RD Sharma Solutions Class 12 Mathematics Chapter 6 FBQ

Updated on Jan 20, 2022 02:41 PM IST

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RD Sharma Class 12 Solutions Chapter 6 FBQ Adjoint & Inverse of Matrix - Other Exercise

Adjoint and Inverese of Matrices Exercise Fill in the blanks Question 1

Answer $\to I_{n}$
Hint

\to A^{- 1} = a d j A / | A |

Given

\to A

is a unit matrix of order

n

then

A (a d j A) =

Explanation: A is a unit matrix of order

n \Rightarrow | A | = 1

Now,

\begin{aligned} A^{- 1} = (adj A) / | A | \\ A A^{- 1} = A (adj A) / | A | \end{aligned}

I_{n \times n} = A (adj A) / | A |

[_{as} | A | = 1]

A (adj A) = I_{n \times n}

.
Hence,

A (adj A) = I_{n \times n}

Remarks:

A (adj A) = I_{n \times n}

where

A

is a unit matrix of order

n

Adjoint and Inverese of Matrices Exercise Fill in the blanks Question 2

Answer

\to A^{2}

Hint

\to

Determinant of non-singular matrix is non-zero
Given

\to A

is a non-singular matrix such that

A^{3} = I

Explanation

\to A

is a non-singular matrix, therefore

| A | \neq 0

\begin{aligned} A^{3} = I \\ A \cdot A^{2} = I \end{aligned}

A^{- 1} A \cdot A^{2} = I \cdot A^{- 1} [A^{- 1} A = A A^{- 1} = I]

Hence,

\begin{aligned} A^{2} & = A^{- 1} \\ A^{- 1} & = A^{2} \end{aligned}

Adjoint and Inverse of Matrix Excercise Fill in the blanks Question 3

Answer →
Given

\to A

and

B

are square matrices of the same order and
Explanation

\to A B = 3 I

\begin{aligned} A^{- 1} A B = A^{- 1} 3 I [A^{- 1} A = A A^{- 1} = I] \\ I B = 3 A^{- 1} I [B I = I = I B^{2}] \\ B = 3 A^{- 1} \end{aligned}

Hence,

A^{- 1} = \frac{1}{3} B

Adjoint and Inverse of Matrix Excercise Fill in the blanks Question 4

Answer
Hint

\to A

matrix is non-invertible if and only if

| A | = 0

Given $\to A = [\begin{matrix} 1 & a & 2 \\ 1 & 2 & 5 \\ 2 & 1 & 1 \end{matrix}]$ is not invertible
Explanation $\to | A | = [\begin{matrix} 1 & a & 2 \\ 1 & 2 & 5 \\ 2 & 1 & 1 \end{matrix}] = 0$
Expanding along first row

\begin{aligned} 1 (2 - 5) - a (1 - 10) + 2 (1 - 4) = 0 \\ - 3 + 9 a - 6 = 0 \\ 9 a = 9 \\ a = 1 \end{aligned}

Hence,

a = 1

Adjoint and Inverse Matrix Exercise Fill in the blanks Question 5

Answer

\to

zero matrix
Hint

\to

Determinant of singular matrix is

0

.
Given

\to A

is a singular matrix
Explanation $\to A^{- 1} = adj A / | A |$

A (adj A) = | A | I

[As

A

is a singular matrix

∴ | A | = 0

]

\begin{aligned} = 0 \cdot I \\ = 0 \end{aligned}

Hence,

A (a d j A)

is a null matrix.

Adjoint and Inverse Matrix Exercise Fill in the blanks Question 6

Answer $\to {(11)}^{4} = 14, 641$
Hint

\to

Determinant of cofactors of elements of

A

is equal to the determinant of adjoint

A

Given

\to

A

is a matrix of order

3

and

| A | = 11

Explanation $\begin{aligned} \to | A | = A (adj A) \\ | A | = A (C)^{T} \end{aligned}$

Where

C

represents the cofactor of elements of

A

\begin{aligned} | C |^{T} | A | = \det (| A | I) \\ \det (C^{T}) = | A |^{3} = | A |^{3 - 1} \\ \det (C^{T}) = | A |^{2} \end{aligned}

| B | = | A |^{2} = (11)^{2} = 121

Hence,

{| B |}^{2} = {(121)}^{2}

= 14641

Adjoint and Inverse of a Matrix exercise Fill in the blanks question 7

Answer $\to 100$
Hint $\to A (adj A) = | A | I$
Given $\to A$ is a square matrix of order

2

such that

A (adj A) = [\begin{array}{cc} 10 & 0 \\ 0 & 10 \end{array}]

Explanation

\begin{aligned} \to A (adj A) = | A | I \\ = | \begin{array}{cc} 10 & 0 \\ 0 & 10 \end{array} | \\ = 100 - 0 \\ = 100 \end{aligned}

