QBManager

Matrix and Determinant

79080 If each element of a $3 \times 3$ matrix is multiplied by 3 , then the determinant of the newly formed matrix is

1

3 | A |

2

9 | A |

3

27 ∣ A

4

| A |^{3}

Explanation:

(C) : Given,
Each element of a $3 \times 3$ matrix is multiples by 3 .
$∵$ Therefore
We can have matrix
$= 3 \times 3 \times 3 | A | = 27 | A |$

Rajasthan PET-2010

Matrix and Determinant

79081 IF $A + B + C = π$ , then
$| \begin{array}{ccc} \sin (A + B + C) & \sin B & \cos C \\ - \sin B & 0 & \tan A \\ \cos (A + B) & - \tan A & 0 \end{array} |$ is equal to

1

\sin A

2

\sin A \cos B

3 0

4 None of these

Explanation:

(C) : Above is skew symmetric determinants of odd order because $\cos (A + B) = - \cos C$ and $\sin (A + B$ $+ C) = 0$ . Therefore its value is zero.
hence, option (c) is correct.

Manipal-2018

Matrix and Determinant

79082 The value of $| \begin{array}{ccc} a & a + b & a + 2 b \\ a + 2 b & a & a + b \\ a + b & a + 2 b & a \end{array} |$ is equal to

1

9 a^{2} (a + b)

2

9 b^{2} (a + b)

3

a^{2} (a + b)

4

b^{2} (a + b)

Explanation:

(B) : Given,
On applying $(R_{1} \to R_{1} + R_{2} + R_{3})$ we get,
$\begin{array}{r} = | \begin{array}{ccc} 3 (a + b) & 3 (a + b) & 3 (a + b) \\ a + 2 b & a & a + b \\ a + b & a + 2 b & a \end{array} | \\ = 3 (a + b) | \begin{array}{ccc} 1 & 1 & 1 \\ a + 2 b & a & a + b \\ a + b & a + 2 b & a \end{array} | \end{array}$
On applying $C_{2} \to C_{2} - C_{1}, C_{3} \to C_{3} - C_{1}$ we get,
$= 3 (a + b) | \begin{array}{ccc} 1 & 0 & 0 \\ a + 2 b & - 2 b & - b \\ a + 2 b & b & - b \end{array} |$
$= 3 (a + b) [1 (- 2 b (- b) - (- b) (b) - 0 + 0)]$
$= 3 (a + b) [2 b^{2} + b^{2}]$
$= 3 (a + b) (3 b^{2}) = 9 b^{2} (a + b)$

Manipal-2018

Matrix and Determinant

79083 $| \begin{array}{ccc} a^{2} + 1 & a b & a c \\ a b & b^{2} + 1 & b c \\ a c & c b & c^{2} + 1 \end{array} | =$

1

1 + b^{2} + c^{2}

2

a^{2} + b^{2} + c^{2}

3

1 + a^{2} + b^{2}

4

1 + a^{2} + b^{2} + c^{2}

Explanation:

(D) : Given,
$| \begin{array}{ccc} a^{2} + 1 & a b & a c \\ a b & b^{2} + 1 & b c \\ a c & c b & c^{2} + 1 \end{array} |$
$= \frac{1}{a b c} | \begin{array}{ccc} a (a^{2} + 1) & a^{2} b & a^{2} c \\ a b^{2} & b (b^{2} + 1) & b^{2} c \\ c^{2} a & c^{2} b & c (c^{2} + 1) \end{array} |$
On applying,
$(R_{1} \to aR R_{1}, R_{2} \to {bR}_{2}, R_{3} \to {cR}_{3})$
$= | \begin{array}{ccc} a^{2} + 1 & a^{2} & a^{2} \\ b^{2} & b^{2} + 1 & b^{2} \\ c^{2} & c^{2} & c^{2} + 1 \end{array} |$
On applying,
$(c_{1} \to \frac{1}{a} c_{1}, c_{2} \to \frac{1}{b} c_{2}, c_{3} \to \frac{1}{3} c_{3})$
And applying $R_{1} \to R_{1} + R_{2} + R_{3}$
$| \begin{array}{ccc} 1 + a^{2} + b^{2} + c^{2} & 1 + a^{2} + b^{2} + c^{2} & 1 + a^{2} + b^{2} + c^{2} \\ b^{2} & b^{2} + 1 & b^{2} \\ c^{2} & c^{2} & c^{2} + 1 \end{array} |$
$= (1 + a^{2} + b^{2} + c^{2}) | \begin{array}{ccc} 1 & 0 & 0 \\ b^{2} & 1 & 0 \\ c^{2} & 0 & 1 \end{array} |$
on applying,
$(c_{2} \to c_{2} - c_{1}, c_{3} \to c_{3} - c_{1})$
$= (1 + a^{2} + b^{2} + c^{2}) 1 [1 - 0]$
$= (1 + a^{2} + b^{2} + c^{2})$

Assam CEE-2022

Matrix and Determinant

79084 If $a + b + c = 0$ , then $a$ root of
$| \begin{array}{ccc} a - x & c & b \\ c & b - x & a \\ b & a & c - x \end{array} | = 0 is$

1 0

2 1

3

a^{2} + b^{2} + c^{2}

4 3

Explanation:

(A) : Given,
$a + b + c = 0$
$| \begin{array}{ccc} a - x & c & b \\ c & b - x & a \\ b & a & c - x \end{array} | = 0$
On applying,
$C_{1} \to C_{1} + C_{2} + C_{3}$
$| \begin{array}{lll} a + b + c - x & c & b \\ a + b + c - x & b - x & a \\ a + b + c - x & a & c - x \end{array} | = 0$
$= (a + b + c - x) | \begin{array}{lll} 1 & c & b \\ 1 & b - x & a \\ 1 & a & c - x \end{array} | = 0$
Given that,
So, $(0 - x) | \begin{array}{ccc} 1 & c & b \\ 1 & b - x & a \\ 1 & a & c - x \end{array} | = 0$
$x | \begin{array}{ccc} 1 & c & b \\ 1 & b - x & a \\ 1 & a & c - x \end{array} | = 0$
Hence, $x = 0$
$| \begin{array}{lll} p & b & c \end{array} |$

Assam CEE-2022

Matrix and Determinant

79080 If each element of a $3 \times 3$ matrix is multiplied by 3 , then the determinant of the newly formed matrix is

1

3 | A |

2

9 | A |

3

27 ∣ A

4

| A |^{3}

Explanation:

(C) : Given,
Each element of a $3 \times 3$ matrix is multiples by 3 .
$∵$ Therefore
We can have matrix
$= 3 \times 3 \times 3 | A | = 27 | A |$

Rajasthan PET-2010

Matrix and Determinant

79081 IF $A + B + C = π$ , then
$| \begin{array}{ccc} \sin (A + B + C) & \sin B & \cos C \\ - \sin B & 0 & \tan A \\ \cos (A + B) & - \tan A & 0 \end{array} |$ is equal to

1

\sin A

2

\sin A \cos B

3 0

4 None of these

Explanation:

(C) : Above is skew symmetric determinants of odd order because $\cos (A + B) = - \cos C$ and $\sin (A + B$ $+ C) = 0$ . Therefore its value is zero.
hence, option (c) is correct.

