QBManager

Matrix and Determinant

78999 The value of $| \begin{array}{lll} 441 & 442 & 443 \\ 445 & 446 & 447 \\ 449 & 450 & 451 \end{array} |$ is :

1

441 \times 446 \times 4510

2 0

3

- 1

4

1

Explanation:

(B) : We have,
$| \begin{array}{lll} 441 & 442 & 443 \\ 445 & 446 & 447 \\ 449 & 450 & 451 \end{array} |$
Applying - $c_{3} \to c_{3} - c_{2}$
We get-
$c_{2} \to c_{2} - c_{1}$
$| \begin{array}{lll} 441 & 1 & 1 \\ 445 & 1 & 1 \\ 450 & 1 & 1 \end{array} | = 0 [∵$ Two column are identical $]$

Karnataka CET-2004

Matrix and Determinant

79000 If $x^{3} - 2 x^{2} - 9 x + 18 = 0$ and $A = | \begin{array}{lll} 1 & 2 & 5 \\ 4 & x & 6 \\ 7 & 8 & 9 \end{array} |$ then
the maximum value of $A$ is

1 96

2 36

3 24

4 120

Explanation:

(A) : Given,
$f (x) = x^{3} - 2 x^{2} + 9 x + 18 = 0$
$x^{2} (x - 2) - 9 (x - 2) = 0$
$(x - 2) (x^{2} - 9) = 0$
$(x - 2) (x + 3) (x - 3) = 0$
$x = 2, - 3, 3$
$A = | \begin{array}{lll} 1 & 2 & 3 \\ 4 & x & 6 \\ 7 & 8 & 9 \end{array} |$
$[\begin{array}{l} R_{2} : R_{2} - 4 R_{1} \\ R_{3} : R_{3} - 7 R_{1} \end{array}]$
$= | \begin{array}{ccc} 1 & 2 & 3 \\ 0 & x - 8 & 6 - 12 \\ 0 & 8 - 14 & 9 - 21 \end{array} |$
$A = | \begin{array}{ccc} 1 & 2 & 3 \\ 0 & x - 8 & - 6 \\ 0 & - 6 & - 12 \end{array} | = 3 | \begin{array}{ccc} 1 & 2 & 1 \\ 0 & x - 8 & - 2 \\ 1 & - 6 & - 4 \end{array} |$
$= 6 | \begin{array}{ccc} 1 & 2 & 1 \\ 0 & x - 8 & - 2 \\ 0 & - 3 & 2 \end{array} |$
$= 6 (- 2 x + 10)$
$= - 12 x + 60$
$A = - 12 x + 60$
of $x = 2 A = 36$
$x = 3 A = 24$
$x = - 3 A = 96$
(A) $max = 96$

Karnataka CET-2021

Matrix and Determinant

79001 If $A$ and $B$ are matrices of order 3 and $| A | = 5$ , $| B | = 3$ , then $| 3 A B |$ is

1 425

2 405

3 565

4 585

Explanation:

(B) : We have,
$| A | = 5 and | B | = 3$
We know that,
$| KA | = K^{n} | A | (where, n = order of matrix)$
$= 27 \times 5 \times 3$
$| 3 AB | = 405$
$∴ | 3 AB | = 3^{3} | A | | B |$

Karnataka CET-2021

Matrix and Determinant

79002 The system of linear equations $x + y + z = 6$ , $x + 2 y + 3 z = 10$ and $x + 2 y + a z = b$ has no solution when

1

a = 3, b \neq 10

2

b = 3, a \neq 10

3

a = 2, b \neq 3

4

b = 2, a = 3

Explanation:

(A) : We have,
$\begin{array}{l} x + y + z = 6 \\ x + 2 y + 3 z = 10 \\ x + 2 y + a z = b \end{array}}$
Equation (i) will have no solution when $D = 0$ and at least one of $D_{1}, D_{2}$ or $D_{3}$ is non-zero.
Here,
$D = | \begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & a \end{array} | = 0$
$1 (2 a - 6) - 1 (a - 3) + 1 (2 - 2) = 0$
$2 a - 6 - a + 3 = 0$
$a - 3 = 0$
$a = 3$
Let, $D_{1} \neq 0$ , then -
$| \begin{array}{ccc} 6 & 1 & 1 \\ 10 & 2 & 3 \\ b & 2 & a \end{array} | \neq 0$
$6 (2 a - 6) - 10 (a - 2) + b (3 - 2) \neq 0$
$2 a + b - 16 \neq 0$
$b - 10 \neq 0$
$b \neq 10$

