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Matrix and Determinant

79131 The constant term in the expansion of
$| \begin{array}{ccc} 3 x + 1 & 2 x - 1 & x + 2 \\ 5 x - 1 & 3 x + 2 & x + 1 \\ 7 x - 2 & 3 x + 1 & 4 x - 1 \end{array} | is$

1 0

2 -10

3 2

4 6

Explanation:

(D) : We have,
$| \begin{array}{ccc} 3 x + 1 & 2 x - 1 & x + 2 \\ 5 x - 1 & 3 x + 2 & x + 1 \\ 7 x - 2 & 3 x + 1 & 4 x - 1 \end{array} |$
In order to find the constant term, putting $x = 0$ , we get
$| \begin{array}{ccc} 1 & - 1 & 2 \\ - 1 & 2 & 1 \\ - 2 & 1 & - 1 \end{array} |$
$1 (- 2 - 1) + 1 (1 + 2) + 2 (- 1 + 4) = - 3 + 3 + 6 = 6$
So, the constant term is 6 .

Karnataka CET-2019

Matrix and Determinant

79132 If $x, y, z \in R$ , then the value of the determinant
$A = [\begin{array}{lll} {(5^{x} + 5^{- x})}^{2} & {(5^{x} - 5^{- x})}^{2} & 1 \\ {(6^{x} + 6^{- x})}^{2} & {(6^{x} - 6^{- x})}^{2} & 1 \\ {(7^{x} + 7^{- x})}^{2} & {(7^{x} - 7^{- x})}^{2} & 1 \end{array}] is$

1 10

2 12

3 1

4 0

Explanation:

(D) : According to given summation,
Let $A = | \begin{array}{lll} {(5^{x} + 5^{- x})}^{2} & {(5^{x} - 5^{- x})}^{2} & 1 \\ {(6^{x} + 6^{- x})}^{2} & {(6^{x} - 6^{- x})}^{2} & 1 \\ {(7^{x} + 7^{- x})}^{2} & {(7^{x} - 7^{- x})}^{2} & 1 \end{array} |$
Operating $C_{1} \to C_{1} - C_{2}$ , we get -
$A = | \begin{array}{lll} 4 & {(5^{x} - 5^{- x})}^{2} & 1 \\ 4 & {(6^{x} - 6^{- x})}^{2} & 1 \\ 4 & {(7^{x} - 7^{- x})}^{2} & 1 \end{array} | \Rightarrow A = 4 | \begin{array}{lll} 1 & {(5^{x} - 5^{- x})}^{2} & 1 \\ 1 & {(6^{x} - 6^{- x})}^{2} & 1 \\ 1 & {(7^{x} - 7^{- x})}^{2} & 1 \end{array} |$
Operating $C_{1} \to C_{1} - C_{3}$ , we get -
Hence, $A = 0$

Karnataka CET-2018

Matrix and Determinant

79134 If $x y z$ are not equal and $\neq 0, \neq 1$ the value of
$| \begin{array}{ccc} \log x & \log y & \log z \\ \log 2 x & \log 2 y & \log 2 z \\ \log 3 x & \log 3 y & \log 3 z \end{array} |$ is equal to

1

\log (x y z)

2

\log (6 x y z)

3 0

4

\log (x + y + z)

Explanation:

(C) : According to given summation,
$Δ = | \begin{array}{ccc} \log x & \log y & \log z \\ \log 2 x & \log 2 y & \log 2 z \\ \log 3 x & \log 3 y & \log 3 z \end{array} |$
Operating $R_{2} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{1}$ , we get -
$Δ = | \begin{array}{lll} \log x & \log y & \log z \\ \log 2 & \log 2 & \log 2 \\ \log 3 & \log 3 & \log 3 \end{array} |$
Taking out $\log 2$ and $\log 3$ from $R_{2}$ and $R_{3}$ respectively.
$Δ = (\log 2) (\log 3) | \begin{array}{ccc} \log x & \log y & \log z \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array} |$
$∴ Δ = 0$