Manipal-2018

Matrix and Determinant

79082 The value of $| \begin{array}{ccc} a & a + b & a + 2 b \\ a + 2 b & a & a + b \\ a + b & a + 2 b & a \end{array} |$ is equal to

1

9 a^{2} (a + b)

2

9 b^{2} (a + b)

3

a^{2} (a + b)

4

b^{2} (a + b)

Explanation:

(B) : Given,
On applying $(R_{1} \to R_{1} + R_{2} + R_{3})$ we get,
$\begin{array}{r} = | \begin{array}{ccc} 3 (a + b) & 3 (a + b) & 3 (a + b) \\ a + 2 b & a & a + b \\ a + b & a + 2 b & a \end{array} | \\ = 3 (a + b) | \begin{array}{ccc} 1 & 1 & 1 \\ a + 2 b & a & a + b \\ a + b & a + 2 b & a \end{array} | \end{array}$
On applying $C_{2} \to C_{2} - C_{1}, C_{3} \to C_{3} - C_{1}$ we get,
$= 3 (a + b) | \begin{array}{ccc} 1 & 0 & 0 \\ a + 2 b & - 2 b & - b \\ a + 2 b & b & - b \end{array} |$
$= 3 (a + b) [1 (- 2 b (- b) - (- b) (b) - 0 + 0)]$
$= 3 (a + b) [2 b^{2} + b^{2}]$
$= 3 (a + b) (3 b^{2}) = 9 b^{2} (a + b)$

Manipal-2018

Matrix and Determinant

79083 $| \begin{array}{ccc} a^{2} + 1 & a b & a c \\ a b & b^{2} + 1 & b c \\ a c & c b & c^{2} + 1 \end{array} | =$

1

1 + b^{2} + c^{2}

2

a^{2} + b^{2} + c^{2}

3

1 + a^{2} + b^{2}

4

1 + a^{2} + b^{2} + c^{2}

Explanation:

(D) : Given,
$| \begin{array}{ccc} a^{2} + 1 & a b & a c \\ a b & b^{2} + 1 & b c \\ a c & c b & c^{2} + 1 \end{array} |$
$= \frac{1}{a b c} | \begin{array}{ccc} a (a^{2} + 1) & a^{2} b & a^{2} c \\ a b^{2} & b (b^{2} + 1) & b^{2} c \\ c^{2} a & c^{2} b & c (c^{2} + 1) \end{array} |$
On applying,
$(R_{1} \to aR R_{1}, R_{2} \to {bR}_{2}, R_{3} \to {cR}_{3})$
$= | \begin{array}{ccc} a^{2} + 1 & a^{2} & a^{2} \\ b^{2} & b^{2} + 1 & b^{2} \\ c^{2} & c^{2} & c^{2} + 1 \end{array} |$
On applying,
$(c_{1} \to \frac{1}{a} c_{1}, c_{2} \to \frac{1}{b} c_{2}, c_{3} \to \frac{1}{3} c_{3})$
And applying $R_{1} \to R_{1} + R_{2} + R_{3}$
$| \begin{array}{ccc} 1 + a^{2} + b^{2} + c^{2} & 1 + a^{2} + b^{2} + c^{2} & 1 + a^{2} + b^{2} + c^{2} \\ b^{2} & b^{2} + 1 & b^{2} \\ c^{2} & c^{2} & c^{2} + 1 \end{array} |$
$= (1 + a^{2} + b^{2} + c^{2}) | \begin{array}{ccc} 1 & 0 & 0 \\ b^{2} & 1 & 0 \\ c^{2} & 0 & 1 \end{array} |$
on applying,
$(c_{2} \to c_{2} - c_{1}, c_{3} \to c_{3} - c_{1})$
$= (1 + a^{2} + b^{2} + c^{2}) 1 [1 - 0]$
$= (1 + a^{2} + b^{2} + c^{2})$

Assam CEE-2022

Matrix and Determinant

79084 If $a + b + c = 0$ , then $a$ root of
$| \begin{array}{ccc} a - x & c & b \\ c & b - x & a \\ b & a & c - x \end{array} | = 0 is$

1 0

2 1

3

a^{2} + b^{2} + c^{2}

4 3

Explanation:

(A) : Given,
$a + b + c = 0$
$| \begin{array}{ccc} a - x & c & b \\ c & b - x & a \\ b & a & c - x \end{array} | = 0$
On applying,
$C_{1} \to C_{1} + C_{2} + C_{3}$
$| \begin{array}{lll} a + b + c - x & c & b \\ a + b + c - x & b - x & a \\ a + b + c - x & a & c - x \end{array} | = 0$
$= (a + b + c - x) | \begin{array}{lll} 1 & c & b \\ 1 & b - x & a \\ 1 & a & c - x \end{array} | = 0$
Given that,
So, $(0 - x) | \begin{array}{ccc} 1 & c & b \\ 1 & b - x & a \\ 1 & a & c - x \end{array} | = 0$
$x | \begin{array}{ccc} 1 & c & b \\ 1 & b - x & a \\ 1 & a & c - x \end{array} | = 0$
Hence, $x = 0$
$| \begin{array}{lll} p & b & c \end{array} |$

Assam CEE-2022

Matrix and Determinant

79080 If each element of a $3 \times 3$ matrix is multiplied by 3 , then the determinant of the newly formed matrix is

1

3 | A |

2

9 | A |

3

27 ∣ A

4

| A |^{3}

Explanation:

(C) : Given,
Each element of a $3 \times 3$ matrix is multiples by 3 .
$∵$ Therefore
We can have matrix
$= 3 \times 3 \times 3 | A | = 27 | A |$

Rajasthan PET-2010

Matrix and Determinant

79081 IF $A + B + C = π$ , then
$| \begin{array}{ccc} \sin (A + B + C) & \sin B & \cos C \\ - \sin B & 0 & \tan A \\ \cos (A + B) & - \tan A & 0 \end{array} |$ is equal to

1

\sin A

2

\sin A \cos B

3 0

4 None of these

Explanation:

(C) : Above is skew symmetric determinants of odd order because $\cos (A + B) = - \cos C$ and $\sin (A + B$ $+ C) = 0$ . Therefore its value is zero.
hence, option (c) is correct.

Manipal-2018

Matrix and Determinant

79082 The value of $| \begin{array}{ccc} a & a + b & a + 2 b \\ a + 2 b & a & a + b \\ a + b & a + 2 b & a \end{array} |$ is equal to

1

9 a^{2} (a + b)

2

9 b^{2} (a + b)

3

a^{2} (a + b)

4

b^{2} (a + b)

Explanation:

(B) : Given,
On applying $(R_{1} \to R_{1} + R_{2} + R_{3})$ we get,
$\begin{array}{r} = | \begin{array}{ccc} 3 (a + b) & 3 (a + b) & 3 (a + b) \\ a + 2 b & a & a + b \\ a + b & a + 2 b & a \end{array} | \\ = 3 (a + b) | \begin{array}{ccc} 1 & 1 & 1 \\ a + 2 b & a & a + b \\ a + b & a + 2 b & a \end{array} | \end{array}$
On applying $C_{2} \to C_{2} - C_{1}, C_{3} \to C_{3} - C_{1}$ we get,
$= 3 (a + b) | \begin{array}{ccc} 1 & 0 & 0 \\ a + 2 b & - 2 b & - b \\ a + 2 b & b & - b \end{array} |$
$= 3 (a + b) [1 (- 2 b (- b) - (- b) (b) - 0 + 0)]$
$= 3 (a + b) [2 b^{2} + b^{2}]$
$= 3 (a + b) (3 b^{2}) = 9 b^{2} (a + b)$

Manipal-2018

Matrix and Determinant

79083 $| \begin{array}{ccc} a^{2} + 1 & a b & a c \\ a b & b^{2} + 1 & b c \\ a c & c b & c^{2} + 1 \end{array} | =$

1

1 + b^{2} + c^{2}

2

a^{2} + b^{2} + c^{2}

3

1 + a^{2} + b^{2}

4

1 + a^{2} + b^{2} + c^{2}

Explanation:

(D) : Given,
$| \begin{array}{ccc} a^{2} + 1 & a b & a c \\ a b & b^{2} + 1 & b c \\ a c & c b & c^{2} + 1 \end{array} |$
$= \frac{1}{a b c} | \begin{array}{ccc} a (a^{2} + 1) & a^{2} b & a^{2} c \\ a b^{2} & b (b^{2} + 1) & b^{2} c \\ c^{2} a & c^{2} b & c (c^{2} + 1) \end{array} |$
On applying,
$(R_{1} \to aR R_{1}, R_{2} \to {bR}_{2}, R_{3} \to {cR}_{3})$
$= | \begin{array}{ccc} a^{2} + 1 & a^{2} & a^{2} \\ b^{2} & b^{2} + 1 & b^{2} \\ c^{2} & c^{2} & c^{2} + 1 \end{array} |$
On applying,
$(c_{1} \to \frac{1}{a} c_{1}, c_{2} \to \frac{1}{b} c_{2}, c_{3} \to \frac{1}{3} c_{3})$
And applying $R_{1} \to R_{1} + R_{2} + R_{3}$
$| \begin{array}{ccc} 1 + a^{2} + b^{2} + c^{2} & 1 + a^{2} + b^{2} + c^{2} & 1 + a^{2} + b^{2} + c^{2} \\ b^{2} & b^{2} + 1 & b^{2} \\ c^{2} & c^{2} & c^{2} + 1 \end{array} |$
$= (1 + a^{2} + b^{2} + c^{2}) | \begin{array}{ccc} 1 & 0 & 0 \\ b^{2} & 1 & 0 \\ c^{2} & 0 & 1 \end{array} |$
on applying,
$(c_{2} \to c_{2} - c_{1}, c_{3} \to c_{3} - c_{1})$
$= (1 + a^{2} + b^{2} + c^{2}) 1 [1 - 0]$
$= (1 + a^{2} + b^{2} + c^{2})$

Assam CEE-2022

Matrix and Determinant

79084 If $a + b + c = 0$ , then $a$ root of
$| \begin{array}{ccc} a - x & c & b \\ c & b - x & a \\ b & a & c - x \end{array} | = 0 is$

1 0

2 1

3

a^{2} + b^{2} + c^{2}

4 3

Explanation:

(A) : Given,
$a + b + c = 0$
$| \begin{array}{ccc} a - x & c & b \\ c & b - x & a \\ b & a & c - x \end{array} | = 0$
On applying,
$C_{1} \to C_{1} + C_{2} + C_{3}$
$| \begin{array}{lll} a + b + c - x & c & b \\ a + b + c - x & b - x & a \\ a + b + c - x & a & c - x \end{array} | = 0$
$= (a + b + c - x) | \begin{array}{lll} 1 & c & b \\ 1 & b - x & a \\ 1 & a & c - x \end{array} | = 0$
Given that,
So, $(0 - x) | \begin{array}{ccc} 1 & c & b \\ 1 & b - x & a \\ 1 & a & c - x \end{array} | = 0$
$x | \begin{array}{ccc} 1 & c & b \\ 1 & b - x & a \\ 1 & a & c - x \end{array} | = 0$
Hence, $x = 0$
$| \begin{array}{lll} p & b & c \end{array} |$

Assam CEE-2022

Matrix and Determinant

79080 If each element of a $3 \times 3$ matrix is multiplied by 3 , then the determinant of the newly formed matrix is

1

3 | A |

2

9 | A |

3

27 ∣ A

4

| A |^{3}

Explanation:

(C) : Given,
Each element of a $3 \times 3$ matrix is multiples by 3 .
$∵$ Therefore
We can have matrix
$= 3 \times 3 \times 3 | A | = 27 | A |$

Rajasthan PET-2010

Matrix and Determinant

79081 IF $A + B + C = π$ , then
$| \begin{array}{ccc} \sin (A + B + C) & \sin B & \cos C \\ - \sin B & 0 & \tan A \\ \cos (A + B) & - \tan A & 0 \end{array} |$ is equal to

1

\sin A

2

\sin A \cos B

3 0

4 None of these

Explanation:

(C) : Above is skew symmetric determinants of odd order because $\cos (A + B) = - \cos C$ and $\sin (A + B$ $+ C) = 0$ . Therefore its value is zero.
hence, option (c) is correct.

