ vanishes
$1 (x - 5) (x - 5)^{2}$
for

1 3 values of

x

2 2 values of

x

3 1 values of

x

4 No value of

x

Explanation:

(D) : The given determinant vanishes, i.e.
$| \begin{array}{ccc} 1 & x - 3 & (x - 3)^{2} \\ 1 & x - 4 & (x - 4)^{2} \\ 1 & x - 5 & (x - 5)^{2} \end{array} | = 0$
Operating,
$R_{1} \to R_{1} - R_{2}$
$R_{2} \to R_{2} - R_{3}$
$| \begin{array}{ccc} 0 & 1 & (x - 3)^{2} - (x - 4)^{2} \\ 0 & 1 & (x - 4)^{2} - (x - 5)^{2} \\ 1 & x - 5 & (x - 5)^{2} \end{array} | = | \begin{array}{ccc} 0 & 1 & 2 x - 7 \\ 0 & 1 & 2 x - 9 \\ 1 & x - 5 & (x - 5)^{2} \end{array} |$
$=$ Expanding along ' $C_{1}$ ', we get -
$1 [2 x - 9 - (2 x - 7)] = 0$
$2 x - 9 - 2 x + 7 \Rightarrow - 2 \neq 0$
Thus the given determinant does not vanish for any value.
So, there is not any value of ' $x$ '

BITSAT-2013

Matrix and Determinant

79149 If [ ] denotes the greatest integer less than or equal to the real number under consideration and $- 1 \leq x < 0; 0 \leq y < 1; 1 \leq z < 2$ , then the value of the determinant
$| \begin{array}{ccc} [x] + 1 & [y] & [z] \\ [x] & [y] + 1 & [z] \\ [x] & [y] & [z] + 1 \end{array} |$ is

1

[z]

2

[y]

3

[x]

4 None of these

Explanation:

(A) : Here, It is given that
$\begin{array}{lll} - 1 \leq x < 0, & 0 \leq y < 1, & 1 \leq z < 2 \\ \Rightarrow [x] = - 1 & \Rightarrow [y] = 0 & \Rightarrow [z] = 1 \end{array}$
Now,
$| \begin{array}{ccc} [x] + 1 & [y] & [z] \\ [x] & [y] + 1 & [z] \\ [x] & [y] & [z] + 1 \end{array} |$
$= | \begin{array}{ccc} 0 & 0 & 1 \\ - 1 & 1 & 1 \\ - 1 & 0 & 2 \end{array} | = 1 = [z]$

BITSAT-2016

Matrix and Determinant

79150 Let $ω = \frac{- 1}{2} + i \frac{\sqrt{3}}{2}$ . Then, the value of the determinant $| \begin{array}{ccc} 1 & 1 & 1 \\ 1 & - 1 - ω^{2} & ω^{2} \\ 1 & ω^{2} & ω^{4} \end{array} |$ is

1

3 ω

2

ω (ω - 1)

3

3 ω

4

3 ω (1 - ω)

Explanation:

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

BITSAT-2016

Matrix and Determinant

79151 The points represented by the complex numbers $1 + i, - 2 + 3 i, 5 / 3 i$ on the argand plane are

1 vertices of an equilateral triangle

2 vertices of an isosceles triangle

3 collinear

4 None of these

Explanation:

(C) : According to given summation,
let $z_{1} = 1 + i, z_{2} = - 2 + 3 i$ and $z_{3} = 0 + \frac{5}{3} i$
Then, $| \begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array} | = | \begin{array}{ccc} 1 & 1 & 1 \\ - 2 & 3 & 1 \\ 0 & 5 / 3 & 1 \end{array} |$
$= 1 [3 - \frac{5}{3}] + 1 (2) + 1 [\frac{- 10}{3}] = \frac{4}{3} + 2 - \frac{10}{3} = \frac{4 + 6 - 10}{3} = 0$
$∴$ The points are collinear.

BITSAT-2018

Matrix and Determinant

79147 If $a > 0, b > 0, c > 0$ are respectively the pth, qth, rth terms of G.P., then the value of the
$determinant | \begin{array}{lll} \log a & p & 1 \\ \log b & q & 1 \end{array} | is$
$\begin{array}{lll} \log c & r & 1 \end{array}$

1 0

2 1

3 -1

4 None of these

Explanation:

(A) : According to given summation,
Let $A$ be the $1^{st}$ term and $R$ the common ratio of G.P. Then, $a = T_{p} = {AR}^{P - 1}$
$∴ \log a = \log A + (p - 1) \log R$
Similarly, $\log b = \log A + (q - 1) \log R$
and $\log c = \log A + (r - 1) \log R$
$∴ Δ = | \begin{array}{lll} \log A + (p - 1) \log R & p & 1 \\ \log A + (q - 1) \log R & q & 1 \\ \log A + (r - 1) \log R & r & 1 \end{array} |$
Split into two determinants and in the first take $\log A$ common and in the second take $\log R$ common.
$Δ = \log A | \begin{array}{lll} 1 & p & 1 \\ 1 & q & 1 \\ 1 & r & 1 \end{array} | + \log R | \begin{array}{ccc} p - 1 & p & 1 \\ q - 1 & q & 1 \\ r - 1 & r & 1 \end{array} |$
Apply $C_{1} \to C_{1} - C_{2} + C_{3}$ in the second part of the above determinant-
$Δ = 0 + \log R | \begin{array}{ccc} 0 & p & 1 \\ 0 & q & 1 \\ 0 & r & 1 \end{array} | \Rightarrow Δ = 0$
$1 (x - 3) (x - 3)^{2}$

BITSAT-2009

Matrix and Determinant

79148 The determinant $1 (x - 4) (x - 4)^{2}$ vanishes
$1 (x - 5) (x - 5)^{2}$
for

1 3 values of

x

2 2 values of

x

3 1 values of

x

4 No value of

x

Explanation:

(D) : The given determinant vanishes, i.e.
$| \begin{array}{ccc} 1 & x - 3 & (x - 3)^{2} \\ 1 & x - 4 & (x - 4)^{2} \\ 1 & x - 5 & (x - 5)^{2} \end{array} | = 0$
Operating,
$R_{1} \to R_{1} - R_{2}$
$R_{2} \to R_{2} - R_{3}$
$| \begin{array}{ccc} 0 & 1 & (x - 3)^{2} - (x - 4)^{2} \\ 0 & 1 & (x - 4)^{2} - (x - 5)^{2} \\ 1 & x - 5 & (x - 5)^{2} \end{array} | = | \begin{array}{ccc} 0 & 1 & 2 x - 7 \\ 0 & 1 & 2 x - 9 \\ 1 & x - 5 & (x - 5)^{2} \end{array} |$
$=$ Expanding along ' $C_{1}$ ', we get -
$1 [2 x - 9 - (2 x - 7)] = 0$
$2 x - 9 - 2 x + 7 \Rightarrow - 2 \neq 0$
Thus the given determinant does not vanish for any value.
So, there is not any value of ' $x$ '

BITSAT-2013

Matrix and Determinant

79149 If [ ] denotes the greatest integer less than or equal to the real number under consideration and $- 1 \leq x < 0; 0 \leq y < 1; 1 \leq z < 2$ , then the value of the determinant
$| \begin{array}{ccc} [x] + 1 & [y] & [z] \\ [x] & [y] + 1 & [z] \\ [x] & [y] & [z] + 1 \end{array} |$ is

1

[z]

2

[y]

3

[x]

4 None of these

Explanation:

(A) : Here, It is given that
$\begin{array}{lll} - 1 \leq x < 0, & 0 \leq y < 1, & 1 \leq z < 2 \\ \Rightarrow [x] = - 1 & \Rightarrow [y] = 0 & \Rightarrow [z] = 1 \end{array}$
Now,
$| \begin{array}{ccc} [x] + 1 & [y] & [z] \\ [x] & [y] + 1 & [z] \\ [x] & [y] & [z] + 1 \end{array} |$
$= | \begin{array}{ccc} 0 & 0 & 1 \\ - 1 & 1 & 1 \\ - 1 & 0 & 2 \end{array} | = 1 = [z]$

BITSAT-2016

Matrix and Determinant

79150 Let $ω = \frac{- 1}{2} + i \frac{\sqrt{3}}{2}$ . Then, the value of the determinant $| \begin{array}{ccc} 1 & 1 & 1 \\ 1 & - 1 - ω^{2} & ω^{2} \\ 1 & ω^{2} & ω^{4} \end{array} |$ is

1

3 ω

2

ω (ω - 1)

3

3 ω

4

3 ω (1 - ω)

Explanation:

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

BITSAT-2016

Matrix and Determinant

79151 The points represented by the complex numbers $1 + i, - 2 + 3 i, 5 / 3 i$ on the argand plane are

1 vertices of an equilateral triangle

2 vertices of an isosceles triangle

3 collinear

4 None of these

Explanation:

(C) : According to given summation,
let $z_{1} = 1 + i, z_{2} = - 2 + 3 i$ and $z_{3} = 0 + \frac{5}{3} i$
Then, $| \begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array} | = | \begin{array}{ccc} 1 & 1 & 1 \\ - 2 & 3 & 1 \\ 0 & 5 / 3 & 1 \end{array} |$
$= 1 [3 - \frac{5}{3}] + 1 (2) + 1 [\frac{- 10}{3}] = \frac{4}{3} + 2 - \frac{10}{3} = \frac{4 + 6 - 10}{3} = 0$
$∴$ The points are collinear.

BITSAT-2018

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