QBManager

Matrix and Determinant

78798 If $ω$ is a complex cube root of unity and $A = [\begin{array}{cc} ω & 0 \\ 0 & ω \end{array}]$ , then $A^{- 1} =$

1

- A

2

2 A

3

A^{2}

4 A

Explanation:

(C) : Given that,
$A = [\begin{array}{ll} ω & 0 \\ 0 & ω \end{array}]$
We know that,
$A^{- 1} = \frac{adj A}{| A |}$
Now, $| A | = (ω^{2} - 0) = ω^{2}$
And $adj A = [\begin{array}{cc} ω & 0 \\ 0 & ω \end{array}]$
$A^{- 1} = \frac{1}{ω^{2}} [\begin{array}{cc} ω & 0 \\ 0 & ω \end{array}]$
$A^{- 1} = [\begin{array}{ll} \frac{1}{ω} & 0 \\ 0 & \frac{1}{ω} \end{array}] = [\begin{array}{cc} ω^{2} & 0 \\ 0 & ω^{2} \end{array}]$
$A^{2} = [\begin{array}{ll} ω & 0 \\ 0 & ω \end{array}] [\begin{array}{ll} ω & 0 \\ 0 & ω \end{array}] = [\begin{array}{cc} ω^{2} & 0 \\ 0 & ω^{2} \end{array}]$
Hence, $A^{- 1} = A^{2}$

MHT CET-2020

Matrix and Determinant

78799 If $A = [\begin{array}{cc} 1 + 2 i & i \\ - i & 1 - 2 i \end{array}]$ where $i = \sqrt{- 1}$ then $A$ $(adj A) =$

1

- 2 I

2

2 I

3

4 I

4

5 I

Explanation:

(C) : Given that,
$A = [\begin{array}{cc} 1 + 2 i & i \\ - i & 1 - 2 i \end{array}]$
Now, $adj A = [\begin{array}{cc} 1 - 2 i & - i \\ i & 1 + 2 i \end{array}]$
$∴ A (adj A) = [\begin{array}{cc} 1 + 2 i & i \\ - i & 1 - 2 i \end{array}] [\begin{array}{cc} 1 - 2 i & - i \\ i & 1 + 2 i \end{array}]$
$A (adj A) = [\begin{array}{cc} 1 - 4 i^{2} + i^{2} & - i - 2 i^{2} + i + 2 i^{2} \\ - i + 2 i^{2} + i - 2 i^{2} & i^{2} + 1 + 4 i^{2} \end{array}]$
$= [\begin{array}{cc} 1 + 4 - 1 & 0 \\ 0 & - 1 + 1 + 4 \end{array}] = [\begin{array}{ll} 4 & 0 \\ 0 & 4 \end{array}] = 4 [\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}]$
$A (adj A) = 4 I$

MHT CET-2019

Matrix and Determinant

78800 If $A$ is non-singular matrix and $(A + I) (A - I)$ $= 0$ , then $A + A^{- 1} =$

1 I

2

2 A

3 0

4

3 I

Explanation:

(B) : According to question, $A$ is non - singular matrix, it means -
$| A | = 0$
Now, $(A + I) (A - I) = 0$
$A^{2} - I^{2} = 0 ∵ a^{2} - b^{2} = (a + b) (a - b)$
$A^{2} = I^{2}$
$A \cdot A = I$
$A = A^{- 1}$
On adding $A$ on both the side we get -
$A + A = A^{- 1} + A$
$A + A^{- 1} = 2 A$

Matrix and Determinant

78801 If $A = [\begin{array}{ll} x & 1 \\ 1 & 0 \end{array}]$ and $A = A^{- 1}$ , then $x =$

1 2

2 1

3

4

4

0

Explanation:

(D) : Given that,
$A = [\begin{array}{ll} x & 1 \\ 1 & 0 \end{array}]$
We know that,
$A^{- 1} = \frac{adj A}{| A |}$
Now, $| A | = 0 - 1 = - 1$
And $adj A = [\begin{array}{cc} 0 & - 1 \\ - 1 & x \end{array}]$
$∴ A^{- 1} = \frac{1}{| A |} adj A = \frac{1}{- 1} [\begin{array}{cc} 0 & - 1 \\ - 1 & x \end{array}] = [\begin{array}{cc} 0 & 1 \\ 1 & - x \end{array}]$
According to question,
$A = A^{- 1}$
$[\begin{array}{ll} x & 1 \\ 1 & 0 \end{array}] = [\begin{array}{cc} 0 & 1 \\ 1 & - x \end{array}]$
On comparing corresponding elements on both side, we get-
$x = 0$

