QBManager

Matrix and Determinant

78822 If $A = [\begin{array}{lll} 1 & 2 & x \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$ and $B = [\begin{array}{ccc} 1 & - 2 & y \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$ and $A B$ $= I_{3}$ , then $x + y$ equals

1 0

2 -1

3 2

4 None of these

Explanation:

(A) : We have,
$AB = [\begin{array}{lll} 1 & 2 & x \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] [\begin{array}{ccc} 1 & - 2 & y \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] = [\begin{array}{ccc} 1 & 0 & x + y \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$
Now, $AB = I_{3}$ (given)
$\Rightarrow [\begin{array}{ccc} 1 & 0 & x + y \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] = [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$
Comparing both sides,
Hence, $x + y = 0$

COMEDK-2019

Matrix and Determinant

78823 The matrix $A$ satisfying the equation
$[\begin{array}{ll} 1 & 3 \\ 0 & 1 \end{array}] A = [\begin{array}{cc} 1 & 1 \\ 0 & - 1 \end{array}]$ is

1

[\begin{array}{rr} 1 & 4 \\ - 1 & 0 \end{array}

2

[\begin{array}{rr} 1 & - 4 \\ 1 & 0 \end{array}]

3

[\begin{array}{rr} 1 & 4 \\ 0 & - 1 \end{array}]

4

[\begin{array}{rr} - 1 & 4 \\ 1 & 0 \end{array}]

Explanation:

(C) : Given that,
$[\begin{array}{ll} 1 & 3 \\ 0 & 1 \end{array}] A = [\begin{array}{cc} 1 & 1 \\ 0 & - 1 \end{array}]$
Let, $B = [\begin{array}{ll} 1 & 3 \\ 0 & 1 \end{array}]$ and $C = [\begin{array}{cc} 1 & 1 \\ 0 & - 1 \end{array}]$
Now,
$BA = C$
$A = B^{- 1} C$
We know that,
$B^{- 1} = \frac{adj (B)}{| B |}$
$| B | = 1 - 0 = 1$
$adj (B) = [\begin{array}{cc} 1 & - 3 \\ 0 & 1 \end{array}]$
$B^{- 1} = \frac{1}{1} [\begin{array}{cc} 1 & - 3 \\ 0 & 1 \end{array}] = [\begin{array}{cc} 1 & - 3 \\ 0 & 1 \end{array}]$
Hence,
$A = B^{- 1} C$
$A = [\begin{array}{cc} 1 & - 3 \\ 0 & 1 \end{array}] [\begin{array}{cc} 1 & 1 \\ 0 & - 1 \end{array}] \Rightarrow A = [\begin{array}{cc} 1 & 4 \\ 0 & - 1 \end{array}]$

SRM JEE-2010

Matrix and Determinant

78824 If $A = [\begin{array}{cc} \cos x & \sin x \\ - \sin x & \cos x \end{array}]$ and $A \cdot Adj A = k [\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}]$ , then the value of $k$ is

1

\sin x \cos x

2 1

3 2

4 8

Explanation:

(B) : Given that,
$A = [\begin{array}{cc} \cos x & \sin x \\ - \sin x & \cos x \end{array}]$
A. $Adj A = k [\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}]$
$Adj (A) = [\begin{array}{cc} \cos x & - \sin x \\ \sin x & \cos x \end{array}]$
$A \cdot Adj A = [\begin{array}{cc} \cos x & \sin x \\ - \sin x & \cos x \end{array}] [\begin{array}{cc} \cos x & - \sin x \\ \sin x & \cos x \end{array}]$
A. $Adj A = [\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}]$
So, A. Adj $A = k [\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}]$ (given)
$[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}] = [\begin{array}{ll} k & 0 \\ 0 & k \end{array}]$
On comparing corresponding elements on the both side we get -
$k = 1$

SRM JEE-2011

Matrix and Determinant

78826 If $A = [\begin{array}{cc} 1 & - 2 \\ 3 & 0 \end{array}], B = [\begin{array}{cc} - 1 & 4 \\ 2 & 3 \end{array}]$ and $C = [\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array}]$ then $5 A - 3 B + 2 C$ is equal to

1

[\begin{array}{cc} 8 & 20 \\ 7 & 9 \end{array}]

2

[\begin{array}{cc} 8 & - 20 \\ 7 & - 9 \end{array}]

3

[\begin{array}{cc} - 8 & 20 \\ - 7 & 9 \end{array}]

4

[\begin{array}{cc} 8 & 7 \\ - 20 & - 9 \end{array}]

