QBManager

Matrix and Determinant

78637 If $A$ and $B$ are two matrices such that $A B = B$ and $B A = A$ then $A^{2} + B^{2} =$

1

2 AB

2

2 BA

3

A + B

4

AB

Explanation:

(C) : Given that,
$A$ and $B$ are two matrices and $AB = B$ and $BA = A$
Now, $AB = B$
On multiply by $A$ on both the side we get -
$\begin{aligned} (AB) A & = BA \\ A (BA) & = BA \\ A (A) & = A \\ A^{2} & = A \end{aligned}$
And $AB = B$
On multiply by $B$ on both the side we get -
$\begin{aligned} B (AB) & = BB \\ (BA) B & = B^{2} & [∵ BA = A] \\ AB & = B^{2} & [∵ AB = B] \\ B & = B^{2} \end{aligned}$
Hence, $A^{2} + B^{2} = A + B$

Karnataka CET-2000

Matrix and Determinant

78638 $G = {(\begin{array}{cc} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{array}) : θ \in R}$ is a group under matrix multiplication. Then which one of the following statements in respect of $G$ is true.

1

(\begin{array}{cc} - 1 & 0 \\ 0 & - 1 \end{array})

is the inverse of itself

2

G

is a finite group

3

(\begin{array}{cc} 1 / 2 & - \sqrt{3} / 2 \\ \sqrt{3} / 2 & 1 / 2 \end{array})

is not an element of

G

4

(\begin{array}{cc} 1 & 1 \\ - 1 & 1 \end{array})

is an element of G.

Explanation:

(A) : Given that,
$G = (\begin{array}{cc} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{array})$
$∵ θ \in R$
For option (a), If the matrix is the inverse of itself when multiplied by itself it should yield identity matrix.
$= [\begin{array}{cc} - 1 & 0 \\ 0 & - 1 \end{array}] [\begin{array}{cc} - 1 & 0 \\ 0 & - 1 \end{array}] = [\begin{array}{ll} 1 + 0 & 0 + 0 \\ 0 + 0 & 0 + 1 \end{array}] = [\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}]$
Hence, option (a) true.

Karnataka CET-2002

Matrix and Determinant

78639 If $A = [\begin{array}{ccc} 1 & - 2 & 1 \\ 2 & 1 & 3 \end{array}], B = [\begin{array}{ll} 2 & 1 \\ 3 & 2 \\ 1 & 1 \end{array}]$ then $(A B)^{T}$ is equal to

1

[\begin{array}{ll} - 3 & - 2 \\ 10 & 7 \end{array}]

2

[\begin{array}{cc} - 3 & 10 \\ - 2 & 7 \end{array}]

3

[\begin{array}{cc} - 3 & 7 \\ 10 & 2 \end{array}]

4

[\begin{array}{cc} - 3 & 7 \\ 10 & - 2 \end{array}]

Explanation:

(B) : Given that,
$A = [\begin{array}{ccc} 1 & - 2 & 1 \\ 2 & 1 & 3 \end{array}] and B = [\begin{array}{ll} 2 & 1 \\ 3 & 2 \\ 1 & 1 \end{array}]$
Then, $A B = [\begin{array}{ccc} 1 & - 2 & 1 \\ 2 & 1 & 3 \end{array}] [\begin{array}{ll} 2 & 1 \\ 3 & 2 \\ 1 & 1 \end{array}]$
$AB = [\begin{array}{cc} 2 - 6 + 1 & 1 - 4 + 1 \\ 4 + 3 + 3 & 2 + 2 + 3 \end{array}] \Rightarrow AB = [\begin{array}{cc} - 3 & - 2 \\ 10 & 7 \end{array}]$
So, $(AB)^{T} = [\begin{array}{cc} - 3 & 10 \\ - 2 & 7 \end{array}]$