Manipal-2018

Matrix and Determinant

79082 The value of $| \begin{array}{ccc} a & a + b & a + 2 b \\ a + 2 b & a & a + b \\ a + b & a + 2 b & a \end{array} |$ is equal to

1

9 a^{2} (a + b)

2

9 b^{2} (a + b)

3

a^{2} (a + b)

4

b^{2} (a + b)

Explanation:

(B) : Given,
On applying $(R_{1} \to R_{1} + R_{2} + R_{3})$ we get,
$\begin{array}{r} = | \begin{array}{ccc} 3 (a + b) & 3 (a + b) & 3 (a + b) \\ a + 2 b & a & a + b \\ a + b & a + 2 b & a \end{array} | \\ = 3 (a + b) | \begin{array}{ccc} 1 & 1 & 1 \\ a + 2 b & a & a + b \\ a + b & a + 2 b & a \end{array} | \end{array}$
On applying $C_{2} \to C_{2} - C_{1}, C_{3} \to C_{3} - C_{1}$ we get,
$= 3 (a + b) | \begin{array}{ccc} 1 & 0 & 0 \\ a + 2 b & - 2 b & - b \\ a + 2 b & b & - b \end{array} |$
$= 3 (a + b) [1 (- 2 b (- b) - (- b) (b) - 0 + 0)]$
$= 3 (a + b) [2 b^{2} + b^{2}]$
$= 3 (a + b) (3 b^{2}) = 9 b^{2} (a + b)$

Manipal-2018

Matrix and Determinant

79083 $| \begin{array}{ccc} a^{2} + 1 & a b & a c \\ a b & b^{2} + 1 & b c \\ a c & c b & c^{2} + 1 \end{array} | =$

1

1 + b^{2} + c^{2}

2

a^{2} + b^{2} + c^{2}

3

1 + a^{2} + b^{2}

4

1 + a^{2} + b^{2} + c^{2}

Explanation:

(D) : Given,
$| \begin{array}{ccc} a^{2} + 1 & a b & a c \\ a b & b^{2} + 1 & b c \\ a c & c b & c^{2} + 1 \end{array} |$
$= \frac{1}{a b c} | \begin{array}{ccc} a (a^{2} + 1) & a^{2} b & a^{2} c \\ a b^{2} & b (b^{2} + 1) & b^{2} c \\ c^{2} a & c^{2} b & c (c^{2} + 1) \end{array} |$
On applying,
$(R_{1} \to aR R_{1}, R_{2} \to {bR}_{2}, R_{3} \to {cR}_{3})$
$= | \begin{array}{ccc} a^{2} + 1 & a^{2} & a^{2} \\ b^{2} & b^{2} + 1 & b^{2} \\ c^{2} & c^{2} & c^{2} + 1 \end{array} |$
On applying,
$(c_{1} \to \frac{1}{a} c_{1}, c_{2} \to \frac{1}{b} c_{2}, c_{3} \to \frac{1}{3} c_{3})$
And applying $R_{1} \to R_{1} + R_{2} + R_{3}$
$| \begin{array}{ccc} 1 + a^{2} + b^{2} + c^{2} & 1 + a^{2} + b^{2} + c^{2} & 1 + a^{2} + b^{2} + c^{2} \\ b^{2} & b^{2} + 1 & b^{2} \\ c^{2} & c^{2} & c^{2} + 1 \end{array} |$
$= (1 + a^{2} + b^{2} + c^{2}) | \begin{array}{ccc} 1 & 0 & 0 \\ b^{2} & 1 & 0 \\ c^{2} & 0 & 1 \end{array} |$
on applying,
$(c_{2} \to c_{2} - c_{1}, c_{3} \to c_{3} - c_{1})$
$= (1 + a^{2} + b^{2} + c^{2}) 1 [1 - 0]$
$= (1 + a^{2} + b^{2} + c^{2})$