Karnataka CET-2016

Matrix and Determinant

79136 If $a x^{4} + b x^{3} + c x^{2} + d x + e =$
$| \begin{array}{ccc} x^{3} + 3 x & x - 1 & x + 3 \\ x + 1 & - 2 x & x - 4 \\ x - 3 & x + 4 & 3 x \end{array} |$ , then $e =$

1 -1

2 1

3 0

4 2

Explanation:

(C) : According to given Summation,
$a x^{4} + b x^{3} + c x^{2} + d x + e = | \begin{array}{ccc} x^{3} + 3 x & x - 1 & x + 3 \\ x + 1 & - 2 x & x - 4 \\ x - 3 & x + 4 & 3 x \end{array} |$ ,
Putting $x = 0$ in both sides, we get -
$e = | \begin{array}{ccc} 0 & - 1 & 3 \\ 1 & 0 & - 4 \\ - 3 & 4 & 0 \end{array} |$
$e = 0 (0) + 16 - (- 1) (0 - 12) + 3 (4 - 0)$
$e = 0 - 12 + 12 \Rightarrow e = 0$

Karnataka CET-2012

Matrix and Determinant

79137 If $| \begin{array}{lll} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 1 & 0 & 2 \end{array} |$ then $| adj A |$

1

1 / 9

2 81

3 0

4 9

Explanation:

(B) : We have, $A = | \begin{array}{lll} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 1 & 0 & 2 \end{array} |$
If $A$ is a matrix of order $n$ ,
Then, $| adj A | = | A |^{n - 1}$
Here, $| A | = 2 (4 - 0) - 1 (0 - 1) + 0 (0 - 2) = 9$
$∴ | adj A | = 9^{3 - 1} = 9^{2} = 81$

Karnataka CET-2009

Matrix and Determinant

79131 The constant term in the expansion of
$| \begin{array}{ccc} 3 x + 1 & 2 x - 1 & x + 2 \\ 5 x - 1 & 3 x + 2 & x + 1 \\ 7 x - 2 & 3 x + 1 & 4 x - 1 \end{array} | is$

1 0

2 -10

3 2

4 6

Explanation:

(D) : We have,
$| \begin{array}{ccc} 3 x + 1 & 2 x - 1 & x + 2 \\ 5 x - 1 & 3 x + 2 & x + 1 \\ 7 x - 2 & 3 x + 1 & 4 x - 1 \end{array} |$
In order to find the constant term, putting $x = 0$ , we get
$| \begin{array}{ccc} 1 & - 1 & 2 \\ - 1 & 2 & 1 \\ - 2 & 1 & - 1 \end{array} |$
$1 (- 2 - 1) + 1 (1 + 2) + 2 (- 1 + 4) = - 3 + 3 + 6 = 6$
So, the constant term is 6 .

Karnataka CET-2019

Matrix and Determinant

79132 If $x, y, z \in R$ , then the value of the determinant
$A = [\begin{array}{lll} {(5^{x} + 5^{- x})}^{2} & {(5^{x} - 5^{- x})}^{2} & 1 \\ {(6^{x} + 6^{- x})}^{2} & {(6^{x} - 6^{- x})}^{2} & 1 \\ {(7^{x} + 7^{- x})}^{2} & {(7^{x} - 7^{- x})}^{2} & 1 \end{array}] is$

1 10

2 12

3 1

4 0

Explanation:

(D) : According to given summation,
Let $A = | \begin{array}{lll} {(5^{x} + 5^{- x})}^{2} & {(5^{x} - 5^{- x})}^{2} & 1 \\ {(6^{x} + 6^{- x})}^{2} & {(6^{x} - 6^{- x})}^{2} & 1 \\ {(7^{x} + 7^{- x})}^{2} & {(7^{x} - 7^{- x})}^{2} & 1 \end{array} |$
Operating $C_{1} \to C_{1} - C_{2}$ , we get -
$A = | \begin{array}{lll} 4 & {(5^{x} - 5^{- x})}^{2} & 1 \\ 4 & {(6^{x} - 6^{- x})}^{2} & 1 \\ 4 & {(7^{x} - 7^{- x})}^{2} & 1 \end{array} | \Rightarrow A = 4 | \begin{array}{lll} 1 & {(5^{x} - 5^{- x})}^{2} & 1 \\ 1 & {(6^{x} - 6^{- x})}^{2} & 1 \\ 1 & {(7^{x} - 7^{- x})}^{2} & 1 \end{array} |$
Operating $C_{1} \to C_{1} - C_{3}$ , we get -
Hence, $A = 0$