MHT CET-2019

Matrix and Determinant

78798 If $ω$ is a complex cube root of unity and $A = [\begin{array}{cc} ω & 0 \\ 0 & ω \end{array}]$ , then $A^{- 1} =$

1

- A

2

2 A

3

A^{2}

4 A

Explanation:

(C) : Given that,
$A = [\begin{array}{ll} ω & 0 \\ 0 & ω \end{array}]$
We know that,
$A^{- 1} = \frac{adj A}{| A |}$
Now, $| A | = (ω^{2} - 0) = ω^{2}$
And $adj A = [\begin{array}{cc} ω & 0 \\ 0 & ω \end{array}]$
$A^{- 1} = \frac{1}{ω^{2}} [\begin{array}{cc} ω & 0 \\ 0 & ω \end{array}]$
$A^{- 1} = [\begin{array}{ll} \frac{1}{ω} & 0 \\ 0 & \frac{1}{ω} \end{array}] = [\begin{array}{cc} ω^{2} & 0 \\ 0 & ω^{2} \end{array}]$
$A^{2} = [\begin{array}{ll} ω & 0 \\ 0 & ω \end{array}] [\begin{array}{ll} ω & 0 \\ 0 & ω \end{array}] = [\begin{array}{cc} ω^{2} & 0 \\ 0 & ω^{2} \end{array}]$
Hence, $A^{- 1} = A^{2}$

MHT CET-2020

Matrix and Determinant

78799 If $A = [\begin{array}{cc} 1 + 2 i & i \\ - i & 1 - 2 i \end{array}]$ where $i = \sqrt{- 1}$ then $A$ $(adj A) =$

1

- 2 I

2

2 I

3

4 I

4

5 I

Explanation:

(C) : Given that,
$A = [\begin{array}{cc} 1 + 2 i & i \\ - i & 1 - 2 i \end{array}]$
Now, $adj A = [\begin{array}{cc} 1 - 2 i & - i \\ i & 1 + 2 i \end{array}]$
$∴ A (adj A) = [\begin{array}{cc} 1 + 2 i & i \\ - i & 1 - 2 i \end{array}] [\begin{array}{cc} 1 - 2 i & - i \\ i & 1 + 2 i \end{array}]$
$A (adj A) = [\begin{array}{cc} 1 - 4 i^{2} + i^{2} & - i - 2 i^{2} + i + 2 i^{2} \\ - i + 2 i^{2} + i - 2 i^{2} & i^{2} + 1 + 4 i^{2} \end{array}]$
$= [\begin{array}{cc} 1 + 4 - 1 & 0 \\ 0 & - 1 + 1 + 4 \end{array}] = [\begin{array}{ll} 4 & 0 \\ 0 & 4 \end{array}] = 4 [\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}]$
$A (adj A) = 4 I$

MHT CET-2019

Matrix and Determinant

78800 If $A$ is non-singular matrix and $(A + I) (A - I)$ $= 0$ , then $A + A^{- 1} =$

1 I

2

2 A

3 0

4

3 I

Explanation:

(B) : According to question, $A$ is non - singular matrix, it means -
$| A | = 0$
Now, $(A + I) (A - I) = 0$
$A^{2} - I^{2} = 0 ∵ a^{2} - b^{2} = (a + b) (a - b)$
$A^{2} = I^{2}$
$A \cdot A = I$
$A = A^{- 1}$
On adding $A$ on both the side we get -
$A + A = A^{- 1} + A$
$A + A^{- 1} = 2 A$

Matrix and Determinant

78801 If $A = [\begin{array}{ll} x & 1 \\ 1 & 0 \end{array}]$ and $A = A^{- 1}$ , then $x =$

1 2

2 1

3

4

4

0

Explanation:

(D) : Given that,
$A = [\begin{array}{ll} x & 1 \\ 1 & 0 \end{array}]$
We know that,
$A^{- 1} = \frac{adj A}{| A |}$
Now, $| A | = 0 - 1 = - 1$
And $adj A = [\begin{array}{cc} 0 & - 1 \\ - 1 & x \end{array}]$
$∴ A^{- 1} = \frac{1}{| A |} adj A = \frac{1}{- 1} [\begin{array}{cc} 0 & - 1 \\ - 1 & x \end{array}] = [\begin{array}{cc} 0 & 1 \\ 1 & - x \end{array}]$
According to question,
$A = A^{- 1}$
$[\begin{array}{ll} x & 1 \\ 1 & 0 \end{array}] = [\begin{array}{cc} 0 & 1 \\ 1 & - x \end{array}]$
On comparing corresponding elements on both side, we get-
$x = 0$