Assam CEE-2022

Matrix and Determinant

79084 If $a + b + c = 0$ , then $a$ root of
$| \begin{array}{ccc} a - x & c & b \\ c & b - x & a \\ b & a & c - x \end{array} | = 0 is$

1 0

2 1

3

a^{2} + b^{2} + c^{2}

4 3

Explanation:

(A) : Given,
$a + b + c = 0$
$| \begin{array}{ccc} a - x & c & b \\ c & b - x & a \\ b & a & c - x \end{array} | = 0$
On applying,
$C_{1} \to C_{1} + C_{2} + C_{3}$
$| \begin{array}{lll} a + b + c - x & c & b \\ a + b + c - x & b - x & a \\ a + b + c - x & a & c - x \end{array} | = 0$
$= (a + b + c - x) | \begin{array}{lll} 1 & c & b \\ 1 & b - x & a \\ 1 & a & c - x \end{array} | = 0$
Given that,
So, $(0 - x) | \begin{array}{ccc} 1 & c & b \\ 1 & b - x & a \\ 1 & a & c - x \end{array} | = 0$
$x | \begin{array}{ccc} 1 & c & b \\ 1 & b - x & a \\ 1 & a & c - x \end{array} | = 0$
Hence, $x = 0$
$| \begin{array}{lll} p & b & c \end{array} |$

Assam CEE-2022

Matrix and Determinant

79080 If each element of a $3 \times 3$ matrix is multiplied by 3 , then the determinant of the newly formed matrix is

1

3 | A |

2

9 | A |

3

27 ∣ A

4

| A |^{3}

Explanation:

(C) : Given,
Each element of a $3 \times 3$ matrix is multiples by 3 .
$∵$ Therefore
We can have matrix
$= 3 \times 3 \times 3 | A | = 27 | A |$

Rajasthan PET-2010

Matrix and Determinant

79081 IF $A + B + C = π$ , then
$| \begin{array}{ccc} \sin (A + B + C) & \sin B & \cos C \\ - \sin B & 0 & \tan A \\ \cos (A + B) & - \tan A & 0 \end{array} |$ is equal to

1

\sin A

2

\sin A \cos B

3 0

4 None of these

Explanation:

(C) : Above is skew symmetric determinants of odd order because $\cos (A + B) = - \cos C$ and $\sin (A + B$ $+ C) = 0$ . Therefore its value is zero.
hence, option (c) is correct.

Manipal-2018

Matrix and Determinant

79082 The value of $| \begin{array}{ccc} a & a + b & a + 2 b \\ a + 2 b & a & a + b \\ a + b & a + 2 b & a \end{array} |$ is equal to

1

9 a^{2} (a + b)

2

9 b^{2} (a + b)

3

a^{2} (a + b)

4

b^{2} (a + b)

Explanation:

(B) : Given,
On applying $(R_{1} \to R_{1} + R_{2} + R_{3})$ we get,
$\begin{array}{r} = | \begin{array}{ccc} 3 (a + b) & 3 (a + b) & 3 (a + b) \\ a + 2 b & a & a + b \\ a + b & a + 2 b & a \end{array} | \\ = 3 (a + b) | \begin{array}{ccc} 1 & 1 & 1 \\ a + 2 b & a & a + b \\ a + b & a + 2 b & a \end{array} | \end{array}$
On applying $C_{2} \to C_{2} - C_{1}, C_{3} \to C_{3} - C_{1}$ we get,
$= 3 (a + b) | \begin{array}{ccc} 1 & 0 & 0 \\ a + 2 b & - 2 b & - b \\ a + 2 b & b & - b \end{array} |$
$= 3 (a + b) [1 (- 2 b (- b) - (- b) (b) - 0 + 0)]$
$= 3 (a + b) [2 b^{2} + b^{2}]$
$= 3 (a + b) (3 b^{2}) = 9 b^{2} (a + b)$