Adjoint and Inverse of a Matrix exercise Fill in the blanks question 8

Answer

\to 9

Hint

\to A (adj A) = \det (A) I_{n \times n}

Given

\to A

is an invertible matrix of order

3

and

| A | = 3

Explanation

\to (adj A) = | A |^{n - 1}

\begin{aligned} = | A |^{3 - 1} [n = 3] \\ = | A |^{2} \\ | adj A | = (3)^{2} \\ = 9 \end{aligned}

Adjoint and Inverse of a Matrix Exercise fill in the blaks Question 9

Answer $\to 5 A$
Hint $\to adj (adj A) = (\det A)^{n - 2} \cdot A$
Given $\to A$ is an invertible matrix of order

3

and

| A | = 5

Explanation

\to A^{- 1} = \frac{adj A}{\det (A)}

…(i)

(adj A)^{- 1} = {(\det (A) A^{- 1})}^{- 1} = A / \det (A)

…(ii)
In eqn (i) replace A by adj A

\begin{aligned} (adj (A))^{- 1} = \frac{adj (adj (A))}{\det (adj A)} \\ adj (adj A) = \det (adj A) (adj A)^{- 1} \end{aligned}

= (\det (A))^{n - 1} \times \frac{A}{\det (A)}

[from (ii)]

= (\det A)^{n - 2} A

adj (adj A) = | A | A = 5 A

[n = 3]

Adjoint and Inverse of a Matrix Exercise fill in the blaks Question 10

Answer $\to 256$
Hint $\to | adj (adj A) | = | A |^{(n - 1)^{2}}$
Given

\to A

is an invertible matrix of order

3

and

| A | = 4

Explanation $\to A (adj A) = | A | I$

| A (adjA) | = ‖ A | I |

| A | | adj A | = | A |^{n}

[| | A | I | = | A |^{n} and | A B | = | A | | B |]

\begin{aligned} | adj A | = | A |^{n - 1} \\ | adj (adj A) | = | adj A |^{n - 1} \end{aligned}

\begin{aligned} {= | A |}^{n - 1) (n - 1)} \\ = | A |^{(n - 1)^{2}} \\ = (4)^{(3 - 1)^{2}} \\ = (4)^{4} \\ = 256 \end{aligned}

Adjoint and Inverse of Matrix Excercise Fill in the Blanks Question 11

Answer

\to 6^{4} = 1296

Hint

\to

In diagonal matrix all elements are zero except diagonal elements and its determinant is equal to product of diagonal elements.
Given $\to A = diag (1, 2, 3)$
Explanation

\to A (adj A) = | A | I

\begin{aligned} | adj (adj A) | = | A |^{n - 1)^{2}} \\ = (6)^{(3 - 1)^{2}} ∣ \\ = (6)^{4} \\ = 1296 \end{aligned}

Hence,

| adj (adj A) | = 1296

Adjoint and Inverse of Matrix Excercise Fill in the Blanks Question 12

Answer $\to 2 / 5$
Hint

\to \det (A^{- 1}) = 1 / \det (A)

Given

\to A

is a square matrix of order

3

such that

| A | = 5 / 2

Explanation

\to | A A^{- 1} | = 1

\begin{aligned} | A | | A^{- 1} | = 1 \\ | A^{- 1} | = \frac{1}{| A |} = \frac{1}{\frac{5}{2}} \\ | A^{- 1} | = 2 / 5 \end{aligned}

Adjoint and Inverse Matrix Exercise Fill in the blanks Question 13

Answer

\to 10

Hint $\to A (a d j A) = | A | . I$
Given

\to A

is a square matrix such that

\to A (a d j A) = 10. I

Explanation $\to A^{- 1} = \frac{1}{| A |} (adj A)$

| A | = A (adj A)

Hence,

A = 10

Adjoint and Inverse Matrix Exercise Fill in the blanks Question 14

Answer $\to 25$
Hint $\to | adj A | = | A |^{n - 1}$
Given

\to A

is a square matrix of order

3

and

B = | A | A^{- 1}

and

| A | = 5

Explanation $\to A (adj A) = | A | I$

adj A = | A | A^{- 1}

B = adj A = | A | A^{- 1}

…(given)

\begin{aligned} | B | = | adj A | \\ = | A |^{n - 1} \\ = (5)^{2} = 25 \end{aligned}

