RD Sharma Class 12 Exercise 6.2 Adjoint and Inverse of Matrix Solutions Maths

RD Sharma Class 12 Exercise 6.2 Adjoint and Inverse of Matrix Solutions Maths - Download PDF Free Online

Updated on Jan 20, 2022 03:00 PM IST

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Adjoint and Inverse of Matrix excercise 6.2 question 1

Answer:

\frac{1}{25} [\begin{matrix} 3 & 1 \\ 4 & - 7 \end{matrix}]

Hint: Here we use the concept of elementary row operation
Given:

[\begin{matrix} 7 & 1 \\ 4 & - 3 \end{matrix}]

Solution:

A = I A

A = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] A

\Rightarrow [\begin{matrix} 7 & 1 \\ 4 & - 3 \end{matrix}] = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] A

Applying

R_{1} \to \frac{1}{7} R_{1}

\Rightarrow [\begin{matrix} 1 & \frac{1}{7} \\ 4 & - 3 \end{matrix}] = [\begin{matrix} \frac{1}{7} & 0 \\ 0 & 1 \end{matrix}] A

Applying

R_{2} \to R_{2} - 4 R_{1}

\Rightarrow [\begin{matrix} 1 & \frac{1}{7} \\ 0 & \frac{- 25}{7} \end{matrix}]

= [\begin{matrix} \frac{1}{7} & 0 \\ \frac{- 4}{7} & 1 \end{matrix}] A

Applying

R_{2} \to \frac{- 7}{25} R_{2}

\Rightarrow [\begin{matrix} 1 & \frac{1}{7} \\ 0 & 1 \end{matrix}] = [\begin{matrix} \frac{1}{7} & 0 \\ \frac{4}{25} & \frac{- 7}{25} \end{matrix}]

Applying

R_{1} \to R_{1} - \frac{1}{7} R_{1}

\Rightarrow [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = [\begin{matrix} \frac{21}{75} & \frac{1}{25} \\ \frac{4}{25} & \frac{- 7}{25} \end{matrix}] A

\Rightarrow [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = [\begin{matrix} \frac{3}{25} & \frac{1}{25} \\ \frac{4}{25} & \frac{- 7}{25} \end{matrix}] A

\Rightarrow A^{- 1} = \frac{1}{25} [\begin{matrix} 3 & 1 \\ 4 & - 7 \end{matrix}]

Adjoint and Inverse of Matrix excercise 6.2 question 2

Answer:

A^{- 1} = [\begin{matrix} 1 & - 2 \\ - 2 & 5 \end{matrix}]

Hint: Here, we use the basic of matrix transpose
Given:

[\begin{matrix} 5 & 2 \\ 2 & 1 \end{matrix}]

Solution: Let

A = I A

A = [\begin{matrix} 5 & 2 \\ 2 & 1 \end{matrix}], I = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]

\Rightarrow [\begin{matrix} 5 & 2 \\ 2 & 1 \end{matrix}] = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] A

Applying

R_{1} = \frac{1}{5} R_{1}

\Rightarrow [\begin{matrix} 1 & \frac{2}{5} \\ 0 & \frac{1}{5} \end{matrix}] = [\begin{matrix} \frac{1}{5} & 0 \\ 0 & 1 \end{matrix}] A

Applying

R_{2} = R_{2} - 2 R_{1}

\Rightarrow [\begin{matrix} 1 & \frac{2}{5} \\ 0 & \frac{1}{5} \end{matrix}] = [\begin{matrix} \frac{1}{5} & 0 \\ \frac{- 2}{5} & 1 \end{matrix}] A

\Rightarrow [\begin{matrix} 1 & \frac{2}{5} \\ 0 & 1 \end{matrix}] = [\begin{matrix} \frac{1}{5} & 0 \\ - 2 & 5 \end{matrix}] A

Applying

R_{2} \to 5 R_{2}

\Rightarrow [\begin{matrix} 1 & \frac{2}{5} \\ 0 & 1 \end{matrix}] = [\begin{matrix} \frac{1}{5} & 0 \\ - 2 & 5 \end{matrix}] A

Applying

R_{1} \to R_{1} - \frac{2}{5} R_{2}

\Rightarrow [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = [\begin{matrix} 1 & - 2 \\ - 2 & 5 \end{matrix}] A

So, Here

A^{- 1} = [\begin{matrix} 1 & - 2 \\ - 2 & 5 \end{matrix}]

Adjoint and Inverse of Matrix excercise 6.2 question 4

Answer:

A^{- 1} = [\begin{array}{cc} 3 & - 5 \\ - 1 & 2 \end{array}]

Hint: Here, we use the concept of matrix inverse using elementary row operation
Given:

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

Solution: Let

A = [\begin{array}{ll} 2 & 5 \\ 1 & 3 \end{array}]

A = I A

A = [\begin{array}{ll} 2 & 5 \\ 1 & 3 \end{array}], I = [\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}]

\Rightarrow [\begin{array}{ll} 2 & 5 \\ 1 & 3 \end{array}] = [\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}] A

Applying

R_{1} \to R_{1} - R_{2}

\Rightarrow [\begin{array}{ll} 1 & 2 \\ 1 & 3 \end{array}] = [\begin{array}{cc} 1 & - 1 \\ 0 & 1 \end{array}] A