Karnataka CET-2015

Matrix and Determinant

79003 If $a e^{x} + b e^{y} = c; p e^{x} + q e^{y} = d$ and
$Δ_{1} = | \begin{array}{ll} a & b \\ p & q \end{array} |; Δ_{2} = | \begin{array}{ll} c & b \\ d & q \end{array} |; Δ_{3} = | \begin{array}{ll} a & c \\ p & d \end{array} |$ the value of $(x, y)$
is

1

(\frac{Δ_{2}}{Δ_{1}}, \frac{Δ_{3}}{Δ_{1}})

2

(\log \frac{Δ_{2}}{Δ_{1}}, \log \frac{Δ_{3}}{Δ_{1}})

3

(\log \frac{Δ_{1}}{Δ_{3}}, \log \frac{Δ_{1}}{Δ_{2}})

4

(\log \frac{Δ_{1}}{Δ_{2}}, \log \frac{Δ_{1}}{Δ_{3}})

Explanation:

(B) : We have,
${ae}^{x} + {be}^{y} = c and {pe}^{x} + {qe}^{y} = d$
Here, by determinant method,
$e^{x} = \frac{Δ_{2}}{Δ_{1}} \Rightarrow x = \log \frac{Δ_{2}}{Δ_{1}}$
Also, $e^{y} = \frac{Δ_{3}}{Δ_{1}} \Rightarrow y = \log \frac{Δ_{3}}{Δ_{1}}$
$∴ (\log \frac{Δ_{2}}{Δ_{1}}, \log \frac{Δ_{3}}{Δ_{1}})$

SRM JEE-2018

Matrix and Determinant

78999 The value of $| \begin{array}{lll} 441 & 442 & 443 \\ 445 & 446 & 447 \\ 449 & 450 & 451 \end{array} |$ is :

1

441 \times 446 \times 4510

2 0

3

- 1

4

1

Explanation:

(B) : We have,
$| \begin{array}{lll} 441 & 442 & 443 \\ 445 & 446 & 447 \\ 449 & 450 & 451 \end{array} |$
Applying - $c_{3} \to c_{3} - c_{2}$
We get-
$c_{2} \to c_{2} - c_{1}$
$| \begin{array}{lll} 441 & 1 & 1 \\ 445 & 1 & 1 \\ 450 & 1 & 1 \end{array} | = 0 [∵$ Two column are identical $]$

Karnataka CET-2004

Matrix and Determinant

79000 If $x^{3} - 2 x^{2} - 9 x + 18 = 0$ and $A = | \begin{array}{lll} 1 & 2 & 5 \\ 4 & x & 6 \\ 7 & 8 & 9 \end{array} |$ then
the maximum value of $A$ is

1 96

2 36

3 24

4 120

Explanation:

(A) : Given,
$f (x) = x^{3} - 2 x^{2} + 9 x + 18 = 0$
$x^{2} (x - 2) - 9 (x - 2) = 0$
$(x - 2) (x^{2} - 9) = 0$
$(x - 2) (x + 3) (x - 3) = 0$
$x = 2, - 3, 3$
$A = | \begin{array}{lll} 1 & 2 & 3 \\ 4 & x & 6 \\ 7 & 8 & 9 \end{array} |$
$[\begin{array}{l} R_{2} : R_{2} - 4 R_{1} \\ R_{3} : R_{3} - 7 R_{1} \end{array}]$
$= | \begin{array}{ccc} 1 & 2 & 3 \\ 0 & x - 8 & 6 - 12 \\ 0 & 8 - 14 & 9 - 21 \end{array} |$
$A = | \begin{array}{ccc} 1 & 2 & 3 \\ 0 & x - 8 & - 6 \\ 0 & - 6 & - 12 \end{array} | = 3 | \begin{array}{ccc} 1 & 2 & 1 \\ 0 & x - 8 & - 2 \\ 1 & - 6 & - 4 \end{array} |$
$= 6 | \begin{array}{ccc} 1 & 2 & 1 \\ 0 & x - 8 & - 2 \\ 0 & - 3 & 2 \end{array} |$
$= 6 (- 2 x + 10)$
$= - 12 x + 60$
$A = - 12 x + 60$
of $x = 2 A = 36$
$x = 3 A = 24$
$x = - 3 A = 96$
(A) $max = 96$