Karnataka CET

Matrix and Determinant

78641 If $(x_{1}, y_{1}), (y_{2}, y_{2})$ and $(x_{3}, y_{3})$ are the vertices of a triangle whose area is $k$ square units, then
${| \begin{array}{lll} x_{1} & y_{1} & 4 \\ x_{2} & y_{2} & 4 \\ x_{3} & y_{3} & 4 \end{array} |}^{2} is$

1

32 k^{2}

2

16 k^{2}

3

64 k^{2}

4

48 k^{2}

Explanation:

(C) : We know that,
Area of triangle $= \frac{1}{2} [\begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}] = k$
From the given question,
${[\begin{array}{lll} x_{1} & y_{1} & 4 \\ x_{2} & y_{2} & 4 \\ x_{3} & y_{3} & 4 \end{array}]}^{2} = 4^{2} {[\begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}]}^{2}$
$= 16 (2 k)^{2} = 64 k^{2}$

Karnataka CET-2018

Matrix and Determinant

78642 If $2 [\begin{array}{ll} 1 & 3 \\ 0 & x \end{array}] + [\begin{array}{ll} y & 0 \\ 1 & 2 \end{array}] = [\begin{array}{ll} 5 & 6 \\ 1 & 8 \end{array}]$ then the value of $x$

1

x = 3, y = 3

2

x = - 3, y = 3

3

x = 3, y = - 3

4

x = - 3, y = - 3

Explanation:

(A) : Given that,
$2 [\begin{array}{ll} 1 & 3 \\ 0 & x \end{array}] + [\begin{array}{ll} y & 0 \\ 1 & 2 \end{array}] = [\begin{array}{ll} 5 & 6 \\ 1 & 8 \end{array}]$
$[\begin{array}{cc} 2 & 6 \\ 0 & 2 x \end{array}] + [\begin{array}{cc} y & 0 \\ 1 & 2 \end{array}] = [\begin{array}{cc} 5 & 6 \\ 1 & 8 \end{array}] \Rightarrow [\begin{array}{cc} 2 + y & 6 + 0 \\ 0 + 1 & 2 x + 2 \end{array}] = [\begin{array}{cc} 5 & 6 \\ 1 & 8 \end{array}]$
On comparing corresponding elements on both the side we get -
$\begin{array}{ll} 2 + y = 5, & 2 x + 2 = 8 \\ y = 5 - 2, & 2 x = 6 \\ y = 3 & x = 3 \end{array}$

Karnataka CET-2017

Matrix and Determinant

78637 If $A$ and $B$ are two matrices such that $A B = B$ and $B A = A$ then $A^{2} + B^{2} =$

1

2 AB

2

2 BA

3

A + B

4

AB

Explanation:

(C) : Given that,
$A$ and $B$ are two matrices and $AB = B$ and $BA = A$
Now, $AB = B$
On multiply by $A$ on both the side we get -
$\begin{aligned} (AB) A & = BA \\ A (BA) & = BA \\ A (A) & = A \\ A^{2} & = A \end{aligned}$
And $AB = B$
On multiply by $B$ on both the side we get -
$\begin{aligned} B (AB) & = BB \\ (BA) B & = B^{2} & [∵ BA = A] \\ AB & = B^{2} & [∵ AB = B] \\ B & = B^{2} \end{aligned}$
Hence, $A^{2} + B^{2} = A + B$

Karnataka CET-2000

Matrix and Determinant

78638 $G = {(\begin{array}{cc} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{array}) : θ \in R}$ is a group under matrix multiplication. Then which one of the following statements in respect of $G$ is true.

1

(\begin{array}{cc} - 1 & 0 \\ 0 & - 1 \end{array})

is the inverse of itself

2

G

is a finite group

3

(\begin{array}{cc} 1 / 2 & - \sqrt{3} / 2 \\ \sqrt{3} / 2 & 1 / 2 \end{array})

is not an element of

G

4

(\begin{array}{cc} 1 & 1 \\ - 1 & 1 \end{array})

is an element of G.

