QBManager

Matrix and Determinant

78787 If $A = [\begin{array}{lll} 1 & 2 & 1 \\ 2 & 1 & 0 \end{array}], B = [\begin{array}{ll} 1 & 2 \\ 2 & 1 \\ 0 & 1 \end{array}]$ , then $(A B)^{- 1}$ is

1

(\frac{1}{5}) [\begin{array}{ll} 5 & - 5 \\ 4 & - 5 \end{array}]

2

(\frac{1}{5}) [\begin{array}{cc} 5 & - 5 \\ 4 & 5 \end{array}]

3

(\frac{1}{5}) [\begin{array}{cc} 5 & - 5 \\ - 4 & 5 \end{array}]

4

(\frac{1}{5}) [\begin{array}{cc} 5 & - 5 \\ - 4 & - 5 \end{array}]

Explanation:

(C) : Given that,
$A = [\begin{array}{lll} 1 & 2 & 1 \\ 2 & 1 & 0 \end{array}] and B = [\begin{array}{ll} 1 & 2 \\ 2 & 1 \\ 0 & 1 \end{array}]$
$∴ AB = [\begin{array}{ll} 1 + 4 + 0 & 2 + 2 + 1 \\ 2 + 2 + 0 & 4 + 1 + 0 \end{array}] = [\begin{array}{cc} 5 & 5 \\ 4 & 5 \end{array}]$
$(AB)^{- 1} = \frac{1}{| AB |} [\begin{array}{cc} 5 & - 5 \\ - 4 & 5 \end{array}] = \frac{1}{25 - 20} [\begin{array}{cc} 5 & - 5 \\ - 4 & 5 \end{array}] = \frac{1}{5} [\begin{array}{cc} 5 & - 5 \\ - 4 & 5 \end{array}]$

MHT CET-2020

Matrix and Determinant

78788 If $A = [\begin{array}{ll} 1 & 1 \\ 1 & 2 \end{array}], B = [\begin{array}{ll} 4 & 1 \\ 3 & 1 \end{array}]$ , then $(A + B)^{- 1} =$

1

\frac{1}{7} [\begin{array}{ll} 3 & 2 \\ 4 & 5 \end{array}]

2

\frac{1}{7} [\begin{array}{cc} 3 & - 2 \\ - 4 & 5 \end{array}]

3

7 [\begin{array}{ll} 3 & 2 \\ 4 & 5 \end{array}]

4

7 [\begin{array}{cc} 3 & - 2 \\ - 4 & 5 \end{array}]

Explanation:

(B) : Given that,
$A = [\begin{array}{ll} 1 & 1 \\ 1 & 2 \end{array}] and B = [\begin{array}{ll} 4 & 1 \\ 3 & 1 \end{array}]$
$A + B = [\begin{array}{ll} 1 & 1 \\ 1 & 2 \end{array}] + [\begin{array}{ll} 4 & 1 \\ 3 & 1 \end{array}] = [\begin{array}{ll} 5 & 2 \\ 4 & 3 \end{array}]$
$| A + B | = 15 - 8 = 7$ and $adj (A + B) = [\begin{array}{cc} 3 & - 2 \\ - 4 & 5 \end{array}]$
$∴ (A + B)^{- 1} = \frac{1}{7} [\begin{array}{cc} 3 & - 2 \\ - 4 & 5 \end{array}]$

MHT CET-2020

Matrix and Determinant

78789 If $AX = B$ , where $A = [\begin{array}{ccc} 1 & - 1 & 1 \\ 2 & - 1 & 0 \\ 3 & 3 & - 4 \end{array}], B = [\begin{array}{l} 1 \\ 1 \\ 2 \end{array}]$
and $X = [\begin{array}{l} x \\ y \\ z \end{array}]$ , then $x + y + z =$

1 6

2 2

3 1

4 3,

Explanation:

(D) : Given that
$A = [\begin{array}{ccc} 1 & - 1 & 1 \\ 2 & - 1 & 0 \\ 3 & 3 & - 4 \end{array}]$ and $B = [\begin{array}{l} 1 \\ 1 \\ 2 X = B \end{array}], X = [\begin{array}{l} x \\ y \\ z \end{array}]$
On multiplying $A^{- 1}$ both the side, we get -
${AA}^{- 1} X = A^{- 1} B$
$IX = A^{- 1} B$
$X = A^{- 1} B$
Now, $| A | = 4 + (- 8) + 9 = 5$ $∴ A^{- 1} = [\begin{array}{lll} 4 & - 1 & 1 \\ 8 & - 7 & 2 \\ 9 & - 6 & 1 \end{array}] \times \frac{1}{5}$
$∴ [\begin{array}{l} x \\ y \\ z \end{array}] = \frac{1}{5} [\begin{array}{lll} 4 & - 1 & 1 \\ 8 & - 7 & 2 \\ 9 & - 6 & 1 \end{array}] [\begin{array}{l} 1 \\ 1 \\ 2 \end{array}]$
$= \frac{1}{5} [\begin{array}{l} 4 - 1 + 2 \\ 8 - 7 + 4 \\ 9 - 6 + 2 \end{array}] = \frac{1}{5} [\begin{array}{l} 5 \\ 5 \\ 5 \end{array}]$
$[\begin{array}{l} x \\ y \\ z \end{array}] = [\begin{array}{l} 1 \\ 1 \\ 1 \end{array}]$
On comparing both side, we get -
$x = y = z = 1$
$x + y + z = 1 + 1 + 1 = 3$

MHT CET-2020

Matrix and Determinant

78790 If $A = [\begin{array}{ccc} 0 & - 1 & 0 \\ - 1 & 0 & 0 \end{array}]$ , then

1

A

is not invertible

2

A^{- 1} = 2 A

3

A^{- 1} = I

4

A = A^{- 1}

Explanation:

(D) : Given that,
$A = [\begin{array}{ccc} 0 & 0 & - 1 \\ 0 & - 1 & 0 \\ - 1 & 0 & 0 \end{array}]$
Therefore, $| A | = - 1 (- 1) = 1$
And, $A^{- 1} = [\begin{array}{ccc} 0 & 0 & - 1 \\ 0 & - 1 & 0 \\ - 1 & 0 & 0 \end{array}] = A$
$A^{- 1} = A$

MHT CET-2020

Matrix and Determinant

78787 If $A = [\begin{array}{lll} 1 & 2 & 1 \\ 2 & 1 & 0 \end{array}], B = [\begin{array}{ll} 1 & 2 \\ 2 & 1 \\ 0 & 1 \end{array}]$ , then $(A B)^{- 1}$ is

1

(\frac{1}{5}) [\begin{array}{ll} 5 & - 5 \\ 4 & - 5 \end{array}]

2

(\frac{1}{5}) [\begin{array}{cc} 5 & - 5 \\ 4 & 5 \end{array}]

3

(\frac{1}{5}) [\begin{array}{cc} 5 & - 5 \\ - 4 & 5 \end{array}]

4

(\frac{1}{5}) [\begin{array}{cc} 5 & - 5 \\ - 4 & - 5 \end{array}]

Explanation:

(C) : Given that,
$A = [\begin{array}{lll} 1 & 2 & 1 \\ 2 & 1 & 0 \end{array}] and B = [\begin{array}{ll} 1 & 2 \\ 2 & 1 \\ 0 & 1 \end{array}]$
$∴ AB = [\begin{array}{ll} 1 + 4 + 0 & 2 + 2 + 1 \\ 2 + 2 + 0 & 4 + 1 + 0 \end{array}] = [\begin{array}{cc} 5 & 5 \\ 4 & 5 \end{array}]$
$(AB)^{- 1} = \frac{1}{| AB |} [\begin{array}{cc} 5 & - 5 \\ - 4 & 5 \end{array}] = \frac{1}{25 - 20} [\begin{array}{cc} 5 & - 5 \\ - 4 & 5 \end{array}] = \frac{1}{5} [\begin{array}{cc} 5 & - 5 \\ - 4 & 5 \end{array}]$

MHT CET-2020

Matrix and Determinant

78788 If $A = [\begin{array}{ll} 1 & 1 \\ 1 & 2 \end{array}], B = [\begin{array}{ll} 4 & 1 \\ 3 & 1 \end{array}]$ , then $(A + B)^{- 1} =$