MHT CET-2019

Matrix and Determinant

78798 If $ω$ is a complex cube root of unity and $A = [\begin{array}{cc} ω & 0 \\ 0 & ω \end{array}]$ , then $A^{- 1} =$

1

- A

2

2 A

3

A^{2}

4 A

Explanation:

(C) : Given that,
$A = [\begin{array}{ll} ω & 0 \\ 0 & ω \end{array}]$
We know that,
$A^{- 1} = \frac{adj A}{| A |}$
Now, $| A | = (ω^{2} - 0) = ω^{2}$
And $adj A = [\begin{array}{cc} ω & 0 \\ 0 & ω \end{array}]$
$A^{- 1} = \frac{1}{ω^{2}} [\begin{array}{cc} ω & 0 \\ 0 & ω \end{array}]$
$A^{- 1} = [\begin{array}{ll} \frac{1}{ω} & 0 \\ 0 & \frac{1}{ω} \end{array}] = [\begin{array}{cc} ω^{2} & 0 \\ 0 & ω^{2} \end{array}]$
$A^{2} = [\begin{array}{ll} ω & 0 \\ 0 & ω \end{array}] [\begin{array}{ll} ω & 0 \\ 0 & ω \end{array}] = [\begin{array}{cc} ω^{2} & 0 \\ 0 & ω^{2} \end{array}]$
Hence, $A^{- 1} = A^{2}$

MHT CET-2020

Matrix and Determinant

78799 If $A = [\begin{array}{cc} 1 + 2 i & i \\ - i & 1 - 2 i \end{array}]$ where $i = \sqrt{- 1}$ then $A$ $(adj A) =$

1

- 2 I

2

2 I

3

4 I

4

5 I

Explanation:

(C) : Given that,
$A = [\begin{array}{cc} 1 + 2 i & i \\ - i & 1 - 2 i \end{array}]$
Now, $adj A = [\begin{array}{cc} 1 - 2 i & - i \\ i & 1 + 2 i \end{array}]$
$∴ A (adj A) = [\begin{array}{cc} 1 + 2 i & i \\ - i & 1 - 2 i \end{array}] [\begin{array}{cc} 1 - 2 i & - i \\ i & 1 + 2 i \end{array}]$
$A (adj A) = [\begin{array}{cc} 1 - 4 i^{2} + i^{2} & - i - 2 i^{2} + i + 2 i^{2} \\ - i + 2 i^{2} + i - 2 i^{2} & i^{2} + 1 + 4 i^{2} \end{array}]$
$= [\begin{array}{cc} 1 + 4 - 1 & 0 \\ 0 & - 1 + 1 + 4 \end{array}] = [\begin{array}{ll} 4 & 0 \\ 0 & 4 \end{array}] = 4 [\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}]$
$A (adj A) = 4 I$

MHT CET-2019

Matrix and Determinant

78800 If $A$ is non-singular matrix and $(A + I) (A - I)$ $= 0$ , then $A + A^{- 1} =$

1 I

2

2 A

3 0

4

3 I

Explanation:

(B) : According to question, $A$ is non - singular matrix, it means -
$| A | = 0$
Now, $(A + I) (A - I) = 0$
$A^{2} - I^{2} = 0 ∵ a^{2} - b^{2} = (a + b) (a - b)$
$A^{2} = I^{2}$
$A \cdot A = I$
$A = A^{- 1}$
On adding $A$ on both the side we get -
$A + A = A^{- 1} + A$
$A + A^{- 1} = 2 A$

Matrix and Determinant

78801 If $A = [\begin{array}{ll} x & 1 \\ 1 & 0 \end{array}]$ and $A = A^{- 1}$ , then $x =$

1 2

2 1

3

4

4

0

Explanation:

(D) : Given that,
$A = [\begin{array}{ll} x & 1 \\ 1 & 0 \end{array}]$
We know that,
$A^{- 1} = \frac{adj A}{| A |}$
Now, $| A | = 0 - 1 = - 1$
And $adj A = [\begin{array}{cc} 0 & - 1 \\ - 1 & x \end{array}]$
$∴ A^{- 1} = \frac{1}{| A |} adj A = \frac{1}{- 1} [\begin{array}{cc} 0 & - 1 \\ - 1 & x \end{array}] = [\begin{array}{cc} 0 & 1 \\ 1 & - x \end{array}]$
According to question,
$A = A^{- 1}$
$[\begin{array}{ll} x & 1 \\ 1 & 0 \end{array}] = [\begin{array}{cc} 0 & 1 \\ 1 & - x \end{array}]$
On comparing corresponding elements on both side, we get-
$x = 0$