Explanation:

(A) : Given that,
$G = (\begin{array}{cc} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{array})$
$∵ θ \in R$
For option (a), If the matrix is the inverse of itself when multiplied by itself it should yield identity matrix.
$= [\begin{array}{cc} - 1 & 0 \\ 0 & - 1 \end{array}] [\begin{array}{cc} - 1 & 0 \\ 0 & - 1 \end{array}] = [\begin{array}{ll} 1 + 0 & 0 + 0 \\ 0 + 0 & 0 + 1 \end{array}] = [\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}]$
Hence, option (a) true.

Karnataka CET-2002

Matrix and Determinant

78639 If $A = [\begin{array}{ccc} 1 & - 2 & 1 \\ 2 & 1 & 3 \end{array}], B = [\begin{array}{ll} 2 & 1 \\ 3 & 2 \\ 1 & 1 \end{array}]$ then $(A B)^{T}$ is equal to

1

[\begin{array}{ll} - 3 & - 2 \\ 10 & 7 \end{array}]

2

[\begin{array}{cc} - 3 & 10 \\ - 2 & 7 \end{array}]

3

[\begin{array}{cc} - 3 & 7 \\ 10 & 2 \end{array}]

4

[\begin{array}{cc} - 3 & 7 \\ 10 & - 2 \end{array}]

Explanation:

(B) : Given that,
$A = [\begin{array}{ccc} 1 & - 2 & 1 \\ 2 & 1 & 3 \end{array}] and B = [\begin{array}{ll} 2 & 1 \\ 3 & 2 \\ 1 & 1 \end{array}]$
Then, $A B = [\begin{array}{ccc} 1 & - 2 & 1 \\ 2 & 1 & 3 \end{array}] [\begin{array}{ll} 2 & 1 \\ 3 & 2 \\ 1 & 1 \end{array}]$
$AB = [\begin{array}{cc} 2 - 6 + 1 & 1 - 4 + 1 \\ 4 + 3 + 3 & 2 + 2 + 3 \end{array}] \Rightarrow AB = [\begin{array}{cc} - 3 & - 2 \\ 10 & 7 \end{array}]$
So, $(AB)^{T} = [\begin{array}{cc} - 3 & 10 \\ - 2 & 7 \end{array}]$

Karnataka CET

Matrix and Determinant

78641 If $(x_{1}, y_{1}), (y_{2}, y_{2})$ and $(x_{3}, y_{3})$ are the vertices of a triangle whose area is $k$ square units, then
${| \begin{array}{lll} x_{1} & y_{1} & 4 \\ x_{2} & y_{2} & 4 \\ x_{3} & y_{3} & 4 \end{array} |}^{2} is$

1

32 k^{2}

2

16 k^{2}

3

64 k^{2}

4

48 k^{2}

Explanation:

(C) : We know that,
Area of triangle $= \frac{1}{2} [\begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}] = k$
From the given question,
${[\begin{array}{lll} x_{1} & y_{1} & 4 \\ x_{2} & y_{2} & 4 \\ x_{3} & y_{3} & 4 \end{array}]}^{2} = 4^{2} {[\begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}]}^{2}$
$= 16 (2 k)^{2} = 64 k^{2}$

Karnataka CET-2018

Matrix and Determinant

78642 If $2 [\begin{array}{ll} 1 & 3 \\ 0 & x \end{array}] + [\begin{array}{ll} y & 0 \\ 1 & 2 \end{array}] = [\begin{array}{ll} 5 & 6 \\ 1 & 8 \end{array}]$ then the value of $x$

1

x = 3, y = 3

2

x = - 3, y = 3

3

x = 3, y = - 3

4

x = - 3, y = - 3

Explanation:

(A) : Given that,
$2 [\begin{array}{ll} 1 & 3 \\ 0 & x \end{array}] + [\begin{array}{ll} y & 0 \\ 1 & 2 \end{array}] = [\begin{array}{ll} 5 & 6 \\ 1 & 8 \end{array}]$
$[\begin{array}{cc} 2 & 6 \\ 0 & 2 x \end{array}] + [\begin{array}{cc} y & 0 \\ 1 & 2 \end{array}] = [\begin{array}{cc} 5 & 6 \\ 1 & 8 \end{array}] \Rightarrow [\begin{array}{cc} 2 + y & 6 + 0 \\ 0 + 1 & 2 x + 2 \end{array}] = [\begin{array}{cc} 5 & 6 \\ 1 & 8 \end{array}]$
On comparing corresponding elements on both the side we get -
$\begin{array}{ll} 2 + y = 5, & 2 x + 2 = 8 \\ y = 5 - 2, & 2 x = 6 \\ y = 3 & x = 3 \end{array}$

Karnataka CET-2017

Matrix and Determinant

78637 If $A$ and $B$ are two matrices such that $A B = B$ and $B A = A$ then $A^{2} + B^{2} =$

1

2 AB

2

2 BA

3

A + B

4

AB

Explanation:

(C) : Given that,
$A$ and $B$ are two matrices and $AB = B$ and $BA = A$
Now, $AB = B$
On multiply by $A$ on both the side we get -
$\begin{aligned} (AB) A & = BA \\ A (BA) & = BA \\ A (A) & = A \\ A^{2} & = A \end{aligned}$
And $AB = B$
On multiply by $B$ on both the side we get -
$\begin{aligned} B (AB) & = BB \\ (BA) B & = B^{2} & [∵ BA = A] \\ AB & = B^{2} & [∵ AB = B] \\ B & = B^{2} \end{aligned}$
Hence, $A^{2} + B^{2} = A + B$

Karnataka CET-2000

Matrix and Determinant

78638 $G = {(\begin{array}{cc} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{array}) : θ \in R}$ is a group under matrix multiplication. Then which one of the following statements in respect of $G$ is true.

1

(\begin{array}{cc} - 1 & 0 \\ 0 & - 1 \end{array})

is the inverse of itself

2

G

is a finite group

3

(\begin{array}{cc} 1 / 2 & - \sqrt{3} / 2 \\ \sqrt{3} / 2 & 1 / 2 \end{array})

is not an element of

G

4

(\begin{array}{cc} 1 & 1 \\ - 1 & 1 \end{array})

is an element of G.

Explanation:

(A) : Given that,
$G = (\begin{array}{cc} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{array})$
$∵ θ \in R$
For option (a), If the matrix is the inverse of itself when multiplied by itself it should yield identity matrix.
$= [\begin{array}{cc} - 1 & 0 \\ 0 & - 1 \end{array}] [\begin{array}{cc} - 1 & 0 \\ 0 & - 1 \end{array}] = [\begin{array}{ll} 1 + 0 & 0 + 0 \\ 0 + 0 & 0 + 1 \end{array}] = [\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}]$
Hence, option (a) true.

Karnataka CET-2002

Matrix and Determinant

78639 If $A = [\begin{array}{ccc} 1 & - 2 & 1 \\ 2 & 1 & 3 \end{array}], B = [\begin{array}{ll} 2 & 1 \\ 3 & 2 \\ 1 & 1 \end{array}]$ then $(A B)^{T}$ is equal to

1

[\begin{array}{ll} - 3 & - 2 \\ 10 & 7 \end{array}]

2

[\begin{array}{cc} - 3 & 10 \\ - 2 & 7 \end{array}]

3

[\begin{array}{cc} - 3 & 7 \\ 10 & 2 \end{array}]

4

[\begin{array}{cc} - 3 & 7 \\ 10 & - 2 \end{array}]

Explanation:

(B) : Given that,
$A = [\begin{array}{ccc} 1 & - 2 & 1 \\ 2 & 1 & 3 \end{array}] and B = [\begin{array}{ll} 2 & 1 \\ 3 & 2 \\ 1 & 1 \end{array}]$
Then, $A B = [\begin{array}{ccc} 1 & - 2 & 1 \\ 2 & 1 & 3 \end{array}] [\begin{array}{ll} 2 & 1 \\ 3 & 2 \\ 1 & 1 \end{array}]$
$AB = [\begin{array}{cc} 2 - 6 + 1 & 1 - 4 + 1 \\ 4 + 3 + 3 & 2 + 2 + 3 \end{array}] \Rightarrow AB = [\begin{array}{cc} - 3 & - 2 \\ 10 & 7 \end{array}]$
So, $(AB)^{T} = [\begin{array}{cc} - 3 & 10 \\ - 2 & 7 \end{array}]$

Karnataka CET

Matrix and Determinant

78641 If $(x_{1}, y_{1}), (y_{2}, y_{2})$ and $(x_{3}, y_{3})$ are the vertices of a triangle whose area is $k$ square units, then
${| \begin{array}{lll} x_{1} & y_{1} & 4 \\ x_{2} & y_{2} & 4 \\ x_{3} & y_{3} & 4 \end{array} |}^{2} is$

1

32 k^{2}

2

16 k^{2}

3

64 k^{2}

4

48 k^{2}

Explanation:

(C) : We know that,
Area of triangle $= \frac{1}{2} [\begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}] = k$
From the given question,
${[\begin{array}{lll} x_{1} & y_{1} & 4 \\ x_{2} & y_{2} & 4 \\ x_{3} & y_{3} & 4 \end{array}]}^{2} = 4^{2} {[\begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}]}^{2}$
$= 16 (2 k)^{2} = 64 k^{2}$

Karnataka CET-2018

Matrix and Determinant

78642 If $2 [\begin{array}{ll} 1 & 3 \\ 0 & x \end{array}] + [\begin{array}{ll} y & 0 \\ 1 & 2 \end{array}] = [\begin{array}{ll} 5 & 6 \\ 1 & 8 \end{array}]$ then the value of $x$

1

x = 3, y = 3

2

x = - 3, y = 3

3

x = 3, y = - 3

4

x = - 3, y = - 3

Explanation:

(A) : Given that,
$2 [\begin{array}{ll} 1 & 3 \\ 0 & x \end{array}] + [\begin{array}{ll} y & 0 \\ 1 & 2 \end{array}] = [\begin{array}{ll} 5 & 6 \\ 1 & 8 \end{array}]$
$[\begin{array}{cc} 2 & 6 \\ 0 & 2 x \end{array}] + [\begin{array}{cc} y & 0 \\ 1 & 2 \end{array}] = [\begin{array}{cc} 5 & 6 \\ 1 & 8 \end{array}] \Rightarrow [\begin{array}{cc} 2 + y & 6 + 0 \\ 0 + 1 & 2 x + 2 \end{array}] = [\begin{array}{cc} 5 & 6 \\ 1 & 8 \end{array}]$
On comparing corresponding elements on both the side we get -
$\begin{array}{ll} 2 + y = 5, & 2 x + 2 = 8 \\ y = 5 - 2, & 2 x = 6 \\ y = 3 & x = 3 \end{array}$

Karnataka CET-2017

Matrix and Determinant

78637 If $A$ and $B$ are two matrices such that $A B = B$ and $B A = A$ then $A^{2} + B^{2} =$

1

2 AB

2

2 BA

3

A + B

4

AB

Explanation:

(C) : Given that,
$A$ and $B$ are two matrices and $AB = B$ and $BA = A$
Now, $AB = B$
On multiply by $A$ on both the side we get -
$\begin{aligned} (AB) A & = BA \\ A (BA) & = BA \\ A (A) & = A \\ A^{2} & = A \end{aligned}$
And $AB = B$
On multiply by $B$ on both the side we get -
$\begin{aligned} B (AB) & = BB \\ (BA) B & = B^{2} & [∵ BA = A] \\ AB & = B^{2} & [∵ AB = B] \\ B & = B^{2} \end{aligned}$
Hence, $A^{2} + B^{2} = A + B$

Karnataka CET-2000

Matrix and Determinant

78638 $G = {(\begin{array}{cc} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{array}) : θ \in R}$ is a group under matrix multiplication. Then which one of the following statements in respect of $G$ is true.