1

\frac{1}{7} [\begin{array}{ll} 3 & 2 \\ 4 & 5 \end{array}]

2

\frac{1}{7} [\begin{array}{cc} 3 & - 2 \\ - 4 & 5 \end{array}]

3

7 [\begin{array}{ll} 3 & 2 \\ 4 & 5 \end{array}]

4

7 [\begin{array}{cc} 3 & - 2 \\ - 4 & 5 \end{array}]

Explanation:

(B) : Given that,
$A = [\begin{array}{ll} 1 & 1 \\ 1 & 2 \end{array}] and B = [\begin{array}{ll} 4 & 1 \\ 3 & 1 \end{array}]$
$A + B = [\begin{array}{ll} 1 & 1 \\ 1 & 2 \end{array}] + [\begin{array}{ll} 4 & 1 \\ 3 & 1 \end{array}] = [\begin{array}{ll} 5 & 2 \\ 4 & 3 \end{array}]$
$| A + B | = 15 - 8 = 7$ and $adj (A + B) = [\begin{array}{cc} 3 & - 2 \\ - 4 & 5 \end{array}]$
$∴ (A + B)^{- 1} = \frac{1}{7} [\begin{array}{cc} 3 & - 2 \\ - 4 & 5 \end{array}]$

MHT CET-2020

Matrix and Determinant

78789 If $AX = B$ , where $A = [\begin{array}{ccc} 1 & - 1 & 1 \\ 2 & - 1 & 0 \\ 3 & 3 & - 4 \end{array}], B = [\begin{array}{l} 1 \\ 1 \\ 2 \end{array}]$
and $X = [\begin{array}{l} x \\ y \\ z \end{array}]$ , then $x + y + z =$

1 6

2 2

3 1

4 3,

Explanation:

(D) : Given that
$A = [\begin{array}{ccc} 1 & - 1 & 1 \\ 2 & - 1 & 0 \\ 3 & 3 & - 4 \end{array}]$ and $B = [\begin{array}{l} 1 \\ 1 \\ 2 X = B \end{array}], X = [\begin{array}{l} x \\ y \\ z \end{array}]$
On multiplying $A^{- 1}$ both the side, we get -
${AA}^{- 1} X = A^{- 1} B$
$IX = A^{- 1} B$
$X = A^{- 1} B$
Now, $| A | = 4 + (- 8) + 9 = 5$ $∴ A^{- 1} = [\begin{array}{lll} 4 & - 1 & 1 \\ 8 & - 7 & 2 \\ 9 & - 6 & 1 \end{array}] \times \frac{1}{5}$
$∴ [\begin{array}{l} x \\ y \\ z \end{array}] = \frac{1}{5} [\begin{array}{lll} 4 & - 1 & 1 \\ 8 & - 7 & 2 \\ 9 & - 6 & 1 \end{array}] [\begin{array}{l} 1 \\ 1 \\ 2 \end{array}]$
$= \frac{1}{5} [\begin{array}{l} 4 - 1 + 2 \\ 8 - 7 + 4 \\ 9 - 6 + 2 \end{array}] = \frac{1}{5} [\begin{array}{l} 5 \\ 5 \\ 5 \end{array}]$
$[\begin{array}{l} x \\ y \\ z \end{array}] = [\begin{array}{l} 1 \\ 1 \\ 1 \end{array}]$
On comparing both side, we get -
$x = y = z = 1$
$x + y + z = 1 + 1 + 1 = 3$

MHT CET-2020

Matrix and Determinant

78790 If $A = [\begin{array}{ccc} 0 & - 1 & 0 \\ - 1 & 0 & 0 \end{array}]$ , then

1

A

is not invertible

2

A^{- 1} = 2 A

3

A^{- 1} = I

4

A = A^{- 1}

Explanation:

(D) : Given that,
$A = [\begin{array}{ccc} 0 & 0 & - 1 \\ 0 & - 1 & 0 \\ - 1 & 0 & 0 \end{array}]$
Therefore, $| A | = - 1 (- 1) = 1$
And, $A^{- 1} = [\begin{array}{ccc} 0 & 0 & - 1 \\ 0 & - 1 & 0 \\ - 1 & 0 & 0 \end{array}] = A$
$A^{- 1} = A$