Hence,

| B | = 25

Adjoint and Inverse of a Matrix exercise Fill in the blanks question 15

Answer $\to k^{2} I$
Hint $\to adj (k A) = k^{n - 1} (adj A)$
Given

\to k

is a scalar and

I

is a unit matrix of order

3

Explanation $\to adj (k I) = k^{n - 1} (a d j I)$

\begin{aligned} adj (k I) = k^{3 - 1} a d j (I) \\ = k^{2} \cdot I \end{aligned}

Hence,

adj (k I) = k^{2} \cdot I

Adjoint and Inverse of a Matrix exercise Fill in the blanks question 16

Answer $\to 1$
Hint $\to A (adj A) = | A | I$
Given

\to A = | \begin{array}{cc} \cos x & \sin x \\ - \sin x & \cos x \end{array} |

and

A (adj A) = [\begin{array}{ll} k & 0 \\ 0 & k \end{array}]

Explanation $\to | A | = | \begin{array}{cc} \cos x & \sin x \\ - \sin x & \cos x \end{array} |$

\begin{aligned} = \cos^{2} x + \sin^{2} x \\ = 1 \end{aligned}

Since we have

\to A (adj A) = | A | I

Applying determinant on both side

| A (adj A) | = | | A | I |

…(i)
Also we have given

| A (a d j A) | = [\begin{matrix} k & 0 \\ 0 & k \end{matrix}] = k^{2} | \begin{matrix} 1 & 0 & 0 & 1 \end{matrix} |

…(ii)
Comparing (i) and (ii) we get

\begin{aligned} | | A | I | = k^{2} \\ | A |^{2} = k^{2} \\ k = 1 \end{aligned}

Adjoint and Inverse of a Matrix Exercise fill in the blaks Question 17

Answer

\to | A | A

Given $\to$ is a non-singular matrix of order

3

Explanation $\to | A | = A . adj A$ …(i)

(adj A)^{- 1} = \frac{A}{| A |}

(adj A)^{- 1} = \frac{adj (adj A)}{\det (adj A)}

[

A = a d j (a d j (A)) and \det (adj A) = (\det \det A)^{n - 1}

]

adj (adj A) = (adj A)^{- 1} \det (adj A)

= (\det A)^{n - 1} \times \frac{A}{\det (A)}

[

\det (adj A) = (\det \det A)^{n - 1}

]

= (\det A)^{n - 2} A

Now,

n = 3

\begin{aligned} adj (adj A) = (\det A) A \\ = | A | A \end{aligned}

Adjoint and Inverse of a Matrix Exercise fill in the blaks Question 18

Answer $\to \frac{1}{9} [\begin{array}{ll} 0 & 3 \\ 3 & 1 \end{array}]$
Given

\to A = {[a_{i j}]}_{2 \times 2} where a_{i j} = {i + j if i \neq j i^{2} - 2 j if i = j

Hint $\to A^{- 1} = \frac{1}{| A |} adj A$
Explanation

\to

find

a_{11}, a_{12}, a_{21}, a_{22}

\begin{aligned} a_{11} = 1^{2} - 2 \times 1 = - 1 \\ a_{12} = 1 + 2 = 3 \\ a_{21} = 2 + 1 = 3 \\ a_{22} = 2^{2} - 2 \times 2 = 0 \\ A = [\begin{array}{cc} - 1 & 3 \\ 3 & 0 \end{array}] \end{aligned}

\begin{aligned} | A | = - 1 \times 0 - 3 \times 3 = - 9 \\ adj A = [\begin{array}{cc} 0 & - 3 \\ - 3 & - 1 \end{array}] \\ A^{- 1} = \frac{1}{- 9} [\begin{array}{cc} 0 & - 3 \\ - 3 & - 1 \end{array}] \end{aligned}

A^{- 1} = \frac{1}{9} [\begin{array}{ll} 0 & 3 \\ 3 & 1 \end{array}]

Adjoint and Inverse Matrix Exercise Fill in the blanks Question 19

Answer

\to λ = - 1 / 6

Given

\to A = [\begin{array}{ll} 0 & 3 \\ 2 & 0 \end{array}], A^{- 1} = λ (adj A)

Explanation $\to adj A = [\begin{array}{cc} 0 & - 3 \\ - 2 & 0 \end{array}]$

λ (adj A) = [\begin{array}{llll} 0 & - 3 λ & - 2 λ & 0 \end{array}]

…(i)
Now,

A^{- 1} = \frac{1}{- 6} | \begin{matrix} 0 & - 3 & - 20 \end{matrix} |

\Rightarrow [\begin{matrix} 0 & \frac{1}{2} & \frac{1}{3} & 0 \end{matrix}] = λ (a d j A) = | \begin{matrix} 0 & - 3 λ & - 2 λ & 0 \end{matrix} |

… [from (i)]

\Rightarrow - 3 λ = \frac{1}{2} \Rightarrow λ = - 1 / 6

Adjoint and Inverse Matrix Exercise Fill in the blanks Question 20

Answer

\to I

Given

\to A A^{T} = A^{T} A and B = A^{- 1} A^{T}

Explanation

\to B = A^{- 1} A^{T}

B^{- 1} = {(A^{- 1} A^{T})}^{- 1}

= {(A^{T})}^{- 1} {(A^{- 1})}^{- 1} [(A B)^{- 1} = B^{- 1} A^{- 1}]