Karnataka CET-2021

Matrix and Determinant

79001 If $A$ and $B$ are matrices of order 3 and $| A | = 5$ , $| B | = 3$ , then $| 3 A B |$ is

1 425

2 405

3 565

4 585

Explanation:

(B) : We have,
$| A | = 5 and | B | = 3$
We know that,
$| KA | = K^{n} | A | (where, n = order of matrix)$
$= 27 \times 5 \times 3$
$| 3 AB | = 405$
$∴ | 3 AB | = 3^{3} | A | | B |$

Karnataka CET-2021

Matrix and Determinant

79002 The system of linear equations $x + y + z = 6$ , $x + 2 y + 3 z = 10$ and $x + 2 y + a z = b$ has no solution when

1

a = 3, b \neq 10

2

b = 3, a \neq 10

3

a = 2, b \neq 3

4

b = 2, a = 3

Explanation:

(A) : We have,
$\begin{array}{l} x + y + z = 6 \\ x + 2 y + 3 z = 10 \\ x + 2 y + a z = b \end{array}}$
Equation (i) will have no solution when $D = 0$ and at least one of $D_{1}, D_{2}$ or $D_{3}$ is non-zero.
Here,
$D = | \begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & a \end{array} | = 0$
$1 (2 a - 6) - 1 (a - 3) + 1 (2 - 2) = 0$
$2 a - 6 - a + 3 = 0$
$a - 3 = 0$
$a = 3$
Let, $D_{1} \neq 0$ , then -
$| \begin{array}{ccc} 6 & 1 & 1 \\ 10 & 2 & 3 \\ b & 2 & a \end{array} | \neq 0$
$6 (2 a - 6) - 10 (a - 2) + b (3 - 2) \neq 0$
$2 a + b - 16 \neq 0$
$b - 10 \neq 0$
$b \neq 10$

Karnataka CET-2015

Matrix and Determinant

79003 If $a e^{x} + b e^{y} = c; p e^{x} + q e^{y} = d$ and
$Δ_{1} = | \begin{array}{ll} a & b \\ p & q \end{array} |; Δ_{2} = | \begin{array}{ll} c & b \\ d & q \end{array} |; Δ_{3} = | \begin{array}{ll} a & c \\ p & d \end{array} |$ the value of $(x, y)$
is

1

(\frac{Δ_{2}}{Δ_{1}}, \frac{Δ_{3}}{Δ_{1}})

2

(\log \frac{Δ_{2}}{Δ_{1}}, \log \frac{Δ_{3}}{Δ_{1}})

3

(\log \frac{Δ_{1}}{Δ_{3}}, \log \frac{Δ_{1}}{Δ_{2}})

4

(\log \frac{Δ_{1}}{Δ_{2}}, \log \frac{Δ_{1}}{Δ_{3}})

Explanation:

(B) : We have,
${ae}^{x} + {be}^{y} = c and {pe}^{x} + {qe}^{y} = d$
Here, by determinant method,
$e^{x} = \frac{Δ_{2}}{Δ_{1}} \Rightarrow x = \log \frac{Δ_{2}}{Δ_{1}}$
Also, $e^{y} = \frac{Δ_{3}}{Δ_{1}} \Rightarrow y = \log \frac{Δ_{3}}{Δ_{1}}$
$∴ (\log \frac{Δ_{2}}{Δ_{1}}, \log \frac{Δ_{3}}{Δ_{1}})$

SRM JEE-2018

Matrix and Determinant

78999 The value of $| \begin{array}{lll} 441 & 442 & 443 \\ 445 & 446 & 447 \\ 449 & 450 & 451 \end{array} |$ is :

1

441 \times 446 \times 4510

2 0

3

- 1

4

1

Explanation:

(B) : We have,
$| \begin{array}{lll} 441 & 442 & 443 \\ 445 & 446 & 447 \\ 449 & 450 & 451 \end{array} |$
Applying - $c_{3} \to c_{3} - c_{2}$
We get-
$c_{2} \to c_{2} - c_{1}$
$| \begin{array}{lll} 441 & 1 & 1 \\ 445 & 1 & 1 \\ 450 & 1 & 1 \end{array} | = 0 [∵$ Two column are identical $]$