Applying

R_{2} \to R_{2} - R_{1}

\Rightarrow [\begin{array}{ll} 1 & 2 \\ 0 & 1 \end{array}] = [\begin{array}{cc} 1 & - 1 \\ - 1 & 2 \end{array}] A

Applying

R_{1} \to R_{1} - 2 R_{2}

\Rightarrow [\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}] = [\begin{array}{cc} 3 & - 5 \\ - 1 & 2 \end{array}] A

Hence,

A^{- 1} = [\begin{array}{cc} 3 & - 5 \\ - 1 & 2 \end{array}]

Adjoint and Inverse of Matrix excercise 6 point 2 question 5

Answer:

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

Hint: Here, we use the concept of matrix inverse using elementary row operation
Given:

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

Solution: Let

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

A = I A

A = [\begin{array}{cc} 3 & 10 \\ 2 & 7 \end{array}], I = [\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}]

\Rightarrow [\begin{array}{cc} 3 & 10 \\ 2 & 7 \end{array}] = [\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}] A

Applying

R_{1} \to \frac{1}{3} R_{1}

\Rightarrow [\begin{array}{ll} 1 & \frac{10}{3} \\ 2 & 7 \end{array}] = [\begin{array}{ll} \frac{1}{3} & 0 \\ 0 & 1 \end{array}] A

Applying

R_{2} \to R_{2} - 2 R_{1}

\Rightarrow [\begin{array}{ll} 1 & \frac{10}{3} \\ 0 & \frac{1}{3} \end{array}] = [\begin{array}{cc} \frac{1}{3} & 0 \\ \frac{- 2}{3} & 1 \end{array}] A

Applying

R_{2} \to 3 R_{2}

\Rightarrow [\begin{array}{ll} 1 & \frac{10}{3} \\ 0 & 1 \end{array}] = [\begin{array}{cc} \frac{1}{3} & 0 \\ - 2 & 3 \end{array}] A

Applying

R_{1} \to R_{1} - \frac{10}{3} R_{2}

\Rightarrow [\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}] = [\begin{array}{cc} 7 & - 10 \\ - 2 & 3 \end{array}]

Hence,

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

Adjoint and Inverse of Matrix excercise 6.2 question 6

Answer:

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

Hint: Here, we use the concept of matrix inverse using elementary row operation
Given:

[\begin{array}{lll} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{array}]

Solution: Let

A = [\begin{array}{lll} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{array}]

A = I A

A = [\begin{array}{lll} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{array}], I = [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]

\Rightarrow [\begin{array}{lll} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{array}] = [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] A

Applying

R_{1} \leftrightarrow R_{2}

\Rightarrow [\begin{array}{lll} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 3 & 1 & 1 \end{array}] = [\begin{array}{lll} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array}] A

Applying

R_{3} \to R_{3} - 3 R_{1}

\Rightarrow [\begin{array}{ccc} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & - 5 & - 8 \end{array}] = [\begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & - 3 & 1 \end{array}] A

Applying

R_{1} \to R_{1} - 2 R_{2}

\Rightarrow [\begin{array}{ccc} 1 & 0 & - 1 \\ 0 & 1 & 2 \\ 0 & - 5 & - 8 \end{array}] = [\begin{array}{ccc} - 2 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & - 3 & 1 \end{array}] A

Applying

R_{3} \to R_{3} + 5 R_{2}

\Rightarrow [\begin{array}{ccc} 1 & 0 & - 1 \\ 0 & 1 & 2 \\ 0 & 0 & 2 \end{array}] = [\begin{array}{ccc} - 2 & 1 & 0 \\ 1 & 0 & 0 \\ 5 & - 3 & 1 \end{array}] A

Applying

R_{3} \to \frac{1}{2} R_{3}

\Rightarrow [\begin{array}{ccc} 1 & 0 & - 1 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{array}] = [\begin{array}{ccc} - 2 & 1 & 0 \\ 1 & 0 & 0 \\ \frac{5}{2} & \frac{- 3}{2} & \frac{1}{2} \end{array}] A

Applying

R_{1} \to R_{1} + R_{3}

\Rightarrow [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{array}] = [\begin{array}{ccc} \frac{1}{2} & - \frac{1}{2} & \frac{1}{2} \\ 1 & 0 & 0 \\ \frac{5}{2} & \frac{- 3}{2} & \frac{1}{2} \end{array}] A

Applying

R_{2} \to R_{2} - 2 R_{3}

\Rightarrow [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] = [\begin{array}{ccc} \frac{1}{2} & - \frac{1}{2} & \frac{1}{2} \\ - 4 & 3 & - 1 \\ \frac{5}{2} & \frac{- 3}{2} & \frac{1}{2} \end{array}] A

So,

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

Adjoint and Inverse of Matrix excercise 6.2 question 7

Answer: $[\begin{array}{ccc} 3 & - 1 & 1 \\ - 15 & 6 & - 5 \\ 5 & - 2 & 2 \end{array}]$

Hint: Here, we use the concept of matrix inverse using elementary row operation

Given: $[\begin{array}{ccc} 2 & 0 & - 1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{array}]$

Solution: Let $A = [\begin{array}{ccc} 2 & 0 & - 1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{array}]$

$A = I A$
$\begin{aligned} A = [\begin{array}{ccc} 2 & 0 & - 1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{array}], I = [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] \\ \Rightarrow [\begin{array}{lll} 2 & 0 & - 1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{array}] = [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] A \end{aligned}$