Karnataka CET-2018

Matrix and Determinant

79134 If $x y z$ are not equal and $\neq 0, \neq 1$ the value of
$| \begin{array}{ccc} \log x & \log y & \log z \\ \log 2 x & \log 2 y & \log 2 z \\ \log 3 x & \log 3 y & \log 3 z \end{array} |$ is equal to

1

\log (x y z)

2

\log (6 x y z)

3 0

4

\log (x + y + z)

Explanation:

(C) : According to given summation,
$Δ = | \begin{array}{ccc} \log x & \log y & \log z \\ \log 2 x & \log 2 y & \log 2 z \\ \log 3 x & \log 3 y & \log 3 z \end{array} |$
Operating $R_{2} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{1}$ , we get -
$Δ = | \begin{array}{lll} \log x & \log y & \log z \\ \log 2 & \log 2 & \log 2 \\ \log 3 & \log 3 & \log 3 \end{array} |$
Taking out $\log 2$ and $\log 3$ from $R_{2}$ and $R_{3}$ respectively.
$Δ = (\log 2) (\log 3) | \begin{array}{ccc} \log x & \log y & \log z \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array} |$
$∴ Δ = 0$

Karnataka CET-2016

Matrix and Determinant

79136 If $a x^{4} + b x^{3} + c x^{2} + d x + e =$
$| \begin{array}{ccc} x^{3} + 3 x & x - 1 & x + 3 \\ x + 1 & - 2 x & x - 4 \\ x - 3 & x + 4 & 3 x \end{array} |$ , then $e =$

1 -1

2 1

3 0

4 2

Explanation:

(C) : According to given Summation,
$a x^{4} + b x^{3} + c x^{2} + d x + e = | \begin{array}{ccc} x^{3} + 3 x & x - 1 & x + 3 \\ x + 1 & - 2 x & x - 4 \\ x - 3 & x + 4 & 3 x \end{array} |$ ,
Putting $x = 0$ in both sides, we get -
$e = | \begin{array}{ccc} 0 & - 1 & 3 \\ 1 & 0 & - 4 \\ - 3 & 4 & 0 \end{array} |$
$e = 0 (0) + 16 - (- 1) (0 - 12) + 3 (4 - 0)$
$e = 0 - 12 + 12 \Rightarrow e = 0$

Karnataka CET-2012

Matrix and Determinant

79137 If $| \begin{array}{lll} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 1 & 0 & 2 \end{array} |$ then $| adj A |$

1

1 / 9

2 81

3 0

4 9

Explanation:

(B) : We have, $A = | \begin{array}{lll} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 1 & 0 & 2 \end{array} |$
If $A$ is a matrix of order $n$ ,
Then, $| adj A | = | A |^{n - 1}$
Here, $| A | = 2 (4 - 0) - 1 (0 - 1) + 0 (0 - 2) = 9$
$∴ | adj A | = 9^{3 - 1} = 9^{2} = 81$

Karnataka CET-2009

Matrix and Determinant

79131 The constant term in the expansion of
$| \begin{array}{ccc} 3 x + 1 & 2 x - 1 & x + 2 \\ 5 x - 1 & 3 x + 2 & x + 1 \\ 7 x - 2 & 3 x + 1 & 4 x - 1 \end{array} | is$

1 0

2 -10

3 2

4 6

Explanation:

(D) : We have,
$| \begin{array}{ccc} 3 x + 1 & 2 x - 1 & x + 2 \\ 5 x - 1 & 3 x + 2 & x + 1 \\ 7 x - 2 & 3 x + 1 & 4 x - 1 \end{array} |$
In order to find the constant term, putting $x = 0$ , we get
$| \begin{array}{ccc} 1 & - 1 & 2 \\ - 1 & 2 & 1 \\ - 2 & 1 & - 1 \end{array} |$
$1 (- 2 - 1) + 1 (1 + 2) + 2 (- 1 + 4) = - 3 + 3 + 6 = 6$
So, the constant term is 6 .