Explanation:

(B) : Given that,
$A = [\begin{array}{cc} 1 & - 2 \\ 3 & 0 \end{array}], B = [\begin{array}{cc} - 1 & 4 \\ 2 & 3 \end{array}]$ and $C = [\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array}]$
Now, $5 A - 3 B + 2 C$
$= 5 [\begin{array}{cc} 1 & - 2 \\ 3 & 0 \end{array}] - 3 [\begin{array}{cc} - 1 & 4 \\ 2 & 3 \end{array}] + 2 [\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array}]$
$= [\begin{array}{cc} 5 & - 10 \\ 15 & 0 \end{array}] - [\begin{array}{cc} - 3 & 12 \\ 6 & 9 \end{array}] + [\begin{array}{cc} 0 & 2 \\ - 2 & 0 \end{array}]$
$= [\begin{array}{cc} 8 & - 22 \\ 9 & - 9 \end{array}] + [\begin{array}{cc} 0 & 2 \\ - 2 & 0 \end{array}] = [\begin{array}{cc} 8 & - 20 \\ 7 & - 9 \end{array}]$

Matrix and Determinant

78822 If $A = [\begin{array}{lll} 1 & 2 & x \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$ and $B = [\begin{array}{ccc} 1 & - 2 & y \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$ and $A B$ $= I_{3}$ , then $x + y$ equals

1 0

2 -1

3 2

4 None of these

Explanation:

(A) : We have,
$AB = [\begin{array}{lll} 1 & 2 & x \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] [\begin{array}{ccc} 1 & - 2 & y \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] = [\begin{array}{ccc} 1 & 0 & x + y \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$
Now, $AB = I_{3}$ (given)
$\Rightarrow [\begin{array}{ccc} 1 & 0 & x + y \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] = [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$
Comparing both sides,
Hence, $x + y = 0$

COMEDK-2019

Matrix and Determinant

78823 The matrix $A$ satisfying the equation
$[\begin{array}{ll} 1 & 3 \\ 0 & 1 \end{array}] A = [\begin{array}{cc} 1 & 1 \\ 0 & - 1 \end{array}]$ is

1

[\begin{array}{rr} 1 & 4 \\ - 1 & 0 \end{array}

2

[\begin{array}{rr} 1 & - 4 \\ 1 & 0 \end{array}]

3

[\begin{array}{rr} 1 & 4 \\ 0 & - 1 \end{array}]

4

[\begin{array}{rr} - 1 & 4 \\ 1 & 0 \end{array}]

Explanation:

(C) : Given that,
$[\begin{array}{ll} 1 & 3 \\ 0 & 1 \end{array}] A = [\begin{array}{cc} 1 & 1 \\ 0 & - 1 \end{array}]$
Let, $B = [\begin{array}{ll} 1 & 3 \\ 0 & 1 \end{array}]$ and $C = [\begin{array}{cc} 1 & 1 \\ 0 & - 1 \end{array}]$
Now,
$BA = C$
$A = B^{- 1} C$
We know that,
$B^{- 1} = \frac{adj (B)}{| B |}$
$| B | = 1 - 0 = 1$
$adj (B) = [\begin{array}{cc} 1 & - 3 \\ 0 & 1 \end{array}]$
$B^{- 1} = \frac{1}{1} [\begin{array}{cc} 1 & - 3 \\ 0 & 1 \end{array}] = [\begin{array}{cc} 1 & - 3 \\ 0 & 1 \end{array}]$
Hence,
$A = B^{- 1} C$
$A = [\begin{array}{cc} 1 & - 3 \\ 0 & 1 \end{array}] [\begin{array}{cc} 1 & 1 \\ 0 & - 1 \end{array}] \Rightarrow A = [\begin{array}{cc} 1 & 4 \\ 0 & - 1 \end{array}]$

SRM JEE-2010

Matrix and Determinant

78824 If $A = [\begin{array}{cc} \cos x & \sin x \\ - \sin x & \cos x \end{array}]$ and $A \cdot Adj A = k [\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}]$ , then the value of $k$ is

1

\sin x \cos x

2 1

3 2

4 8

Explanation:

(B) : Given that,
$A = [\begin{array}{cc} \cos x & \sin x \\ - \sin x & \cos x \end{array}]$
A. $Adj A = k [\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}]$
$Adj (A) = [\begin{array}{cc} \cos x & - \sin x \\ \sin x & \cos x \end{array}]$
$A \cdot Adj A = [\begin{array}{cc} \cos x & \sin x \\ - \sin x & \cos x \end{array}] [\begin{array}{cc} \cos x & - \sin x \\ \sin x & \cos x \end{array}]$
A. $Adj A = [\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}]$
So, A. Adj $A = k [\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}]$ (given)
$[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}] = [\begin{array}{ll} k & 0 \\ 0 & k \end{array}]$
On comparing corresponding elements on the both side we get -
$k = 1$

SRM JEE-2011

Matrix and Determinant

78826 If $A = [\begin{array}{cc} 1 & - 2 \\ 3 & 0 \end{array}], B = [\begin{array}{cc} - 1 & 4 \\ 2 & 3 \end{array}]$ and $C = [\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array}]$ then $5 A - 3 B + 2 C$ is equal to

1

[\begin{array}{cc} 8 & 20 \\ 7 & 9 \end{array}]

2

[\begin{array}{cc} 8 & - 20 \\ 7 & - 9 \end{array}]

3

[\begin{array}{cc} - 8 & 20 \\ - 7 & 9 \end{array}]

4

[\begin{array}{cc} 8 & 7 \\ - 20 & - 9 \end{array}]

Explanation:

(B) : Given that,
$A = [\begin{array}{cc} 1 & - 2 \\ 3 & 0 \end{array}], B = [\begin{array}{cc} - 1 & 4 \\ 2 & 3 \end{array}]$ and $C = [\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array}]$
Now, $5 A - 3 B + 2 C$
$= 5 [\begin{array}{cc} 1 & - 2 \\ 3 & 0 \end{array}] - 3 [\begin{array}{cc} - 1 & 4 \\ 2 & 3 \end{array}] + 2 [\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array}]$
$= [\begin{array}{cc} 5 & - 10 \\ 15 & 0 \end{array}] - [\begin{array}{cc} - 3 & 12 \\ 6 & 9 \end{array}] + [\begin{array}{cc} 0 & 2 \\ - 2 & 0 \end{array}]$
$= [\begin{array}{cc} 8 & - 22 \\ 9 & - 9 \end{array}] + [\begin{array}{cc} 0 & 2 \\ - 2 & 0 \end{array}] = [\begin{array}{cc} 8 & - 20 \\ 7 & - 9 \end{array}]$

Matrix and Determinant

78822 If $A = [\begin{array}{lll} 1 & 2 & x \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$ and $B = [\begin{array}{ccc} 1 & - 2 & y \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$ and $A B$ $= I_{3}$ , then $x + y$ equals

1 0

2 -1

3 2

4 None of these

Explanation:

(A) : We have,
$AB = [\begin{array}{lll} 1 & 2 & x \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] [\begin{array}{ccc} 1 & - 2 & y \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] = [\begin{array}{ccc} 1 & 0 & x + y \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$
Now, $AB = I_{3}$ (given)
$\Rightarrow [\begin{array}{ccc} 1 & 0 & x + y \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] = [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$
Comparing both sides,
Hence, $x + y = 0$

COMEDK-2019

Matrix and Determinant

78823 The matrix $A$ satisfying the equation
$[\begin{array}{ll} 1 & 3 \\ 0 & 1 \end{array}] A = [\begin{array}{cc} 1 & 1 \\ 0 & - 1 \end{array}]$ is

1

[\begin{array}{rr} 1 & 4 \\ - 1 & 0 \end{array}

2

[\begin{array}{rr} 1 & - 4 \\ 1 & 0 \end{array}]

3

[\begin{array}{rr} 1 & 4 \\ 0 & - 1 \end{array}]

4

[\begin{array}{rr} - 1 & 4 \\ 1 & 0 \end{array}]

Explanation:

(C) : Given that,
$[\begin{array}{ll} 1 & 3 \\ 0 & 1 \end{array}] A = [\begin{array}{cc} 1 & 1 \\ 0 & - 1 \end{array}]$
Let, $B = [\begin{array}{ll} 1 & 3 \\ 0 & 1 \end{array}]$ and $C = [\begin{array}{cc} 1 & 1 \\ 0 & - 1 \end{array}]$
Now,
$BA = C$
$A = B^{- 1} C$
We know that,
$B^{- 1} = \frac{adj (B)}{| B |}$
$| B | = 1 - 0 = 1$
$adj (B) = [\begin{array}{cc} 1 & - 3 \\ 0 & 1 \end{array}]$
$B^{- 1} = \frac{1}{1} [\begin{array}{cc} 1 & - 3 \\ 0 & 1 \end{array}] = [\begin{array}{cc} 1 & - 3 \\ 0 & 1 \end{array}]$
Hence,
$A = B^{- 1} C$
$A = [\begin{array}{cc} 1 & - 3 \\ 0 & 1 \end{array}] [\begin{array}{cc} 1 & 1 \\ 0 & - 1 \end{array}] \Rightarrow A = [\begin{array}{cc} 1 & 4 \\ 0 & - 1 \end{array}]$