MHT CET-2019

Matrix and Determinant

78798 If $ω$ is a complex cube root of unity and $A = [\begin{array}{cc} ω & 0 \\ 0 & ω \end{array}]$ , then $A^{- 1} =$

1

- A

2

2 A

3

A^{2}

4 A

Explanation:

(C) : Given that,
$A = [\begin{array}{ll} ω & 0 \\ 0 & ω \end{array}]$
We know that,
$A^{- 1} = \frac{adj A}{| A |}$
Now, $| A | = (ω^{2} - 0) = ω^{2}$
And $adj A = [\begin{array}{cc} ω & 0 \\ 0 & ω \end{array}]$
$A^{- 1} = \frac{1}{ω^{2}} [\begin{array}{cc} ω & 0 \\ 0 & ω \end{array}]$
$A^{- 1} = [\begin{array}{ll} \frac{1}{ω} & 0 \\ 0 & \frac{1}{ω} \end{array}] = [\begin{array}{cc} ω^{2} & 0 \\ 0 & ω^{2} \end{array}]$
$A^{2} = [\begin{array}{ll} ω & 0 \\ 0 & ω \end{array}] [\begin{array}{ll} ω & 0 \\ 0 & ω \end{array}] = [\begin{array}{cc} ω^{2} & 0 \\ 0 & ω^{2} \end{array}]$
Hence, $A^{- 1} = A^{2}$

MHT CET-2020

Matrix and Determinant

78799 If $A = [\begin{array}{cc} 1 + 2 i & i \\ - i & 1 - 2 i \end{array}]$ where $i = \sqrt{- 1}$ then $A$ $(adj A) =$

1

- 2 I

2

2 I

3

4 I

4

5 I

Explanation:

(C) : Given that,
$A = [\begin{array}{cc} 1 + 2 i & i \\ - i & 1 - 2 i \end{array}]$
Now, $adj A = [\begin{array}{cc} 1 - 2 i & - i \\ i & 1 + 2 i \end{array}]$
$∴ A (adj A) = [\begin{array}{cc} 1 + 2 i & i \\ - i & 1 - 2 i \end{array}] [\begin{array}{cc} 1 - 2 i & - i \\ i & 1 + 2 i \end{array}]$
$A (adj A) = [\begin{array}{cc} 1 - 4 i^{2} + i^{2} & - i - 2 i^{2} + i + 2 i^{2} \\ - i + 2 i^{2} + i - 2 i^{2} & i^{2} + 1 + 4 i^{2} \end{array}]$
$= [\begin{array}{cc} 1 + 4 - 1 & 0 \\ 0 & - 1 + 1 + 4 \end{array}] = [\begin{array}{ll} 4 & 0 \\ 0 & 4 \end{array}] = 4 [\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}]$
$A (adj A) = 4 I$

MHT CET-2019

Matrix and Determinant

78800 If $A$ is non-singular matrix and $(A + I) (A - I)$ $= 0$ , then $A + A^{- 1} =$

1 I

2

2 A

3 0

4

3 I

Explanation:

(B) : According to question, $A$ is non - singular matrix, it means -
$| A | = 0$
Now, $(A + I) (A - I) = 0$
$A^{2} - I^{2} = 0 ∵ a^{2} - b^{2} = (a + b) (a - b)$
$A^{2} = I^{2}$
$A \cdot A = I$
$A = A^{- 1}$
On adding $A$ on both the side we get -
$A + A = A^{- 1} + A$
$A + A^{- 1} = 2 A$

Matrix and Determinant

78801 If $A = [\begin{array}{ll} x & 1 \\ 1 & 0 \end{array}]$ and $A = A^{- 1}$ , then $x =$

1 2

2 1

3

4

4

0

Explanation:

(D) : Given that,
$A = [\begin{array}{ll} x & 1 \\ 1 & 0 \end{array}]$
We know that,
$A^{- 1} = \frac{adj A}{| A |}$
Now, $| A | = 0 - 1 = - 1$
And $adj A = [\begin{array}{cc} 0 & - 1 \\ - 1 & x \end{array}]$
$∴ A^{- 1} = \frac{1}{| A |} adj A = \frac{1}{- 1} [\begin{array}{cc} 0 & - 1 \\ - 1 & x \end{array}] = [\begin{array}{cc} 0 & 1 \\ 1 & - x \end{array}]$
According to question,
$A = A^{- 1}$
$[\begin{array}{ll} x & 1 \\ 1 & 0 \end{array}] = [\begin{array}{cc} 0 & 1 \\ 1 & - x \end{array}]$
On comparing corresponding elements on both side, we get-
$x = 0$

MHT CET-2019

Question Bank Series

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