Karnataka CET-2019

Matrix and Determinant

79132 If $x, y, z \in R$ , then the value of the determinant
$A = [\begin{array}{lll} {(5^{x} + 5^{- x})}^{2} & {(5^{x} - 5^{- x})}^{2} & 1 \\ {(6^{x} + 6^{- x})}^{2} & {(6^{x} - 6^{- x})}^{2} & 1 \\ {(7^{x} + 7^{- x})}^{2} & {(7^{x} - 7^{- x})}^{2} & 1 \end{array}] is$

1 10

2 12

3 1

4 0

Explanation:

(D) : According to given summation,
Let $A = | \begin{array}{lll} {(5^{x} + 5^{- x})}^{2} & {(5^{x} - 5^{- x})}^{2} & 1 \\ {(6^{x} + 6^{- x})}^{2} & {(6^{x} - 6^{- x})}^{2} & 1 \\ {(7^{x} + 7^{- x})}^{2} & {(7^{x} - 7^{- x})}^{2} & 1 \end{array} |$
Operating $C_{1} \to C_{1} - C_{2}$ , we get -
$A = | \begin{array}{lll} 4 & {(5^{x} - 5^{- x})}^{2} & 1 \\ 4 & {(6^{x} - 6^{- x})}^{2} & 1 \\ 4 & {(7^{x} - 7^{- x})}^{2} & 1 \end{array} | \Rightarrow A = 4 | \begin{array}{lll} 1 & {(5^{x} - 5^{- x})}^{2} & 1 \\ 1 & {(6^{x} - 6^{- x})}^{2} & 1 \\ 1 & {(7^{x} - 7^{- x})}^{2} & 1 \end{array} |$
Operating $C_{1} \to C_{1} - C_{3}$ , we get -
Hence, $A = 0$

Karnataka CET-2018

Matrix and Determinant

79134 If $x y z$ are not equal and $\neq 0, \neq 1$ the value of
$| \begin{array}{ccc} \log x & \log y & \log z \\ \log 2 x & \log 2 y & \log 2 z \\ \log 3 x & \log 3 y & \log 3 z \end{array} |$ is equal to

1

\log (x y z)

2

\log (6 x y z)

3 0

4

\log (x + y + z)

Explanation:

(C) : According to given summation,
$Δ = | \begin{array}{ccc} \log x & \log y & \log z \\ \log 2 x & \log 2 y & \log 2 z \\ \log 3 x & \log 3 y & \log 3 z \end{array} |$
Operating $R_{2} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{1}$ , we get -
$Δ = | \begin{array}{lll} \log x & \log y & \log z \\ \log 2 & \log 2 & \log 2 \\ \log 3 & \log 3 & \log 3 \end{array} |$
Taking out $\log 2$ and $\log 3$ from $R_{2}$ and $R_{3}$ respectively.
$Δ = (\log 2) (\log 3) | \begin{array}{ccc} \log x & \log y & \log z \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array} |$
$∴ Δ = 0$

Karnataka CET-2016

Matrix and Determinant

79136 If $a x^{4} + b x^{3} + c x^{2} + d x + e =$
$| \begin{array}{ccc} x^{3} + 3 x & x - 1 & x + 3 \\ x + 1 & - 2 x & x - 4 \\ x - 3 & x + 4 & 3 x \end{array} |$ , then $e =$

1 -1

2 1

3 0

4 2

Explanation:

(C) : According to given Summation,
$a x^{4} + b x^{3} + c x^{2} + d x + e = | \begin{array}{ccc} x^{3} + 3 x & x - 1 & x + 3 \\ x + 1 & - 2 x & x - 4 \\ x - 3 & x + 4 & 3 x \end{array} |$ ,
Putting $x = 0$ in both sides, we get -
$e = | \begin{array}{ccc} 0 & - 1 & 3 \\ 1 & 0 & - 4 \\ - 3 & 4 & 0 \end{array} |$
$e = 0 (0) + 16 - (- 1) (0 - 12) + 3 (4 - 0)$
$e = 0 - 12 + 12 \Rightarrow e = 0$