Manipal-2018

Matrix and Determinant

79083 $| \begin{array}{ccc} a^{2} + 1 & a b & a c \\ a b & b^{2} + 1 & b c \\ a c & c b & c^{2} + 1 \end{array} | =$

1

1 + b^{2} + c^{2}

2

a^{2} + b^{2} + c^{2}

3

1 + a^{2} + b^{2}

4

1 + a^{2} + b^{2} + c^{2}

Explanation:

(D) : Given,
$| \begin{array}{ccc} a^{2} + 1 & a b & a c \\ a b & b^{2} + 1 & b c \\ a c & c b & c^{2} + 1 \end{array} |$
$= \frac{1}{a b c} | \begin{array}{ccc} a (a^{2} + 1) & a^{2} b & a^{2} c \\ a b^{2} & b (b^{2} + 1) & b^{2} c \\ c^{2} a & c^{2} b & c (c^{2} + 1) \end{array} |$
On applying,
$(R_{1} \to aR R_{1}, R_{2} \to {bR}_{2}, R_{3} \to {cR}_{3})$
$= | \begin{array}{ccc} a^{2} + 1 & a^{2} & a^{2} \\ b^{2} & b^{2} + 1 & b^{2} \\ c^{2} & c^{2} & c^{2} + 1 \end{array} |$
On applying,
$(c_{1} \to \frac{1}{a} c_{1}, c_{2} \to \frac{1}{b} c_{2}, c_{3} \to \frac{1}{3} c_{3})$
And applying $R_{1} \to R_{1} + R_{2} + R_{3}$
$| \begin{array}{ccc} 1 + a^{2} + b^{2} + c^{2} & 1 + a^{2} + b^{2} + c^{2} & 1 + a^{2} + b^{2} + c^{2} \\ b^{2} & b^{2} + 1 & b^{2} \\ c^{2} & c^{2} & c^{2} + 1 \end{array} |$
$= (1 + a^{2} + b^{2} + c^{2}) | \begin{array}{ccc} 1 & 0 & 0 \\ b^{2} & 1 & 0 \\ c^{2} & 0 & 1 \end{array} |$
on applying,
$(c_{2} \to c_{2} - c_{1}, c_{3} \to c_{3} - c_{1})$
$= (1 + a^{2} + b^{2} + c^{2}) 1 [1 - 0]$
$= (1 + a^{2} + b^{2} + c^{2})$

Assam CEE-2022

Matrix and Determinant

79084 If $a + b + c = 0$ , then $a$ root of
$| \begin{array}{ccc} a - x & c & b \\ c & b - x & a \\ b & a & c - x \end{array} | = 0 is$

1 0

2 1

3

a^{2} + b^{2} + c^{2}

4 3

Explanation:

(A) : Given,
$a + b + c = 0$
$| \begin{array}{ccc} a - x & c & b \\ c & b - x & a \\ b & a & c - x \end{array} | = 0$
On applying,
$C_{1} \to C_{1} + C_{2} + C_{3}$
$| \begin{array}{lll} a + b + c - x & c & b \\ a + b + c - x & b - x & a \\ a + b + c - x & a & c - x \end{array} | = 0$
$= (a + b + c - x) | \begin{array}{lll} 1 & c & b \\ 1 & b - x & a \\ 1 & a & c - x \end{array} | = 0$
Given that,
So, $(0 - x) | \begin{array}{ccc} 1 & c & b \\ 1 & b - x & a \\ 1 & a & c - x \end{array} | = 0$
$x | \begin{array}{ccc} 1 & c & b \\ 1 & b - x & a \\ 1 & a & c - x \end{array} | = 0$
Hence, $x = 0$
$| \begin{array}{lll} p & b & c \end{array} |$

Assam CEE-2022

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