MHT CET-2020

Matrix and Determinant

78787 If $A = [\begin{array}{lll} 1 & 2 & 1 \\ 2 & 1 & 0 \end{array}], B = [\begin{array}{ll} 1 & 2 \\ 2 & 1 \\ 0 & 1 \end{array}]$ , then $(A B)^{- 1}$ is

1

(\frac{1}{5}) [\begin{array}{ll} 5 & - 5 \\ 4 & - 5 \end{array}]

2

(\frac{1}{5}) [\begin{array}{cc} 5 & - 5 \\ 4 & 5 \end{array}]

3

(\frac{1}{5}) [\begin{array}{cc} 5 & - 5 \\ - 4 & 5 \end{array}]

4

(\frac{1}{5}) [\begin{array}{cc} 5 & - 5 \\ - 4 & - 5 \end{array}]

Explanation:

(C) : Given that,
$A = [\begin{array}{lll} 1 & 2 & 1 \\ 2 & 1 & 0 \end{array}] and B = [\begin{array}{ll} 1 & 2 \\ 2 & 1 \\ 0 & 1 \end{array}]$
$∴ AB = [\begin{array}{ll} 1 + 4 + 0 & 2 + 2 + 1 \\ 2 + 2 + 0 & 4 + 1 + 0 \end{array}] = [\begin{array}{cc} 5 & 5 \\ 4 & 5 \end{array}]$
$(AB)^{- 1} = \frac{1}{| AB |} [\begin{array}{cc} 5 & - 5 \\ - 4 & 5 \end{array}] = \frac{1}{25 - 20} [\begin{array}{cc} 5 & - 5 \\ - 4 & 5 \end{array}] = \frac{1}{5} [\begin{array}{cc} 5 & - 5 \\ - 4 & 5 \end{array}]$

MHT CET-2020

Matrix and Determinant

78788 If $A = [\begin{array}{ll} 1 & 1 \\ 1 & 2 \end{array}], B = [\begin{array}{ll} 4 & 1 \\ 3 & 1 \end{array}]$ , then $(A + B)^{- 1} =$

1

\frac{1}{7} [\begin{array}{ll} 3 & 2 \\ 4 & 5 \end{array}]

2

\frac{1}{7} [\begin{array}{cc} 3 & - 2 \\ - 4 & 5 \end{array}]

3

7 [\begin{array}{ll} 3 & 2 \\ 4 & 5 \end{array}]

4

7 [\begin{array}{cc} 3 & - 2 \\ - 4 & 5 \end{array}]

Explanation:

(B) : Given that,
$A = [\begin{array}{ll} 1 & 1 \\ 1 & 2 \end{array}] and B = [\begin{array}{ll} 4 & 1 \\ 3 & 1 \end{array}]$
$A + B = [\begin{array}{ll} 1 & 1 \\ 1 & 2 \end{array}] + [\begin{array}{ll} 4 & 1 \\ 3 & 1 \end{array}] = [\begin{array}{ll} 5 & 2 \\ 4 & 3 \end{array}]$
$| A + B | = 15 - 8 = 7$ and $adj (A + B) = [\begin{array}{cc} 3 & - 2 \\ - 4 & 5 \end{array}]$
$∴ (A + B)^{- 1} = \frac{1}{7} [\begin{array}{cc} 3 & - 2 \\ - 4 & 5 \end{array}]$

MHT CET-2020

Matrix and Determinant

78789 If $AX = B$ , where $A = [\begin{array}{ccc} 1 & - 1 & 1 \\ 2 & - 1 & 0 \\ 3 & 3 & - 4 \end{array}], B = [\begin{array}{l} 1 \\ 1 \\ 2 \end{array}]$
and $X = [\begin{array}{l} x \\ y \\ z \end{array}]$ , then $x + y + z =$

1 6

2 2

3 1

4 3,

Explanation:

(D) : Given that
$A = [\begin{array}{ccc} 1 & - 1 & 1 \\ 2 & - 1 & 0 \\ 3 & 3 & - 4 \end{array}]$ and $B = [\begin{array}{l} 1 \\ 1 \\ 2 X = B \end{array}], X = [\begin{array}{l} x \\ y \\ z \end{array}]$
On multiplying $A^{- 1}$ both the side, we get -
${AA}^{- 1} X = A^{- 1} B$
$IX = A^{- 1} B$
$X = A^{- 1} B$
Now, $| A | = 4 + (- 8) + 9 = 5$ $∴ A^{- 1} = [\begin{array}{lll} 4 & - 1 & 1 \\ 8 & - 7 & 2 \\ 9 & - 6 & 1 \end{array}] \times \frac{1}{5}$
$∴ [\begin{array}{l} x \\ y \\ z \end{array}] = \frac{1}{5} [\begin{array}{lll} 4 & - 1 & 1 \\ 8 & - 7 & 2 \\ 9 & - 6 & 1 \end{array}] [\begin{array}{l} 1 \\ 1 \\ 2 \end{array}]$
$= \frac{1}{5} [\begin{array}{l} 4 - 1 + 2 \\ 8 - 7 + 4 \\ 9 - 6 + 2 \end{array}] = \frac{1}{5} [\begin{array}{l} 5 \\ 5 \\ 5 \end{array}]$
$[\begin{array}{l} x \\ y \\ z \end{array}] = [\begin{array}{l} 1 \\ 1 \\ 1 \end{array}]$
On comparing both side, we get -
$x = y = z = 1$
$x + y + z = 1 + 1 + 1 = 3$

MHT CET-2020

Matrix and Determinant

78790 If $A = [\begin{array}{ccc} 0 & - 1 & 0 \\ - 1 & 0 & 0 \end{array}]$ , then

1

A

is not invertible

2

A^{- 1} = 2 A

3

A^{- 1} = I

4

A = A^{- 1}

Explanation:

(D) : Given that,
$A = [\begin{array}{ccc} 0 & 0 & - 1 \\ 0 & - 1 & 0 \\ - 1 & 0 & 0 \end{array}]$
Therefore, $| A | = - 1 (- 1) = 1$
And, $A^{- 1} = [\begin{array}{ccc} 0 & 0 & - 1 \\ 0 & - 1 & 0 \\ - 1 & 0 & 0 \end{array}] = A$
$A^{- 1} = A$

MHT CET-2020

Matrix and Determinant

78787 If $A = [\begin{array}{lll} 1 & 2 & 1 \\ 2 & 1 & 0 \end{array}], B = [\begin{array}{ll} 1 & 2 \\ 2 & 1 \\ 0 & 1 \end{array}]$ , then $(A B)^{- 1}$ is

1

(\frac{1}{5}) [\begin{array}{ll} 5 & - 5 \\ 4 & - 5 \end{array}]

2

(\frac{1}{5}) [\begin{array}{cc} 5 & - 5 \\ 4 & 5 \end{array}]

3

(\frac{1}{5}) [\begin{array}{cc} 5 & - 5 \\ - 4 & 5 \end{array}]

4

(\frac{1}{5}) [\begin{array}{cc} 5 & - 5 \\ - 4 & - 5 \end{array}]

Explanation:

(C) : Given that,
$A = [\begin{array}{lll} 1 & 2 & 1 \\ 2 & 1 & 0 \end{array}] and B = [\begin{array}{ll} 1 & 2 \\ 2 & 1 \\ 0 & 1 \end{array}]$
$∴ AB = [\begin{array}{ll} 1 + 4 + 0 & 2 + 2 + 1 \\ 2 + 2 + 0 & 4 + 1 + 0 \end{array}] = [\begin{array}{cc} 5 & 5 \\ 4 & 5 \end{array}]$
$(AB)^{- 1} = \frac{1}{| AB |} [\begin{array}{cc} 5 & - 5 \\ - 4 & 5 \end{array}] = \frac{1}{25 - 20} [\begin{array}{cc} 5 & - 5 \\ - 4 & 5 \end{array}] = \frac{1}{5} [\begin{array}{cc} 5 & - 5 \\ - 4 & 5 \end{array}]$

MHT CET-2020

Matrix and Determinant

78788 If $A = [\begin{array}{ll} 1 & 1 \\ 1 & 2 \end{array}], B = [\begin{array}{ll} 4 & 1 \\ 3 & 1 \end{array}]$ , then $(A + B)^{- 1} =$