Karnataka CET-2004

Matrix and Determinant

79000 If $x^{3} - 2 x^{2} - 9 x + 18 = 0$ and $A = | \begin{array}{lll} 1 & 2 & 5 \\ 4 & x & 6 \\ 7 & 8 & 9 \end{array} |$ then
the maximum value of $A$ is

1 96

2 36

3 24

4 120

Explanation:

(A) : Given,
$f (x) = x^{3} - 2 x^{2} + 9 x + 18 = 0$
$x^{2} (x - 2) - 9 (x - 2) = 0$
$(x - 2) (x^{2} - 9) = 0$
$(x - 2) (x + 3) (x - 3) = 0$
$x = 2, - 3, 3$
$A = | \begin{array}{lll} 1 & 2 & 3 \\ 4 & x & 6 \\ 7 & 8 & 9 \end{array} |$
$[\begin{array}{l} R_{2} : R_{2} - 4 R_{1} \\ R_{3} : R_{3} - 7 R_{1} \end{array}]$
$= | \begin{array}{ccc} 1 & 2 & 3 \\ 0 & x - 8 & 6 - 12 \\ 0 & 8 - 14 & 9 - 21 \end{array} |$
$A = | \begin{array}{ccc} 1 & 2 & 3 \\ 0 & x - 8 & - 6 \\ 0 & - 6 & - 12 \end{array} | = 3 | \begin{array}{ccc} 1 & 2 & 1 \\ 0 & x - 8 & - 2 \\ 1 & - 6 & - 4 \end{array} |$
$= 6 | \begin{array}{ccc} 1 & 2 & 1 \\ 0 & x - 8 & - 2 \\ 0 & - 3 & 2 \end{array} |$
$= 6 (- 2 x + 10)$
$= - 12 x + 60$
$A = - 12 x + 60$
of $x = 2 A = 36$
$x = 3 A = 24$
$x = - 3 A = 96$
(A) $max = 96$

Karnataka CET-2021

Matrix and Determinant

79001 If $A$ and $B$ are matrices of order 3 and $| A | = 5$ , $| B | = 3$ , then $| 3 A B |$ is

1 425

2 405

3 565

4 585

Explanation:

(B) : We have,
$| A | = 5 and | B | = 3$
We know that,
$| KA | = K^{n} | A | (where, n = order of matrix)$
$= 27 \times 5 \times 3$
$| 3 AB | = 405$
$∴ | 3 AB | = 3^{3} | A | | B |$

Karnataka CET-2021

Matrix and Determinant

79002 The system of linear equations $x + y + z = 6$ , $x + 2 y + 3 z = 10$ and $x + 2 y + a z = b$ has no solution when

1

a = 3, b \neq 10

2

b = 3, a \neq 10

3

a = 2, b \neq 3

4

b = 2, a = 3

Explanation:

(A) : We have,
$\begin{array}{l} x + y + z = 6 \\ x + 2 y + 3 z = 10 \\ x + 2 y + a z = b \end{array}}$
Equation (i) will have no solution when $D = 0$ and at least one of $D_{1}, D_{2}$ or $D_{3}$ is non-zero.
Here,
$D = | \begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & a \end{array} | = 0$
$1 (2 a - 6) - 1 (a - 3) + 1 (2 - 2) = 0$
$2 a - 6 - a + 3 = 0$
$a - 3 = 0$
$a = 3$
Let, $D_{1} \neq 0$ , then -
$| \begin{array}{ccc} 6 & 1 & 1 \\ 10 & 2 & 3 \\ b & 2 & a \end{array} | \neq 0$
$6 (2 a - 6) - 10 (a - 2) + b (3 - 2) \neq 0$
$2 a + b - 16 \neq 0$
$b - 10 \neq 0$
$b \neq 10$

Karnataka CET-2015

Matrix and Determinant

79003 If $a e^{x} + b e^{y} = c; p e^{x} + q e^{y} = d$ and
$Δ_{1} = | \begin{array}{ll} a & b \\ p & q \end{array} |; Δ_{2} = | \begin{array}{ll} c & b \\ d & q \end{array} |; Δ_{3} = | \begin{array}{ll} a & c \\ p & d \end{array} |$ the value of $(x, y)$
is

1

(\frac{Δ_{2}}{Δ_{1}}, \frac{Δ_{3}}{Δ_{1}})

2

(\log \frac{Δ_{2}}{Δ_{1}}, \log \frac{Δ_{3}}{Δ_{1}})