SRM JEE-2010

Matrix and Determinant

78824 If $A = [\begin{array}{cc} \cos x & \sin x \\ - \sin x & \cos x \end{array}]$ and $A \cdot Adj A = k [\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}]$ , then the value of $k$ is

1

\sin x \cos x

2 1

3 2

4 8

Explanation:

(B) : Given that,
$A = [\begin{array}{cc} \cos x & \sin x \\ - \sin x & \cos x \end{array}]$
A. $Adj A = k [\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}]$
$Adj (A) = [\begin{array}{cc} \cos x & - \sin x \\ \sin x & \cos x \end{array}]$
$A \cdot Adj A = [\begin{array}{cc} \cos x & \sin x \\ - \sin x & \cos x \end{array}] [\begin{array}{cc} \cos x & - \sin x \\ \sin x & \cos x \end{array}]$
A. $Adj A = [\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}]$
So, A. Adj $A = k [\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}]$ (given)
$[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}] = [\begin{array}{ll} k & 0 \\ 0 & k \end{array}]$
On comparing corresponding elements on the both side we get -
$k = 1$

SRM JEE-2011

Matrix and Determinant

78826 If $A = [\begin{array}{cc} 1 & - 2 \\ 3 & 0 \end{array}], B = [\begin{array}{cc} - 1 & 4 \\ 2 & 3 \end{array}]$ and $C = [\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array}]$ then $5 A - 3 B + 2 C$ is equal to

1

[\begin{array}{cc} 8 & 20 \\ 7 & 9 \end{array}]

2

[\begin{array}{cc} 8 & - 20 \\ 7 & - 9 \end{array}]

3

[\begin{array}{cc} - 8 & 20 \\ - 7 & 9 \end{array}]

4

[\begin{array}{cc} 8 & 7 \\ - 20 & - 9 \end{array}]

Explanation:

(B) : Given that,
$A = [\begin{array}{cc} 1 & - 2 \\ 3 & 0 \end{array}], B = [\begin{array}{cc} - 1 & 4 \\ 2 & 3 \end{array}]$ and $C = [\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array}]$
Now, $5 A - 3 B + 2 C$
$= 5 [\begin{array}{cc} 1 & - 2 \\ 3 & 0 \end{array}] - 3 [\begin{array}{cc} - 1 & 4 \\ 2 & 3 \end{array}] + 2 [\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array}]$
$= [\begin{array}{cc} 5 & - 10 \\ 15 & 0 \end{array}] - [\begin{array}{cc} - 3 & 12 \\ 6 & 9 \end{array}] + [\begin{array}{cc} 0 & 2 \\ - 2 & 0 \end{array}]$
$= [\begin{array}{cc} 8 & - 22 \\ 9 & - 9 \end{array}] + [\begin{array}{cc} 0 & 2 \\ - 2 & 0 \end{array}] = [\begin{array}{cc} 8 & - 20 \\ 7 & - 9 \end{array}]$

Matrix and Determinant

78822 If $A = [\begin{array}{lll} 1 & 2 & x \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$ and $B = [\begin{array}{ccc} 1 & - 2 & y \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$ and $A B$ $= I_{3}$ , then $x + y$ equals

1 0

2 -1

3 2

4 None of these

Explanation:

(A) : We have,
$AB = [\begin{array}{lll} 1 & 2 & x \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] [\begin{array}{ccc} 1 & - 2 & y \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] = [\begin{array}{ccc} 1 & 0 & x + y \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$
Now, $AB = I_{3}$ (given)
$\Rightarrow [\begin{array}{ccc} 1 & 0 & x + y \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] = [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$
Comparing both sides,
Hence, $x + y = 0$

COMEDK-2019

Matrix and Determinant

78823 The matrix $A$ satisfying the equation
$[\begin{array}{ll} 1 & 3 \\ 0 & 1 \end{array}] A = [\begin{array}{cc} 1 & 1 \\ 0 & - 1 \end{array}]$ is

1

[\begin{array}{rr} 1 & 4 \\ - 1 & 0 \end{array}

2

[\begin{array}{rr} 1 & - 4 \\ 1 & 0 \end{array}]