1

(\begin{array}{cc} - 1 & 0 \\ 0 & - 1 \end{array})

is the inverse of itself

2

G

is a finite group

3

(\begin{array}{cc} 1 / 2 & - \sqrt{3} / 2 \\ \sqrt{3} / 2 & 1 / 2 \end{array})

is not an element of

G

4

(\begin{array}{cc} 1 & 1 \\ - 1 & 1 \end{array})

is an element of G.

Explanation:

(A) : Given that,
$G = (\begin{array}{cc} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{array})$
$∵ θ \in R$
For option (a), If the matrix is the inverse of itself when multiplied by itself it should yield identity matrix.
$= [\begin{array}{cc} - 1 & 0 \\ 0 & - 1 \end{array}] [\begin{array}{cc} - 1 & 0 \\ 0 & - 1 \end{array}] = [\begin{array}{ll} 1 + 0 & 0 + 0 \\ 0 + 0 & 0 + 1 \end{array}] = [\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}]$
Hence, option (a) true.

Karnataka CET-2002

Matrix and Determinant

78639 If $A = [\begin{array}{ccc} 1 & - 2 & 1 \\ 2 & 1 & 3 \end{array}], B = [\begin{array}{ll} 2 & 1 \\ 3 & 2 \\ 1 & 1 \end{array}]$ then $(A B)^{T}$ is equal to

1

[\begin{array}{ll} - 3 & - 2 \\ 10 & 7 \end{array}]

2

[\begin{array}{cc} - 3 & 10 \\ - 2 & 7 \end{array}]

3

[\begin{array}{cc} - 3 & 7 \\ 10 & 2 \end{array}]

4

[\begin{array}{cc} - 3 & 7 \\ 10 & - 2 \end{array}]

Explanation:

(B) : Given that,
$A = [\begin{array}{ccc} 1 & - 2 & 1 \\ 2 & 1 & 3 \end{array}] and B = [\begin{array}{ll} 2 & 1 \\ 3 & 2 \\ 1 & 1 \end{array}]$
Then, $A B = [\begin{array}{ccc} 1 & - 2 & 1 \\ 2 & 1 & 3 \end{array}] [\begin{array}{ll} 2 & 1 \\ 3 & 2 \\ 1 & 1 \end{array}]$
$AB = [\begin{array}{cc} 2 - 6 + 1 & 1 - 4 + 1 \\ 4 + 3 + 3 & 2 + 2 + 3 \end{array}] \Rightarrow AB = [\begin{array}{cc} - 3 & - 2 \\ 10 & 7 \end{array}]$
So, $(AB)^{T} = [\begin{array}{cc} - 3 & 10 \\ - 2 & 7 \end{array}]$

Karnataka CET

Matrix and Determinant

78641 If $(x_{1}, y_{1}), (y_{2}, y_{2})$ and $(x_{3}, y_{3})$ are the vertices of a triangle whose area is $k$ square units, then
${| \begin{array}{lll} x_{1} & y_{1} & 4 \\ x_{2} & y_{2} & 4 \\ x_{3} & y_{3} & 4 \end{array} |}^{2} is$

1

32 k^{2}

2

16 k^{2}

3

64 k^{2}

4

48 k^{2}

Explanation:

(C) : We know that,
Area of triangle $= \frac{1}{2} [\begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}] = k$
From the given question,
${[\begin{array}{lll} x_{1} & y_{1} & 4 \\ x_{2} & y_{2} & 4 \\ x_{3} & y_{3} & 4 \end{array}]}^{2} = 4^{2} {[\begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}]}^{2}$
$= 16 (2 k)^{2} = 64 k^{2}$