Now,

\begin{aligned} B B^{- 1} = A^{- 1} A^{T} {(A^{T})}^{- 1} {(A^{- 1})}^{- 1} \\ = A^{- 1} A^{T} {(A^{T})}^{- 1} A \\ = A^{- 1} {(A A^{- 1})}^{T} A \\ = A^{- 1} \cdot I \cdot A \end{aligned}

= A^{- 1} A = 1

B B^{- 1} = I

Adjoint and Inverse of a Matrix Exercise fill in the blaks Question 21

Answer $\to A^{2} + B^{2}$
Hint $\to A B = - B A$
Given $\to B = - A^{- 1} B A$
Explanation

\to A

B

are square matrices of same order

\begin{aligned} B = - A^{- 1} B A \\ A B = - A A^{- 1} B A \end{aligned}

A B = - I . B A

A B = - B A

...(i)
Now,

(A + B)^{2} = A^{2} + B^{2} + A B + B A

= A^{2} + B^{2} - B A + B A

from (i)

(A + B)^{2} = A^{2} + B^{2}

Adjoint and Inverse of a Matrix Exercise fill in the blaks Question 22

Answer $\to {(A^{- 1})}^{3}$
Hint

\to A

is a non-singular matrix hence

A

is non-zero
Given

\to A

is a non-singular matrix of order

3 \times 3

Explanation $\to {(A^{3})}^{- 1} = \frac{1}{A^{3}}$

\begin{aligned} = {(\frac{1}{A})}^{3} \\ {(A^{3})}^{- 1} = {(A^{- 1})}^{3} \end{aligned}

Adjoint and Inverse of a Matrix exercise Fill in the blanks question 23

Answer $\to 3 \times 3$
Hint $\to (adj A) = | A |^{n - 1}$
Given

\to | (adj A) | = | A |^{2}

Explanation $\to adj A = | A | A^{- 1}$

| adj A | = | A |^{n - 1}

| A |^{2} = | A |^{n - 1}

… [ given

| (adj A) | = | A |^{2}

]

\begin{aligned} n - 1 = 2 \\ n = 3 \end{aligned}

Adjoint and Inverse of a Matrix exercise Fill in the blanks question 24

Answer $\to 79$
Hint $\to Aadj A = | A | I$
Given

\to A = [\begin{array}{lll} x & 5 & 2 \\ 2 & y & 3 \\ 1 & 1 & z \end{array}] x y z = 80, 3 x + 2 y + 10 z = 20

Explanation

\to | A | = | \begin{array}{lll} x & 5 & 2 \\ 2 & y & 3 \\ 1 & 1 & z \end{array} |

\begin{aligned} = x (y z - 3) - 5 (2 z - 3) + 2 (2 - y) \\ = x y z - 3 x - 10 z + 15 + 4 - 2 y \\ = x y z - (3 x + 2 y + 10 z) + 19 \\ = 80 - 20 + 19 \\ = 79 \end{aligned}

Adjoint and Inverse of Matrix Excercise Fill in the Blanks Question 25

Answer $\to [\begin{array}{ll} 3 & - 4 \\ 1 & - 1 \end{array}]$
Hint $\to A^{- 1} = \frac{1}{| A |} adj A$
Given $\to A = [\begin{array}{ll} 3 & - 4 \\ 1 & - 1 \end{array}]$
Explanation $\to | A | = [\begin{array}{ll} 3 & - 4 \\ 1 & - 1 \end{array}]$

\begin{aligned} = - 3 + 4 = 1 \\ adj A = [\begin{array}{rr} - 1 & 4 \\ - 1 & 3 \end{array}] \end{aligned}

\begin{aligned} A^{- 1} = \frac{1}{| A |} adj A \\ A^{- 1} = [\begin{array}{ll} - 1 & 4 \\ - 1 & 3 \end{array}] \end{aligned}

RD Sharma Class 12th Chapter 6 FBQ is specifically designed for students who require extra content to score better marks in exams. This material contains in-detail explanations about all concepts that will help students better understand the subject.

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4. What is a symmetric matrix?

If the transpose of a matrix is equal to itself, then it is called a symmetric matrix. For more details, check RD Sharma class 12 solutions chapter 6 FBQ.

5. What is an Identity matrix?

A Matrix that contains elements only in its principal diagonal is called an identity Matrix. The rest of the elements are zeros. To know more about Matrices check, RD Sharma class 12 solutions chapter 6 FBQ.

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RD Sharma Solutions Class 12 Mathematics Chapter 6 FBQ

RD Sharma Class 12 Solutions Chapter 6 FBQ Adjoint & Inverse of Matrix - Other Exercise

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