Applying $R_{1} \to \frac{1}{2} R_{1}$

$\Rightarrow [\begin{array}{lll} 1 & 0 & - \frac{1}{2} \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{array}] = [\begin{array}{lll} \frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] A$

Applying $R_{2} \to R_{2} - 5 R_{1}$

$\Rightarrow [\begin{array}{ccc} 1 & 0 & - \frac{1}{2} \\ 0 & 1 & \frac{5}{2} \\ 0 & 1 & 3 \end{array}] = [\begin{array}{lll} \frac{1}{2} & 0 & 0 \\ \frac{- 5}{2} & 1 & 0 \\ 0 & 0 & 1 \end{array}] A$

Applying $R_{3} \to R_{3} - R_{2}$

$\Rightarrow [\begin{array}{ccc} 1 & 0 & - \frac{1}{2} \\ 0 & 1 & \frac{5}{2} \\ 0 & 0 & \frac{1}{2} \end{array}] = [\begin{array}{ccc} \frac{1}{2} & 0 & 0 \\ \frac{- 5}{2} & 1 & 0 \\ \frac{5}{2} & - 1 & 1 \end{array}] A$

Applying $R_{3} \to 2 R_{3}$

$\Rightarrow [\begin{array}{ccc} 1 & 0 & - \frac{1}{2} \\ 0 & 1 & \frac{5}{2} \\ 0 & 0 & 1 \end{array}] = [\begin{array}{ccc} \frac{1}{2} & 0 & 0 \\ \frac{- 5}{2} & 1 & 0 \\ 5 & - 2 & 2 \end{array}] A$

Applying

$\begin{aligned} R_{1} \to R_{1} + \frac{1}{2} R_{3} \\ R_{2} \to R_{2} - \frac{5}{2} R_{3} \\ \Rightarrow [\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] = [\begin{array}{ccc} 3 & - 1 & 1 \\ - 15 & 6 & - 5 \\ 5 & - 2 & 2 \end{array}] A \end{aligned}$

So, $A^{- 1} = [\begin{array}{ccc} 3 & - 1 & 1 \\ - 15 & 6 & - 5 \\ 5 & - 2 & 2 \end{array}]$

Adjoint and inverse of matrix exercise 6 point 2 question 8

Answer: $[\begin{array}{ccc} 1 & 1 & - 1 \\ - 1 & 1 & 0 \\ 2 & - 5 & 2 \end{array}]$

Hint: Here, we use the concept of matrix inverse using elementary row operation

Given: $[\begin{array}{lll} 2 & 3 & 1 \\ 2 & 4 & 1 \\ 3 & 7 & 2 \end{array}]$

Solution: Let $A = [\begin{array}{lll} 2 & 3 & 1 \\ 2 & 4 & 1 \\ 3 & 7 & 2 \end{array}]$

$A = I A$

$\begin{aligned} A = [\begin{array}{lll} 2 & 3 & 1 \\ 2 & 4 & 1 \\ 3 & 7 & 2 \end{array}], I = [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] \\ \Rightarrow [\begin{array}{lll} 2 & 3 & 1 \\ 2 & 4 & 1 \\ 3 & 7 & 2 \end{array}] = [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] A \end{aligned}$

Applying $R_{1} \to \frac{1}{2} R_{1}$

$\Rightarrow [\begin{array}{lll} 1 & \frac{3}{2} & \frac{1}{2} \\ 2 & 4 & 1 \\ 3 & 7 & 2 \end{array}] = [\begin{array}{lll} \frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] A$

Applying

$\begin{aligned} R_{2} \to R_{2} - 2 R_{1} \\ R_{3} & \to R_{3} - 3 R_{1} \\ \Rightarrow [\begin{array}{lll} 1 & \frac{3}{2} & \frac{1}{2} \\ 0 & 1 & 0 \\ 0 & \frac{5}{2} & \frac{1}{2} \end{array}] & = [\begin{array}{ccc} \frac{1}{2} & 0 & 0 \\ - 1 & 1 & 0 \\ \frac{- 3}{2} & 0 & 1 \end{array}] A \end{aligned}$

Applying

$\begin{aligned} R_{1} \to R_{1} - \frac{3}{2} R_{2} \\ R_{3} \to R_{3} - \frac{5}{2} R_{2} \\ \Rightarrow [\begin{array}{lll} 1 & 0 & \frac{1}{2} \\ 0 & 1 & 0 \\ 0 & 0 & \frac{1}{2} \end{array}] = [\begin{array}{ccc} 2 & \frac{- 3}{2} & 0 \\ - 1 & 1 & 0 \\ 1 & \frac{- 5}{2} & 1 \end{array}] A \end{aligned}$

Applying $R_{3} \to 2 R_{3}$

$\Rightarrow [\begin{array}{lll} 1 & 0 & \frac{1}{2} \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] = [\begin{array}{ccc} 2 & \frac{- 3}{2} & 0 \\ - 1 & 1 & 0 \\ 2 & - 5 & 2 \end{array}] A$

Applying $R_{1} \to R_{1} - \frac{1}{2} R_{3}$

$\Rightarrow [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] = [\begin{array}{ccc} 1 & 1 & - 1 \\ - 1 & 1 & 0 \\ 2 & - 5 & 2 \end{array}] A$

So, $A^{- 1} =$ $[\begin{array}{ccc} 1 & 1 & - 1 \\ - 1 & 1 & 0 \\ 2 & - 5 & 2 \end{array}]$