Karnataka CET-2018

Matrix and Determinant

78642 If $2 [\begin{array}{ll} 1 & 3 \\ 0 & x \end{array}] + [\begin{array}{ll} y & 0 \\ 1 & 2 \end{array}] = [\begin{array}{ll} 5 & 6 \\ 1 & 8 \end{array}]$ then the value of $x$

1

x = 3, y = 3

2

x = - 3, y = 3

3

x = 3, y = - 3

4

x = - 3, y = - 3

Explanation:

(A) : Given that,
$2 [\begin{array}{ll} 1 & 3 \\ 0 & x \end{array}] + [\begin{array}{ll} y & 0 \\ 1 & 2 \end{array}] = [\begin{array}{ll} 5 & 6 \\ 1 & 8 \end{array}]$
$[\begin{array}{cc} 2 & 6 \\ 0 & 2 x \end{array}] + [\begin{array}{cc} y & 0 \\ 1 & 2 \end{array}] = [\begin{array}{cc} 5 & 6 \\ 1 & 8 \end{array}] \Rightarrow [\begin{array}{cc} 2 + y & 6 + 0 \\ 0 + 1 & 2 x + 2 \end{array}] = [\begin{array}{cc} 5 & 6 \\ 1 & 8 \end{array}]$
On comparing corresponding elements on both the side we get -
$\begin{array}{ll} 2 + y = 5, & 2 x + 2 = 8 \\ y = 5 - 2, & 2 x = 6 \\ y = 3 & x = 3 \end{array}$

Karnataka CET-2017

Matrix and Determinant

78637 If $A$ and $B$ are two matrices such that $A B = B$ and $B A = A$ then $A^{2} + B^{2} =$

1

2 AB

2

2 BA

3

A + B

4

AB

Explanation:

(C) : Given that,
$A$ and $B$ are two matrices and $AB = B$ and $BA = A$
Now, $AB = B$
On multiply by $A$ on both the side we get -
$\begin{aligned} (AB) A & = BA \\ A (BA) & = BA \\ A (A) & = A \\ A^{2} & = A \end{aligned}$
And $AB = B$
On multiply by $B$ on both the side we get -
$\begin{aligned} B (AB) & = BB \\ (BA) B & = B^{2} & [∵ BA = A] \\ AB & = B^{2} & [∵ AB = B] \\ B & = B^{2} \end{aligned}$
Hence, $A^{2} + B^{2} = A + B$

Karnataka CET-2000

Matrix and Determinant

78638 $G = {(\begin{array}{cc} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{array}) : θ \in R}$ is a group under matrix multiplication. Then which one of the following statements in respect of $G$ is true.

1

(\begin{array}{cc} - 1 & 0 \\ 0 & - 1 \end{array})

is the inverse of itself

2

G

is a finite group

3

(\begin{array}{cc} 1 / 2 & - \sqrt{3} / 2 \\ \sqrt{3} / 2 & 1 / 2 \end{array})

is not an element of

G

4

(\begin{array}{cc} 1 & 1 \\ - 1 & 1 \end{array})

is an element of G.

Explanation:

(A) : Given that,
$G = (\begin{array}{cc} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{array})$
$∵ θ \in R$
For option (a), If the matrix is the inverse of itself when multiplied by itself it should yield identity matrix.
$= [\begin{array}{cc} - 1 & 0 \\ 0 & - 1 \end{array}] [\begin{array}{cc} - 1 & 0 \\ 0 & - 1 \end{array}] = [\begin{array}{ll} 1 + 0 & 0 + 0 \\ 0 + 0 & 0 + 1 \end{array}] = [\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}]$
Hence, option (a) true.