3

[\begin{array}{rr} 1 & 4 \\ 0 & - 1 \end{array}]

4

[\begin{array}{rr} - 1 & 4 \\ 1 & 0 \end{array}]

Explanation:

(C) : Given that,
$[\begin{array}{ll} 1 & 3 \\ 0 & 1 \end{array}] A = [\begin{array}{cc} 1 & 1 \\ 0 & - 1 \end{array}]$
Let, $B = [\begin{array}{ll} 1 & 3 \\ 0 & 1 \end{array}]$ and $C = [\begin{array}{cc} 1 & 1 \\ 0 & - 1 \end{array}]$
Now,
$BA = C$
$A = B^{- 1} C$
We know that,
$B^{- 1} = \frac{adj (B)}{| B |}$
$| B | = 1 - 0 = 1$
$adj (B) = [\begin{array}{cc} 1 & - 3 \\ 0 & 1 \end{array}]$
$B^{- 1} = \frac{1}{1} [\begin{array}{cc} 1 & - 3 \\ 0 & 1 \end{array}] = [\begin{array}{cc} 1 & - 3 \\ 0 & 1 \end{array}]$
Hence,
$A = B^{- 1} C$
$A = [\begin{array}{cc} 1 & - 3 \\ 0 & 1 \end{array}] [\begin{array}{cc} 1 & 1 \\ 0 & - 1 \end{array}] \Rightarrow A = [\begin{array}{cc} 1 & 4 \\ 0 & - 1 \end{array}]$

SRM JEE-2010

Matrix and Determinant

78824 If $A = [\begin{array}{cc} \cos x & \sin x \\ - \sin x & \cos x \end{array}]$ and $A \cdot Adj A = k [\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}]$ , then the value of $k$ is

1

\sin x \cos x

2 1

3 2

4 8

Explanation:

(B) : Given that,
$A = [\begin{array}{cc} \cos x & \sin x \\ - \sin x & \cos x \end{array}]$
A. $Adj A = k [\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}]$
$Adj (A) = [\begin{array}{cc} \cos x & - \sin x \\ \sin x & \cos x \end{array}]$
$A \cdot Adj A = [\begin{array}{cc} \cos x & \sin x \\ - \sin x & \cos x \end{array}] [\begin{array}{cc} \cos x & - \sin x \\ \sin x & \cos x \end{array}]$
A. $Adj A = [\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}]$
So, A. Adj $A = k [\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}]$ (given)
$[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}] = [\begin{array}{ll} k & 0 \\ 0 & k \end{array}]$
On comparing corresponding elements on the both side we get -
$k = 1$

SRM JEE-2011

Matrix and Determinant

78826 If $A = [\begin{array}{cc} 1 & - 2 \\ 3 & 0 \end{array}], B = [\begin{array}{cc} - 1 & 4 \\ 2 & 3 \end{array}]$ and $C = [\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array}]$ then $5 A - 3 B + 2 C$ is equal to

1

[\begin{array}{cc} 8 & 20 \\ 7 & 9 \end{array}]

2

[\begin{array}{cc} 8 & - 20 \\ 7 & - 9 \end{array}]

3

[\begin{array}{cc} - 8 & 20 \\ - 7 & 9 \end{array}]

4

[\begin{array}{cc} 8 & 7 \\ - 20 & - 9 \end{array}]

Explanation:

(B) : Given that,
$A = [\begin{array}{cc} 1 & - 2 \\ 3 & 0 \end{array}], B = [\begin{array}{cc} - 1 & 4 \\ 2 & 3 \end{array}]$ and $C = [\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array}]$
Now, $5 A - 3 B + 2 C$
$= 5 [\begin{array}{cc} 1 & - 2 \\ 3 & 0 \end{array}] - 3 [\begin{array}{cc} - 1 & 4 \\ 2 & 3 \end{array}] + 2 [\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array}]$
$= [\begin{array}{cc} 5 & - 10 \\ 15 & 0 \end{array}] - [\begin{array}{cc} - 3 & 12 \\ 6 & 9 \end{array}] + [\begin{array}{cc} 0 & 2 \\ - 2 & 0 \end{array}]$
$= [\begin{array}{cc} 8 & - 22 \\ 9 & - 9 \end{array}] + [\begin{array}{cc} 0 & 2 \\ - 2 & 0 \end{array}] = [\begin{array}{cc} 8 & - 20 \\ 7 & - 9 \end{array}]$

Question Bank Series

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