Adjoint and Inverse of Matrix excercise 6.2 question 9

Answer:

[\begin{array}{ccc} 1 & - 1 & 0 \\ - 2 & 3 & - 4 \\ - 2 & 3 & - 3 \end{array}]

Hint: Here, we use the concept of matrix inverse using elementary row operation
Given:

[\begin{array}{rrr} 3 & - 3 & 4 \\ 2 & - 3 & 4 \\ 0 & - 1 & 1 \end{array}]

Solution: Let

A = [\begin{array}{lll} 3 & - 3 & 4 \\ 2 & - 3 & 4 \\ 0 & - 1 & 1 \end{array}]

\begin{aligned} A = I A \\ A = [\begin{array}{lll} 3 & - 3 & 4 \\ 2 & - 3 & 4 \\ 0 & - 1 & 1 \end{array}], I = [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] \\ \Rightarrow [\begin{array}{lll} 3 & - 3 & 4 \\ 2 & - 3 & 4 \\ 0 & - 1 & 1 \end{array}] = [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] A \end{aligned}

Applying

R_{1} \to \frac{1}{3} R_{1}

\Rightarrow [\begin{array}{lll} 1 & - 1 & \frac{4}{3} \\ 2 & - 3 & 4 \\ 0 & - 1 & 1 \end{array}] = [\begin{array}{lll} \frac{1}{3} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] A

Applying

R_{2} \to R_{2} - 2 R_{1}

\Rightarrow [\begin{array}{ccc} 1 & - 1 & \frac{4}{3} \\ 0 & - 1 & \frac{4}{3} \\ 0 & - 1 & 1 \end{array}] = [\begin{array}{ccc} \frac{1}{3} & 0 & 0 \\ \frac{- 2}{3} & 1 & 0 \\ 0 & 0 & 1 \end{array}] A

Applying

R_{2} \to - R_{2}

\Rightarrow [\begin{array}{ccc} 1 & - 1 & \frac{4}{3} \\ 0 & 1 & \frac{- 4}{3} \\ 0 & - 1 & 1 \end{array}] = [\begin{array}{ccc} \frac{1}{3} & 0 & 0 \\ \frac{2}{3} & - 1 & 0 \\ 0 & 0 & 1 \end{array}] A

Applying

\begin{aligned} R_{1} \to R_{1} + R_{2} \\ R_{3} \to R_{3} + R_{2} \\ \Rightarrow [\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & \frac{- 4}{3} \\ 0 & 0 & \frac{- 1}{3} \end{array}] = [\begin{array}{ccc} 1 & - 1 & 0 \\ \frac{2}{3} & - 1 & 0 \\ \frac{2}{3} & - 1 & 1 \end{array}] A \end{aligned}

Applying

R_{3} \to - 3 R_{3}

\Rightarrow [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] = [\begin{array}{ccc} 1 & - 1 & 0 \\ - 2 & 3 & - 4 \\ - 2 & 3 & - 3 \end{array}] A

Applying

R_{2} \to R_{2} + \frac{4}{3} R_{3}

\Rightarrow [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] = [\begin{array}{ccc} 1 & - 1 & 0 \\ - 2 & 3 & - 4 \\ - 2 & 3 & - 3 \end{array}] A

So,

A^{- 1} = [\begin{array}{ccc} 1 & - 1 & 0 \\ - 2 & 3 & - 4 \\ - 2 & 3 & - 3 \end{array}]

Adjoint and Inverse of Matrix excercise 6 point 2 question 2

Answer:

[\begin{array}{ccc} \frac{- 4}{3} & 1 & \frac{1}{3} \\ \frac{7}{6} & \frac{- 1}{2} & \frac{- 1}{6} \\ \frac{5}{6} & \frac{- 1}{2} & \frac{1}{6} \end{array}]

Hint: Here, we use the concept of matrix inverse using elementary row operation
Given:

[\begin{array}{ccc} 1 & 2 & 0 \\ 2 & 3 & - 1 \\ 1 & - 1 & 3 \end{array}]

Solution: Let

A = [\begin{array}{ccc} 1 & 2 & 0 \\ 2 & 3 & - 1 \\ 1 & - 1 & 3 \end{array}]

\begin{aligned} A = I A \\ A = [\begin{array}{ccc} 1 & 2 & 0 \\ 2 & 3 & - 1 \\ 1 & - 1 & 3 \end{array}], I = [\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] \\ \Rightarrow [\begin{array}{ccc} 1 & 2 & 0 \\ 2 & 3 & - 1 \\ 1 & - 1 & 3 \end{array}] = [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] A \end{aligned}

Applying

\begin{matrix} R_{2} \to R_{2} - 2 R_{1} \\ R_{3} \to R_{3} - R_{1} \\ \Rightarrow [\begin{array}{ccc} 1 & 2 & 0 \\ 0 & - 1 & - 1 \\ 0 & - 3 & 3 \end{array}] = [\begin{array}{ccc} 1 & 0 & 0 \\ - 2 & 1 & 0 \\ - 1 & 0 & 1 \end{array}] A \end{matrix}

Applying

R_{2} \to - R_{2}

\Rightarrow [\begin{array}{ccc} 1 & 2 & 0 \\ 0 & 1 & 1 \\ 0 & - 3 & 3 \end{array}] = [\begin{array}{ccc} 1 & 0 & 0 \\ 2 & - 1 & 0 \\ - 1 & 0 & 1 \end{array}] A