1

\frac{1}{7} [\begin{array}{ll} 3 & 2 \\ 4 & 5 \end{array}]

2

\frac{1}{7} [\begin{array}{cc} 3 & - 2 \\ - 4 & 5 \end{array}]

3

7 [\begin{array}{ll} 3 & 2 \\ 4 & 5 \end{array}]

4

7 [\begin{array}{cc} 3 & - 2 \\ - 4 & 5 \end{array}]

Explanation:

(B) : Given that,
$A = [\begin{array}{ll} 1 & 1 \\ 1 & 2 \end{array}] and B = [\begin{array}{ll} 4 & 1 \\ 3 & 1 \end{array}]$
$A + B = [\begin{array}{ll} 1 & 1 \\ 1 & 2 \end{array}] + [\begin{array}{ll} 4 & 1 \\ 3 & 1 \end{array}] = [\begin{array}{ll} 5 & 2 \\ 4 & 3 \end{array}]$
$| A + B | = 15 - 8 = 7$ and $adj (A + B) = [\begin{array}{cc} 3 & - 2 \\ - 4 & 5 \end{array}]$
$∴ (A + B)^{- 1} = \frac{1}{7} [\begin{array}{cc} 3 & - 2 \\ - 4 & 5 \end{array}]$

MHT CET-2020

Matrix and Determinant

78789 If $AX = B$ , where $A = [\begin{array}{ccc} 1 & - 1 & 1 \\ 2 & - 1 & 0 \\ 3 & 3 & - 4 \end{array}], B = [\begin{array}{l} 1 \\ 1 \\ 2 \end{array}]$
and $X = [\begin{array}{l} x \\ y \\ z \end{array}]$ , then $x + y + z =$

1 6

2 2

3 1

4 3,

Explanation:

(D) : Given that
$A = [\begin{array}{ccc} 1 & - 1 & 1 \\ 2 & - 1 & 0 \\ 3 & 3 & - 4 \end{array}]$ and $B = [\begin{array}{l} 1 \\ 1 \\ 2 X = B \end{array}], X = [\begin{array}{l} x \\ y \\ z \end{array}]$
On multiplying $A^{- 1}$ both the side, we get -
${AA}^{- 1} X = A^{- 1} B$
$IX = A^{- 1} B$
$X = A^{- 1} B$
Now, $| A | = 4 + (- 8) + 9 = 5$ $∴ A^{- 1} = [\begin{array}{lll} 4 & - 1 & 1 \\ 8 & - 7 & 2 \\ 9 & - 6 & 1 \end{array}] \times \frac{1}{5}$
$∴ [\begin{array}{l} x \\ y \\ z \end{array}] = \frac{1}{5} [\begin{array}{lll} 4 & - 1 & 1 \\ 8 & - 7 & 2 \\ 9 & - 6 & 1 \end{array}] [\begin{array}{l} 1 \\ 1 \\ 2 \end{array}]$
$= \frac{1}{5} [\begin{array}{l} 4 - 1 + 2 \\ 8 - 7 + 4 \\ 9 - 6 + 2 \end{array}] = \frac{1}{5} [\begin{array}{l} 5 \\ 5 \\ 5 \end{array}]$
$[\begin{array}{l} x \\ y \\ z \end{array}] = [\begin{array}{l} 1 \\ 1 \\ 1 \end{array}]$
On comparing both side, we get -
$x = y = z = 1$
$x + y + z = 1 + 1 + 1 = 3$

MHT CET-2020

Matrix and Determinant

78790 If $A = [\begin{array}{ccc} 0 & - 1 & 0 \\ - 1 & 0 & 0 \end{array}]$ , then

1

A

is not invertible

2

A^{- 1} = 2 A

3

A^{- 1} = I

4

A = A^{- 1}

Explanation:

(D) : Given that,
$A = [\begin{array}{ccc} 0 & 0 & - 1 \\ 0 & - 1 & 0 \\ - 1 & 0 & 0 \end{array}]$
Therefore, $| A | = - 1 (- 1) = 1$
And, $A^{- 1} = [\begin{array}{ccc} 0 & 0 & - 1 \\ 0 & - 1 & 0 \\ - 1 & 0 & 0 \end{array}] = A$
$A^{- 1} = A$

MHT CET-2020

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