Karnataka CET-2002

Matrix and Determinant

78639 If $A = [\begin{array}{ccc} 1 & - 2 & 1 \\ 2 & 1 & 3 \end{array}], B = [\begin{array}{ll} 2 & 1 \\ 3 & 2 \\ 1 & 1 \end{array}]$ then $(A B)^{T}$ is equal to

1

[\begin{array}{ll} - 3 & - 2 \\ 10 & 7 \end{array}]

2

[\begin{array}{cc} - 3 & 10 \\ - 2 & 7 \end{array}]

3

[\begin{array}{cc} - 3 & 7 \\ 10 & 2 \end{array}]

4

[\begin{array}{cc} - 3 & 7 \\ 10 & - 2 \end{array}]

Explanation:

(B) : Given that,
$A = [\begin{array}{ccc} 1 & - 2 & 1 \\ 2 & 1 & 3 \end{array}] and B = [\begin{array}{ll} 2 & 1 \\ 3 & 2 \\ 1 & 1 \end{array}]$
Then, $A B = [\begin{array}{ccc} 1 & - 2 & 1 \\ 2 & 1 & 3 \end{array}] [\begin{array}{ll} 2 & 1 \\ 3 & 2 \\ 1 & 1 \end{array}]$
$AB = [\begin{array}{cc} 2 - 6 + 1 & 1 - 4 + 1 \\ 4 + 3 + 3 & 2 + 2 + 3 \end{array}] \Rightarrow AB = [\begin{array}{cc} - 3 & - 2 \\ 10 & 7 \end{array}]$
So, $(AB)^{T} = [\begin{array}{cc} - 3 & 10 \\ - 2 & 7 \end{array}]$

Karnataka CET

Matrix and Determinant

78641 If $(x_{1}, y_{1}), (y_{2}, y_{2})$ and $(x_{3}, y_{3})$ are the vertices of a triangle whose area is $k$ square units, then
${| \begin{array}{lll} x_{1} & y_{1} & 4 \\ x_{2} & y_{2} & 4 \\ x_{3} & y_{3} & 4 \end{array} |}^{2} is$

1

32 k^{2}

2

16 k^{2}

3

64 k^{2}

4

48 k^{2}

Explanation:

(C) : We know that,
Area of triangle $= \frac{1}{2} [\begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}] = k$
From the given question,
${[\begin{array}{lll} x_{1} & y_{1} & 4 \\ x_{2} & y_{2} & 4 \\ x_{3} & y_{3} & 4 \end{array}]}^{2} = 4^{2} {[\begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}]}^{2}$
$= 16 (2 k)^{2} = 64 k^{2}$

Karnataka CET-2018

Matrix and Determinant

78642 If $2 [\begin{array}{ll} 1 & 3 \\ 0 & x \end{array}] + [\begin{array}{ll} y & 0 \\ 1 & 2 \end{array}] = [\begin{array}{ll} 5 & 6 \\ 1 & 8 \end{array}]$ then the value of $x$

1

x = 3, y = 3

2

x = - 3, y = 3

3

x = 3, y = - 3

4

x = - 3, y = - 3

Explanation:

(A) : Given that,
$2 [\begin{array}{ll} 1 & 3 \\ 0 & x \end{array}] + [\begin{array}{ll} y & 0 \\ 1 & 2 \end{array}] = [\begin{array}{ll} 5 & 6 \\ 1 & 8 \end{array}]$
$[\begin{array}{cc} 2 & 6 \\ 0 & 2 x \end{array}] + [\begin{array}{cc} y & 0 \\ 1 & 2 \end{array}] = [\begin{array}{cc} 5 & 6 \\ 1 & 8 \end{array}] \Rightarrow [\begin{array}{cc} 2 + y & 6 + 0 \\ 0 + 1 & 2 x + 2 \end{array}] = [\begin{array}{cc} 5 & 6 \\ 1 & 8 \end{array}]$
On comparing corresponding elements on both the side we get -
$\begin{array}{ll} 2 + y = 5, & 2 x + 2 = 8 \\ y = 5 - 2, & 2 x = 6 \\ y = 3 & x = 3 \end{array}$

Karnataka CET-2017