Applying

\begin{aligned} R_{1} \to R_{1} - 2 R_{2} \\ R_{3} \to R_{3} + 3 R_{2} \\ \Rightarrow [\begin{array}{ccc} 1 & 0 & - 2 \\ 0 & 1 & 1 \\ 0 & 0 & 6 \end{array}] = [\begin{array}{ccc} - 3 & 2 & 0 \\ 2 & - 1 & 0 \\ 5 & - 3 & 1 \end{array}] A \end{aligned}

Applying

R_{3} \to \frac{1}{6} R_{3}

\Rightarrow [\begin{array}{ccc} 1 & 0 & - 2 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array}] = [\begin{array}{ccc} - 3 & 2 & 0 \\ 2 & - 1 & 0 \\ \frac{5}{6} & - \frac{1}{2} & \frac{1}{6} \end{array}] A

Applying

\begin{aligned} R_{1} \to R_{1} + 2 R_{3} \\ R_{2} \to R_{2} - R_{3} \\ \Rightarrow [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] = [\begin{array}{ccc} - \frac{4}{3} & 1 & \frac{1}{3} \\ \frac{7}{6} & \frac{- 1}{2} & \frac{- 1}{6} \\ \frac{5}{6} & - \frac{1}{2} & \frac{1}{6} \end{array}] A \end{aligned}

So,

A^{- 1} = [\begin{array}{ccc} - \frac{4}{3} & 1 & \frac{1}{3} \\ \frac{7}{6} & \frac{- 1}{2} & \frac{- 1}{6} \\ \frac{5}{6} & - \frac{1}{2} & \frac{1}{6} \end{array}]

Adjoint and inverse of matrix exercise 6.2 question 11

Answer:

[\begin{array}{ccc} \frac{1}{15} & \frac{- 2}{15} & \frac{- 1}{3} \\ \frac{- 11}{30} & \frac{7}{30} & \frac{1}{6} \\ \frac{1}{6} & \frac{1}{6} & \frac{- 1}{6} \end{array}]

Hint: Here, we use the concept of matrix inverse using elementary row operation
Given:

[\begin{array}{ccc} 2 & - 1 & 3 \\ 1 & 2 & 4 \\ 3 & 1 & 1 \end{array}]

Solution:
Let,

\begin{aligned} A = [\begin{array}{ccc} 2 & - 1 & 3 \\ 1 & 2 & 4 \\ 3 & 1 & 1 \end{array}] \\ A = I A \\ A = [\begin{array}{ccc} 2 & - 1 & 3 \\ 1 & 2 & 4 \\ 3 & 1 & 1 \end{array}], I = [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] \\ \Rightarrow [\begin{array}{lll} 2 & - 1 & 3 \\ 1 & 2 & 4 \\ 3 & 1 & 1 \end{array}] = [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] A \end{aligned}

Applying

R_{1} \to \frac{1}{2} R_{1}

\Rightarrow [\begin{array}{ccc} 1 & - \frac{1}{2} & \frac{3}{2} \\ 1 & 2 & 4 \\ 3 & 1 & 1 \end{array}] = [\begin{array}{lll} \frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] A

Applying

\begin{aligned} R_{2} \to R_{2} - R_{1} \\ R_{3} \to R_{3} - 3 R_{1} \\ \Rightarrow [\begin{array}{ccc} 1 & - \frac{1}{2} & \frac{3}{2} \\ 0 & \frac{5}{2} & \frac{5}{2} \\ 0 & \frac{5}{2} & \frac{- 7}{2} \end{array}] = [\begin{array}{ccc} \frac{1}{2} & 0 & 0 \\ \frac{- 1}{2} & 1 & 0 \\ \frac{- 3}{2} & 0 & 1 \end{array}] A \end{aligned}

Applying

R_{2} \to \frac{2}{5} R_{2}

\Rightarrow [\begin{array}{ccc} 1 & - \frac{1}{2} & \frac{3}{2} \\ 0 & 1 & 1 \\ 0 & \frac{5}{2} & \frac{- 7}{2} \end{array}] = [\begin{array}{ccc} \frac{1}{2} & 0 & 0 \\ \frac{- 1}{5} & \frac{2}{5} & 0 \\ \frac{- 3}{2} & 0 & 1 \end{array}] A

Applying

\begin{aligned} R_{1} \to R_{1} + \frac{1}{2} R_{2} \\ R_{3} \to R_{3} - \frac{5}{2} R_{2} \\ \Rightarrow [\begin{array}{ccc} 1 & 0 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & - 6 \end{array}] = [\begin{array}{ccc} \frac{2}{5} & \frac{1}{5} & 0 \\ \frac{- 1}{5} & \frac{2}{5} & 0 \\ - 1 & - 1 & 1 \end{array}] A \end{aligned}

Applying

R_{3} \to \frac{- 1}{6} R_{3}

\Rightarrow [\begin{array}{lll} 1 & 0 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array}] = [\begin{array}{ccc} \frac{2}{5} & \frac{1}{5} & 0 \\ \frac{- 1}{5} & \frac{2}{5} & 0 \\ \frac{1}{6} & \frac{1}{6} & \frac{- 1}{6} \end{array}] A