Karnataka CET-2012

Matrix and Determinant

79137 If $| \begin{array}{lll} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 1 & 0 & 2 \end{array} |$ then $| adj A |$

1

1 / 9

2 81

3 0

4 9

Explanation:

(B) : We have, $A = | \begin{array}{lll} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 1 & 0 & 2 \end{array} |$
If $A$ is a matrix of order $n$ ,
Then, $| adj A | = | A |^{n - 1}$
Here, $| A | = 2 (4 - 0) - 1 (0 - 1) + 0 (0 - 2) = 9$
$∴ | adj A | = 9^{3 - 1} = 9^{2} = 81$

Karnataka CET-2009

Matrix and Determinant

79131 The constant term in the expansion of
$| \begin{array}{ccc} 3 x + 1 & 2 x - 1 & x + 2 \\ 5 x - 1 & 3 x + 2 & x + 1 \\ 7 x - 2 & 3 x + 1 & 4 x - 1 \end{array} | is$

1 0

2 -10

3 2

4 6

Explanation:

(D) : We have,
$| \begin{array}{ccc} 3 x + 1 & 2 x - 1 & x + 2 \\ 5 x - 1 & 3 x + 2 & x + 1 \\ 7 x - 2 & 3 x + 1 & 4 x - 1 \end{array} |$
In order to find the constant term, putting $x = 0$ , we get
$| \begin{array}{ccc} 1 & - 1 & 2 \\ - 1 & 2 & 1 \\ - 2 & 1 & - 1 \end{array} |$
$1 (- 2 - 1) + 1 (1 + 2) + 2 (- 1 + 4) = - 3 + 3 + 6 = 6$
So, the constant term is 6 .

Karnataka CET-2019

Matrix and Determinant

79132 If $x, y, z \in R$ , then the value of the determinant
$A = [\begin{array}{lll} {(5^{x} + 5^{- x})}^{2} & {(5^{x} - 5^{- x})}^{2} & 1 \\ {(6^{x} + 6^{- x})}^{2} & {(6^{x} - 6^{- x})}^{2} & 1 \\ {(7^{x} + 7^{- x})}^{2} & {(7^{x} - 7^{- x})}^{2} & 1 \end{array}] is$

1 10

2 12

3 1

4 0

Explanation:

(D) : According to given summation,
Let $A = | \begin{array}{lll} {(5^{x} + 5^{- x})}^{2} & {(5^{x} - 5^{- x})}^{2} & 1 \\ {(6^{x} + 6^{- x})}^{2} & {(6^{x} - 6^{- x})}^{2} & 1 \\ {(7^{x} + 7^{- x})}^{2} & {(7^{x} - 7^{- x})}^{2} & 1 \end{array} |$
Operating $C_{1} \to C_{1} - C_{2}$ , we get -
$A = | \begin{array}{lll} 4 & {(5^{x} - 5^{- x})}^{2} & 1 \\ 4 & {(6^{x} - 6^{- x})}^{2} & 1 \\ 4 & {(7^{x} - 7^{- x})}^{2} & 1 \end{array} | \Rightarrow A = 4 | \begin{array}{lll} 1 & {(5^{x} - 5^{- x})}^{2} & 1 \\ 1 & {(6^{x} - 6^{- x})}^{2} & 1 \\ 1 & {(7^{x} - 7^{- x})}^{2} & 1 \end{array} |$
Operating $C_{1} \to C_{1} - C_{3}$ , we get -
Hence, $A = 0$

Karnataka CET-2018

Matrix and Determinant

79134 If $x y z$ are not equal and $\neq 0, \neq 1$ the value of
$| \begin{array}{ccc} \log x & \log y & \log z \\ \log 2 x & \log 2 y & \log 2 z \\ \log 3 x & \log 3 y & \log 3 z \end{array} |$ is equal to

1

\log (x y z)

2

\log (6 x y z)

3 0

4

\log (x + y + z)

Explanation:

(C) : According to given summation,
$Δ = | \begin{array}{ccc} \log x & \log y & \log z \\ \log 2 x & \log 2 y & \log 2 z \\ \log 3 x & \log 3 y & \log 3 z \end{array} |$
Operating $R_{2} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{1}$ , we get -
$Δ = | \begin{array}{lll} \log x & \log y & \log z \\ \log 2 & \log 2 & \log 2 \\ \log 3 & \log 3 & \log 3 \end{array} |$
Taking out $\log 2$ and $\log 3$ from $R_{2}$ and $R_{3}$ respectively.
$Δ = (\log 2) (\log 3) | \begin{array}{ccc} \log x & \log y & \log z \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array} |$
$∴ Δ = 0$

Karnataka CET-2016

Matrix and Determinant

79136 If $a x^{4} + b x^{3} + c x^{2} + d x + e =$
$| \begin{array}{ccc} x^{3} + 3 x & x - 1 & x + 3 \\ x + 1 & - 2 x & x - 4 \\ x - 3 & x + 4 & 3 x \end{array} |$ , then $e =$

1 -1

2 1

3 0

4 2

Explanation:

(C) : According to given Summation,
$a x^{4} + b x^{3} + c x^{2} + d x + e = | \begin{array}{ccc} x^{3} + 3 x & x - 1 & x + 3 \\ x + 1 & - 2 x & x - 4 \\ x - 3 & x + 4 & 3 x \end{array} |$ ,
Putting $x = 0$ in both sides, we get -
$e = | \begin{array}{ccc} 0 & - 1 & 3 \\ 1 & 0 & - 4 \\ - 3 & 4 & 0 \end{array} |$
$e = 0 (0) + 16 - (- 1) (0 - 12) + 3 (4 - 0)$
$e = 0 - 12 + 12 \Rightarrow e = 0$

Karnataka CET-2012

Matrix and Determinant

79137 If $| \begin{array}{lll} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 1 & 0 & 2 \end{array} |$ then $| adj A |$

1

1 / 9

2 81

3 0

4 9

Explanation:

(B) : We have, $A = | \begin{array}{lll} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 1 & 0 & 2 \end{array} |$
If $A$ is a matrix of order $n$ ,
Then, $| adj A | = | A |^{n - 1}$
Here, $| A | = 2 (4 - 0) - 1 (0 - 1) + 0 (0 - 2) = 9$
$∴ | adj A | = 9^{3 - 1} = 9^{2} = 81$

Karnataka CET-2009

Matrix and Determinant

79131 The constant term in the expansion of
$| \begin{array}{ccc} 3 x + 1 & 2 x - 1 & x + 2 \\ 5 x - 1 & 3 x + 2 & x + 1 \\ 7 x - 2 & 3 x + 1 & 4 x - 1 \end{array} | is$

1 0

2 -10

3 2

4 6

Explanation:

(D) : We have,
$| \begin{array}{ccc} 3 x + 1 & 2 x - 1 & x + 2 \\ 5 x - 1 & 3 x + 2 & x + 1 \\ 7 x - 2 & 3 x + 1 & 4 x - 1 \end{array} |$
In order to find the constant term, putting $x = 0$ , we get
$| \begin{array}{ccc} 1 & - 1 & 2 \\ - 1 & 2 & 1 \\ - 2 & 1 & - 1 \end{array} |$
$1 (- 2 - 1) + 1 (1 + 2) + 2 (- 1 + 4) = - 3 + 3 + 6 = 6$
So, the constant term is 6 .

Karnataka CET-2019

Matrix and Determinant

79132 If $x, y, z \in R$ , then the value of the determinant
$A = [\begin{array}{lll} {(5^{x} + 5^{- x})}^{2} & {(5^{x} - 5^{- x})}^{2} & 1 \\ {(6^{x} + 6^{- x})}^{2} & {(6^{x} - 6^{- x})}^{2} & 1 \\ {(7^{x} + 7^{- x})}^{2} & {(7^{x} - 7^{- x})}^{2} & 1 \end{array}] is$

1 10

2 12

3 1

4 0

Explanation:

(D) : According to given summation,
Let $A = | \begin{array}{lll} {(5^{x} + 5^{- x})}^{2} & {(5^{x} - 5^{- x})}^{2} & 1 \\ {(6^{x} + 6^{- x})}^{2} & {(6^{x} - 6^{- x})}^{2} & 1 \\ {(7^{x} + 7^{- x})}^{2} & {(7^{x} - 7^{- x})}^{2} & 1 \end{array} |$
Operating $C_{1} \to C_{1} - C_{2}$ , we get -
$A = | \begin{array}{lll} 4 & {(5^{x} - 5^{- x})}^{2} & 1 \\ 4 & {(6^{x} - 6^{- x})}^{2} & 1 \\ 4 & {(7^{x} - 7^{- x})}^{2} & 1 \end{array} | \Rightarrow A = 4 | \begin{array}{lll} 1 & {(5^{x} - 5^{- x})}^{2} & 1 \\ 1 & {(6^{x} - 6^{- x})}^{2} & 1 \\ 1 & {(7^{x} - 7^{- x})}^{2} & 1 \end{array} |$
Operating $C_{1} \to C_{1} - C_{3}$ , we get -
Hence, $A = 0$

Karnataka CET-2018

Matrix and Determinant

79134 If $x y z$ are not equal and $\neq 0, \neq 1$ the value of
$| \begin{array}{ccc} \log x & \log y & \log z \\ \log 2 x & \log 2 y & \log 2 z \\ \log 3 x & \log 3 y & \log 3 z \end{array} |$ is equal to

1

\log (x y z)

2

\log (6 x y z)

3 0

4

\log (x + y + z)

Explanation:

(C) : According to given summation,
$Δ = | \begin{array}{ccc} \log x & \log y & \log z \\ \log 2 x & \log 2 y & \log 2 z \\ \log 3 x & \log 3 y & \log 3 z \end{array} |$
Operating $R_{2} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{1}$ , we get -
$Δ = | \begin{array}{lll} \log x & \log y & \log z \\ \log 2 & \log 2 & \log 2 \\ \log 3 & \log 3 & \log 3 \end{array} |$
Taking out $\log 2$ and $\log 3$ from $R_{2}$ and $R_{3}$ respectively.
$Δ = (\log 2) (\log 3) | \begin{array}{ccc} \log x & \log y & \log z \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array} |$
$∴ Δ = 0$

Karnataka CET-2016

Matrix and Determinant

79136 If $a x^{4} + b x^{3} + c x^{2} + d x + e =$
$| \begin{array}{ccc} x^{3} + 3 x & x - 1 & x + 3 \\ x + 1 & - 2 x & x - 4 \\ x - 3 & x + 4 & 3 x \end{array} |$ , then $e =$

1 -1

2 1

3 0

4 2

Explanation:

(C) : According to given Summation,
$a x^{4} + b x^{3} + c x^{2} + d x + e = | \begin{array}{ccc} x^{3} + 3 x & x - 1 & x + 3 \\ x + 1 & - 2 x & x - 4 \\ x - 3 & x + 4 & 3 x \end{array} |$ ,
Putting $x = 0$ in both sides, we get -
$e = | \begin{array}{ccc} 0 & - 1 & 3 \\ 1 & 0 & - 4 \\ - 3 & 4 & 0 \end{array} |$
$e = 0 (0) + 16 - (- 1) (0 - 12) + 3 (4 - 0)$
$e = 0 - 12 + 12 \Rightarrow e = 0$

Karnataka CET-2012

Matrix and Determinant

79137 If $| \begin{array}{lll} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 1 & 0 & 2 \end{array} |$ then $| adj A |$

1

1 / 9

2 81

3 0

4 9

Explanation:

(B) : We have, $A = | \begin{array}{lll} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 1 & 0 & 2 \end{array} |$
If $A$ is a matrix of order $n$ ,
Then, $| adj A | = | A |^{n - 1}$
Here, $| A | = 2 (4 - 0) - 1 (0 - 1) + 0 (0 - 2) = 9$
$∴ | adj A | = 9^{3 - 1} = 9^{2} = 81$

Karnataka CET-2009