3

(\log \frac{Δ_{1}}{Δ_{3}}, \log \frac{Δ_{1}}{Δ_{2}})

4

(\log \frac{Δ_{1}}{Δ_{2}}, \log \frac{Δ_{1}}{Δ_{3}})

Explanation:

(B) : We have,
${ae}^{x} + {be}^{y} = c and {pe}^{x} + {qe}^{y} = d$
Here, by determinant method,
$e^{x} = \frac{Δ_{2}}{Δ_{1}} \Rightarrow x = \log \frac{Δ_{2}}{Δ_{1}}$
Also, $e^{y} = \frac{Δ_{3}}{Δ_{1}} \Rightarrow y = \log \frac{Δ_{3}}{Δ_{1}}$
$∴ (\log \frac{Δ_{2}}{Δ_{1}}, \log \frac{Δ_{3}}{Δ_{1}})$

SRM JEE-2018

Matrix and Determinant

78999 The value of $| \begin{array}{lll} 441 & 442 & 443 \\ 445 & 446 & 447 \\ 449 & 450 & 451 \end{array} |$ is :

1

441 \times 446 \times 4510

2 0

3

- 1

4

1

Explanation:

(B) : We have,
$| \begin{array}{lll} 441 & 442 & 443 \\ 445 & 446 & 447 \\ 449 & 450 & 451 \end{array} |$
Applying - $c_{3} \to c_{3} - c_{2}$
We get-
$c_{2} \to c_{2} - c_{1}$
$| \begin{array}{lll} 441 & 1 & 1 \\ 445 & 1 & 1 \\ 450 & 1 & 1 \end{array} | = 0 [∵$ Two column are identical $]$

Karnataka CET-2004

Matrix and Determinant

79000 If $x^{3} - 2 x^{2} - 9 x + 18 = 0$ and $A = | \begin{array}{lll} 1 & 2 & 5 \\ 4 & x & 6 \\ 7 & 8 & 9 \end{array} |$ then
the maximum value of $A$ is

1 96

2 36

3 24

4 120

Explanation:

(A) : Given,
$f (x) = x^{3} - 2 x^{2} + 9 x + 18 = 0$
$x^{2} (x - 2) - 9 (x - 2) = 0$
$(x - 2) (x^{2} - 9) = 0$
$(x - 2) (x + 3) (x - 3) = 0$
$x = 2, - 3, 3$
$A = | \begin{array}{lll} 1 & 2 & 3 \\ 4 & x & 6 \\ 7 & 8 & 9 \end{array} |$
$[\begin{array}{l} R_{2} : R_{2} - 4 R_{1} \\ R_{3} : R_{3} - 7 R_{1} \end{array}]$
$= | \begin{array}{ccc} 1 & 2 & 3 \\ 0 & x - 8 & 6 - 12 \\ 0 & 8 - 14 & 9 - 21 \end{array} |$
$A = | \begin{array}{ccc} 1 & 2 & 3 \\ 0 & x - 8 & - 6 \\ 0 & - 6 & - 12 \end{array} | = 3 | \begin{array}{ccc} 1 & 2 & 1 \\ 0 & x - 8 & - 2 \\ 1 & - 6 & - 4 \end{array} |$
$= 6 | \begin{array}{ccc} 1 & 2 & 1 \\ 0 & x - 8 & - 2 \\ 0 & - 3 & 2 \end{array} |$
$= 6 (- 2 x + 10)$
$= - 12 x + 60$
$A = - 12 x + 60$
of $x = 2 A = 36$
$x = 3 A = 24$
$x = - 3 A = 96$
(A) $max = 96$

Karnataka CET-2021

Matrix and Determinant

79001 If $A$ and $B$ are matrices of order 3 and $| A | = 5$ , $| B | = 3$ , then $| 3 A B |$ is

1 425

2 405

3 565

4 585

Explanation:

(B) : We have,
$| A | = 5 and | B | = 3$
We know that,
$| KA | = K^{n} | A | (where, n = order of matrix)$
$= 27 \times 5 \times 3$
$| 3 AB | = 405$
$∴ | 3 AB | = 3^{3} | A | | B |$

Karnataka CET-2021

Matrix and Determinant

79002 The system of linear equations $x + y + z = 6$ , $x + 2 y + 3 z = 10$ and $x + 2 y + a z = b$ has no solution when

1

a = 3, b \neq 10

2

b = 3, a \neq 10

3

a = 2, b \neq 3

4

b = 2, a = 3

Explanation:

(A) : We have,
$\begin{array}{l} x + y + z = 6 \\ x + 2 y + 3 z = 10 \\ x + 2 y + a z = b \end{array}}$
Equation (i) will have no solution when $D = 0$ and at least one of $D_{1}, D_{2}$ or $D_{3}$ is non-zero.
Here,
$D = | \begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & a \end{array} | = 0$
$1 (2 a - 6) - 1 (a - 3) + 1 (2 - 2) = 0$
$2 a - 6 - a + 3 = 0$
$a - 3 = 0$
$a = 3$
Let, $D_{1} \neq 0$ , then -
$| \begin{array}{ccc} 6 & 1 & 1 \\ 10 & 2 & 3 \\ b & 2 & a \end{array} | \neq 0$
$6 (2 a - 6) - 10 (a - 2) + b (3 - 2) \neq 0$
$2 a + b - 16 \neq 0$
$b - 10 \neq 0$
$b \neq 10$

Karnataka CET-2015

Matrix and Determinant

79003 If $a e^{x} + b e^{y} = c; p e^{x} + q e^{y} = d$ and
$Δ_{1} = | \begin{array}{ll} a & b \\ p & q \end{array} |; Δ_{2} = | \begin{array}{ll} c & b \\ d & q \end{array} |; Δ_{3} = | \begin{array}{ll} a & c \\ p & d \end{array} |$ the value of $(x, y)$
is

1

(\frac{Δ_{2}}{Δ_{1}}, \frac{Δ_{3}}{Δ_{1}})

2

(\log \frac{Δ_{2}}{Δ_{1}}, \log \frac{Δ_{3}}{Δ_{1}})

3

(\log \frac{Δ_{1}}{Δ_{3}}, \log \frac{Δ_{1}}{Δ_{2}})

4

(\log \frac{Δ_{1}}{Δ_{2}}, \log \frac{Δ_{1}}{Δ_{3}})

Explanation:

(B) : We have,
${ae}^{x} + {be}^{y} = c and {pe}^{x} + {qe}^{y} = d$
Here, by determinant method,
$e^{x} = \frac{Δ_{2}}{Δ_{1}} \Rightarrow x = \log \frac{Δ_{2}}{Δ_{1}}$
Also, $e^{y} = \frac{Δ_{3}}{Δ_{1}} \Rightarrow y = \log \frac{Δ_{3}}{Δ_{1}}$
$∴ (\log \frac{Δ_{2}}{Δ_{1}}, \log \frac{Δ_{3}}{Δ_{1}})$

SRM JEE-2018

Matrix and Determinant

78999 The value of $| \begin{array}{lll} 441 & 442 & 443 \\ 445 & 446 & 447 \\ 449 & 450 & 451 \end{array} |$ is :

1

441 \times 446 \times 4510

2 0

3

- 1

4

1

Explanation:

(B) : We have,
$| \begin{array}{lll} 441 & 442 & 443 \\ 445 & 446 & 447 \\ 449 & 450 & 451 \end{array} |$
Applying - $c_{3} \to c_{3} - c_{2}$
We get-
$c_{2} \to c_{2} - c_{1}$
$| \begin{array}{lll} 441 & 1 & 1 \\ 445 & 1 & 1 \\ 450 & 1 & 1 \end{array} | = 0 [∵$ Two column are identical $]$

Karnataka CET-2004

Matrix and Determinant

79000 If $x^{3} - 2 x^{2} - 9 x + 18 = 0$ and $A = | \begin{array}{lll} 1 & 2 & 5 \\ 4 & x & 6 \\ 7 & 8 & 9 \end{array} |$ then
the maximum value of $A$ is

1 96

2 36

3 24

4 120

Explanation:

(A) : Given,
$f (x) = x^{3} - 2 x^{2} + 9 x + 18 = 0$
$x^{2} (x - 2) - 9 (x - 2) = 0$
$(x - 2) (x^{2} - 9) = 0$
$(x - 2) (x + 3) (x - 3) = 0$
$x = 2, - 3, 3$
$A = | \begin{array}{lll} 1 & 2 & 3 \\ 4 & x & 6 \\ 7 & 8 & 9 \end{array} |$
$[\begin{array}{l} R_{2} : R_{2} - 4 R_{1} \\ R_{3} : R_{3} - 7 R_{1} \end{array}]$
$= | \begin{array}{ccc} 1 & 2 & 3 \\ 0 & x - 8 & 6 - 12 \\ 0 & 8 - 14 & 9 - 21 \end{array} |$
$A = | \begin{array}{ccc} 1 & 2 & 3 \\ 0 & x - 8 & - 6 \\ 0 & - 6 & - 12 \end{array} | = 3 | \begin{array}{ccc} 1 & 2 & 1 \\ 0 & x - 8 & - 2 \\ 1 & - 6 & - 4 \end{array} |$
$= 6 | \begin{array}{ccc} 1 & 2 & 1 \\ 0 & x - 8 & - 2 \\ 0 & - 3 & 2 \end{array} |$
$= 6 (- 2 x + 10)$
$= - 12 x + 60$
$A = - 12 x + 60$
of $x = 2 A = 36$
$x = 3 A = 24$
$x = - 3 A = 96$
(A) $max = 96$

Karnataka CET-2021

Matrix and Determinant

79001 If $A$ and $B$ are matrices of order 3 and $| A | = 5$ , $| B | = 3$ , then $| 3 A B |$ is

1 425

2 405

3 565

4 585

Explanation:

(B) : We have,
$| A | = 5 and | B | = 3$
We know that,
$| KA | = K^{n} | A | (where, n = order of matrix)$
$= 27 \times 5 \times 3$
$| 3 AB | = 405$
$∴ | 3 AB | = 3^{3} | A | | B |$

Karnataka CET-2021

Matrix and Determinant

79002 The system of linear equations $x + y + z = 6$ , $x + 2 y + 3 z = 10$ and $x + 2 y + a z = b$ has no solution when

1

a = 3, b \neq 10

2

b = 3, a \neq 10

3

a = 2, b \neq 3

4

b = 2, a = 3

Explanation:

(A) : We have,
$\begin{array}{l} x + y + z = 6 \\ x + 2 y + 3 z = 10 \\ x + 2 y + a z = b \end{array}}$
Equation (i) will have no solution when $D = 0$ and at least one of $D_{1}, D_{2}$ or $D_{3}$ is non-zero.
Here,
$D = | \begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & a \end{array} | = 0$
$1 (2 a - 6) - 1 (a - 3) + 1 (2 - 2) = 0$
$2 a - 6 - a + 3 = 0$
$a - 3 = 0$
$a = 3$
Let, $D_{1} \neq 0$ , then -
$| \begin{array}{ccc} 6 & 1 & 1 \\ 10 & 2 & 3 \\ b & 2 & a \end{array} | \neq 0$
$6 (2 a - 6) - 10 (a - 2) + b (3 - 2) \neq 0$
$2 a + b - 16 \neq 0$
$b - 10 \neq 0$
$b \neq 10$

Karnataka CET-2015

Matrix and Determinant

79003 If $a e^{x} + b e^{y} = c; p e^{x} + q e^{y} = d$ and
$Δ_{1} = | \begin{array}{ll} a & b \\ p & q \end{array} |; Δ_{2} = | \begin{array}{ll} c & b \\ d & q \end{array} |; Δ_{3} = | \begin{array}{ll} a & c \\ p & d \end{array} |$ the value of $(x, y)$
is

1

(\frac{Δ_{2}}{Δ_{1}}, \frac{Δ_{3}}{Δ_{1}})

2

(\log \frac{Δ_{2}}{Δ_{1}}, \log \frac{Δ_{3}}{Δ_{1}})

3

(\log \frac{Δ_{1}}{Δ_{3}}, \log \frac{Δ_{1}}{Δ_{2}})

4

(\log \frac{Δ_{1}}{Δ_{2}}, \log \frac{Δ_{1}}{Δ_{3}})

Explanation:

(B) : We have,
${ae}^{x} + {be}^{y} = c and {pe}^{x} + {qe}^{y} = d$
Here, by determinant method,
$e^{x} = \frac{Δ_{2}}{Δ_{1}} \Rightarrow x = \log \frac{Δ_{2}}{Δ_{1}}$
Also, $e^{y} = \frac{Δ_{3}}{Δ_{1}} \Rightarrow y = \log \frac{Δ_{3}}{Δ_{1}}$
$∴ (\log \frac{Δ_{2}}{Δ_{1}}, \log \frac{Δ_{3}}{Δ_{1}})$

SRM JEE-2018