Applying

\begin{aligned} R_{2} \to R_{2} - R_{3} \\ R_{1} \to R_{1} - 2 R_{3} \\ \Rightarrow [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array}] = [\begin{array}{ccc} \frac{1}{15} & \frac{- 2}{15} & \frac{- 1}{3} \\ \frac{- 11}{30} & \frac{7}{30} & \frac{1}{6} \\ \frac{1}{6} & \frac{1}{6} & \frac{- 1}{6} \end{array}] A \end{aligned}

So,

A^{- 1} = [\begin{array}{ccc} \frac{1}{15} & \frac{- 2}{15} & \frac{- 1}{3} \\ \frac{- 11}{30} & \frac{7}{30} & \frac{1}{6} \\ \frac{1}{6} & \frac{1}{6} & \frac{- 1}{6} \end{array}]

Adjoint and Inverse of Matrix excercise 6.2 question 12

Answer:

\frac{1}{11} [\begin{array}{ccc} - 2 & 5 & - 1 \\ - 1 & - 3 & 5 \\ 7 & - 1 & - 2 \end{array}]

Hint: Here, we use the concept of matrix inverse using elementary row operation
Given:

[\begin{array}{lll} 1 & 1 & 2 \\ 3 & 1 & 1 \\ 2 & 3 & 1 \end{array}]

Solution: Let

A = I A

\begin{aligned} A = [\begin{array}{lll} 1 & 1 & 2 \\ 3 & 1 & 1 \\ 2 & 3 & 1 \end{array}], I = [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] \\ \Rightarrow [\begin{array}{lll} 1 & 1 & 2 \\ 3 & 1 & 1 \\ 2 & 3 & 1 \end{array}] = [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] A \end{aligned}

Applying

\begin{aligned} R_{2} \to R_{2} - 3 R_{1} \\ R_{3} \to R_{3} - 2 R_{1} \\ \Rightarrow [\begin{array}{ccc} 1 & 1 & 2 \\ 0 & - 2 & - 5 \\ 0 & 1 & - 3 \end{array}] = [\begin{array}{ccc} 1 & 0 & 0 \\ - 3 & 1 & 0 \\ - 2 & 0 & 1 \end{array}] A \end{aligned}

Applying

\begin{aligned} R_{2} \to R_{2} - 3 R_{1} \\ R_{3} \to R_{3} - 2 R_{1} \\ \Rightarrow [\begin{array}{ccc} 1 & 1 & 2 \\ 0 & - 2 & - 5 \\ 0 & 1 & - 3 \end{array}] = [\begin{array}{ccc} 1 & 0 & 0 \\ - 3 & 1 & 0 \\ - 2 & 0 & 1 \end{array}] A \end{aligned}

Applying

R_{2} \to \frac{- 1}{2} R_{2}

\Rightarrow [\begin{array}{ccc} 1 & 1 & 2 \\ 0 & 1 & \frac{5}{2} \\ 0 & 1 & - 3 \end{array}] = [\begin{array}{ccc} 1 & 0 & 0 \\ \frac{3}{2} & \frac{- 1}{2} & 0 \\ - 2 & 0 & 1 \end{array}] A

Applying

\begin{aligned} R_{1} \to R_{1} - R_{2} \\ R_{3} \to R_{3} - R_{2} \\ \Rightarrow [\begin{array}{ccc} 1 & 0 & \frac{- 1}{2} \\ 0 & 1 & \frac{5}{2} \\ 0 & 0 & - \frac{11}{2} \end{array}] = [\begin{array}{ccc} \frac{- 1}{2} & \frac{1}{2} & 0 \\ \frac{3}{2} & \frac{- 1}{2} & 0 \\ - \frac{7}{2} & \frac{1}{2} & 1 \end{array}] A \end{aligned}

Applying

R_{3} \to \frac{- 2}{11} R_{3}

\Rightarrow [\begin{array}{ccc} 1 & 0 & \frac{- 1}{2} \\ 0 & 1 & \frac{5}{2} \\ 0 & 0 & 1 \end{array}] = [\begin{array}{ccc} - \frac{1}{2} & \frac{1}{2} & 0 \\ \frac{3}{2} & \frac{- 1}{2} & 0 \\ \frac{7}{11} & \frac{- 1}{11} & \frac{- 2}{11} \end{array}] A

Applying ,

R_{1} \to R_{1} + \frac{1}{2} R_{3}

R_{2} \to R_{2} - \frac{5}{2} R_{3}

\Rightarrow [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] = [\begin{array}{ccc} - \frac{2}{11} & \frac{5}{11} & \frac{- 1}{11} \\ \frac{- 1}{11} & \frac{- 3}{11} & \frac{5}{11} \\ \frac{7}{11} & \frac{- 1}{11} & \frac{- 2}{11} \end{array}] A

So,

A^{- 1} = \frac{1}{11} [\begin{array}{ccc} - 2 & 5 & - 1 \\ - 1 & - 3 & 5 \\ 7 & - 1 & - 2 \end{array}]

Adjoint and Inverse of Matrix excercise 6.2 question 13

Answer:

[\begin{array}{ccc} - 2 & \frac{1}{2} & 1 \\ 11 & - 1 & - 6 \\ 4 & \frac{- 1}{2} & - 2 \end{array}]

Hint: Here, we use the concept of matrix inverse using elementary row operation
Given:

[\begin{array}{ccc} 2 & - 1 & 4 \\ 4 & 0 & 2 \\ 3 & - 2 & 7 \end{array}]

Solution: Let

A = [\begin{array}{ccc} 2 & - 1 & 4 \\ 4 & 0 & 2 \\ 3 & - 2 & 7 \end{array}]

\begin{aligned} A = I A \\ A = [\begin{array}{ccc} 2 & - 1 & 4 \\ 4 & 0 & 2 \\ 3 & - 2 & 7 \end{array}], I = [\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] \\ \Rightarrow [\begin{array}{ccc} 2 & - 1 & 4 \\ 4 & 0 & 2 \\ 3 & - 2 & 7 \end{array}] = [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] A \end{aligned}

Applying

R_{1} \to \frac{1}{2} R_{1}

\Rightarrow [\begin{array}{ccc} 1 & - \frac{1}{2} & 2 \\ 4 & 0 & 2 \\ 3 & - 2 & 7 \end{array}] = [\begin{array}{ccc} \frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] A

Applying

\begin{aligned} R_{2} \to R_{2} - 4 R_{1} \\ R_{3} \to R_{3} - 3 R_{1} \\ \Rightarrow [\begin{array}{ccc} 1 & - \frac{1}{2} & 2 \\ 0 & 2 & - 6 \\ 0 & - \frac{1}{2} & 1 \end{array}] = [\begin{array}{ccc} \frac{1}{2} & 0 & 0 \\ - 2 & 1 & 0 \\ - \frac{3}{2} & 0 & 1 \end{array}] A \end{aligned}

Applying

R_{2} \to \frac{1}{2} R_{2}

\Rightarrow [\begin{array}{ccc} 1 & - \frac{1}{2} & 2 \\ 0 & 1 & - 3 \\ 0 & - \frac{1}{2} & 1 \end{array}] = [\begin{array}{ccc} \frac{1}{2} & 0 & 0 \\ - 1 & \frac{1}{2} & 0 \\ - \frac{3}{2} & 0 & 1 \end{array}] A

Applying

\begin{aligned} R_{1} \to R_{1} + \frac{1}{2} R_{2} \\ \begin{matrix} R_{3} \to R_{3} + \frac{1}{2} R_{2} \\ \Rightarrow [\begin{array}{ccc} 1 & 0 & \frac{1}{2} \\ 0 & 1 & - 3 \\ 0 & 0 & \frac{- 1}{2} \end{array}] = [\begin{array}{ccc} 0 & \frac{1}{4} & 0 \\ - 1 & \frac{1}{2} & 0 \\ - 2 & \frac{1}{4} & 1 \end{array}] A \end{matrix} \end{aligned}

Applying

R_{3} \to - 2 R_{3}

\Rightarrow [\begin{array}{ccc} 1 & 0 & \frac{1}{2} \\ 0 & 1 & - 3 \\ 0 & 0 & 1 \end{array}] = [\begin{array}{ccc} 0 & \frac{1}{4} & 0 \\ - 1 & \frac{1}{2} & 0 \\ 4 & \frac{- 1}{2} & - 2 \end{array}] A

Applying

\begin{aligned} R_{1} \to R_{1} - \frac{1}{2} R_{3} \\ R_{2} \to R_{2} + 3 R_{3} \\ \Rightarrow [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] = [\begin{array}{ccc} - 2 & \frac{1}{2} & 1 \\ 11 & - 1 & - 6 \\ 4 & \frac{- 1}{2} & - 2 \end{array}] A \end{aligned}

So,

A^{- 1} = [\begin{array}{ccc} - 2 & \frac{1}{2} & 1 \\ 11 & - 1 & - 6 \\ 4 & \frac{- 1}{2} & - 2 \end{array}]

Adjoint and Inverse of Matrix excercise 6.2 question 14

Answer:

[\begin{array}{ccc} 3 & - 4 & 3 \\ - 2 & 3 & - 2 \\ 8 & - 12 & 9 \end{array}]

Hint: Here, we use the concept of matrix inverse using elementary row operation
Given:

[\begin{array}{ccc} 3 & 0 & - 1 \\ 2 & 3 & 0 \\ 0 & 4 & 1 \end{array}]

Solution: Let

A = [\begin{array}{ccc} 3 & 0 & - 1 \\ 2 & 3 & 0 \\ 0 & 4 & 1 \end{array}]

\begin{aligned} A = I A \\ A = [\begin{array}{ccc} 3 & 0 & - 1 \\ 2 & 3 & 0 \\ 0 & 4 & 1 \end{array}], I = [\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] \\ \Rightarrow [\begin{array}{lll} 3 & 0 & - 1 \\ 2 & 3 & 0 \\ 0 & 4 & 1 \end{array}] = [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] A \end{aligned}

Applying

R_{1} \to \frac{1}{3} R_{1}

\Rightarrow [\begin{array}{lll} 1 & 0 & \frac{- 1}{3} \\ 2 & 3 & 0 \\ 0 & 4 & 1 \end{array}] = [\begin{array}{lll} \frac{1}{3} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] A

Applying

R_{2} \to R_{2} - 2 R_{1}

\Rightarrow [\begin{array}{ccc} 1 & 0 & \frac{- 1}{3} \\ 0 & 3 & \frac{2}{3} \\ 0 & 4 & 1 \end{array}] = [\begin{array}{ccc} \frac{1}{3} & 0 & 0 \\ \frac{- 2}{3} & 1 & 0 \\ 0 & 0 & 1 \end{array}] A

Applying

R_{2} \to \frac{1}{3} R_{2}

\Rightarrow [\begin{array}{ccc} 1 & 0 & \frac{- 1}{3} \\ 0 & 1 & \frac{2}{9} \\ 0 & 4 & 1 \end{array}] = [\begin{array}{ccc} \frac{1}{3} & 0 & 0 \\ \frac{- 2}{9} & \frac{1}{3} & 0 \\ 0 & 0 & 1 \end{array}] A

Applying

R_{3} \to R_{3} - 4 R_{2}

\Rightarrow [\begin{array}{ccc} 1 & 0 & \frac{- 1}{3} \\ 0 & 1 & \frac{2}{9} \\ 0 & 0 & \frac{1}{9} \end{array}] = [\begin{array}{ccc} \frac{1}{3} & 0 & 0 \\ \frac{- 2}{9} & \frac{1}{3} & 0 \\ \frac{8}{9} & \frac{- 4}{3} & 1 \end{array}] A

Applying

R_{3} \to 9 R_{3}

\Rightarrow [\begin{array}{ccc} 1 & 0 & \frac{- 1}{3} \\ 0 & 1 & \frac{2}{9} \\ 0 & 0 & 1 \end{array}] = [\begin{array}{ccc} \frac{1}{3} & 0 & 0 \\ \frac{- 2}{9} & \frac{1}{3} & 0 \\ 8 & - 12 & 9 \end{array}] A

Applying

\begin{aligned} R_{1} \to R_{1} + \frac{1}{3} R_{3} \\ R_{2} \to R_{2} - \frac{2}{9} R_{3} \end{aligned}

\Rightarrow [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] = [\begin{array}{ccc} 3 & - 4 & 3 \\ - 2 & 3 & - 2 \\ 8 & - 12 & 9 \end{array}] A

So,

A^{- 1} = [\begin{array}{ccc} 3 & - 4 & 3 \\ - 2 & 3 & - 2 \\ 8 & - 12 & 9 \end{array}]

Adjoint and inverse of matrix exercise 6.2 question 15

Answer:

A^{- 1} = [\begin{array}{ccc} 1 & - 2 & - 3 \\ - 2 & 4 & 7 \\ 3 & 5 & 9 \end{array}]

Hint :Here we use basic inverse elementary operation
Given :

[\begin{array}{ccc} 1 & 3 & - 2 \\ - 3 & 0 & - 1 \\ 2 & 1 & 0 \end{array}]

Solution:

A = I A

[\begin{array}{ccc} 1 & 3 & - 2 \\ - 3 & 0 & - 1 \\ 2 & 1 & 0 \end{array}] = [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] A

Applying

R_{2} \Rightarrow R_{2} + 3 R_{1}

and

R_{3} \Rightarrow R_{3} - 2 R_{1}

we get

[\begin{array}{ccc} 1 & 3 & - 2 \\ 0 & 9 & - 7 \\ 0 & - 5 & 4 \end{array}] = [\begin{array}{ccc} 1 & 0 & 0 \\ 3 & 1 & 0 \\ - 2 & 0 & 1 \end{array}] A

Applying

R_{2} \Rightarrow 1 / 9 R_{2}

[\begin{array}{ccc} 1 & 3 & - 2 \\ 0 & 1 & - 7 / 9 \\ 0 & - 5 & 4 \end{array}] = [\begin{array}{ccc} 1 & 0 & 0 \\ 1 / 3 & 1 / 9 & 0 \\ - 2 & 0 & 1 \end{array}] A

Applying

R_{3} \Rightarrow 9 R_{3}

[\begin{array}{ccc} 1 & 0 & 1 / 3 \\ 0 & 1 & - 7 / 9 \\ 0 & 0 & 1 \end{array}] = [\begin{array}{ccc} 0 & - 1 / 3 & 0 \\ 1 / 3 & 1 / 9 & 0 \\ - 3 & 5 & 9 \end{array}] A

Applying

R_{1} \Rightarrow R_{1} - 1 / 3 R_{3}

and

R_{2} \Rightarrow R_{2} + 7 / 9 R_{3}

[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] = [\begin{array}{ccc} 1 & - 2 & - 3 \\ - 2 & 4 & 7 \\ - 3 & 5 & 9 \end{array}] A

I = [\begin{array}{ccc} 1 & - 2 & - 3 \\ - 2 & 4 & 7 \\ - 3 & 5 & 9 \end{array}] A

A^{- 1} = [\begin{array}{ccc} 1 & - 2 & - 3 \\ - 2 & 4 & 7 \\ - 3 & 5 & 9 \end{array}]

Adjoint and Inverse of Matrix excercise 6.2 question 16

Answer:

[\begin{array}{ccc} 1 & - 1 & 1 \\ - 8 & 7 & - 5 \\ 5 & - 4 & 3 \end{array}]

Hint :Here we use basic inverse elementary operation
Given :

[\begin{array}{ccc} - 1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{array}]

Solution: Let

A = [\begin{array}{ccc} - 1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{array}]

For applying elementary row operation we write,

\begin{aligned} A = I A \\ : [\begin{array}{ccc} - 1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{array}] = [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] A \end{aligned}

Applying

R_{1} \leftrightarrow R_{2}

we get

[\begin{array}{ccc} 1 & 2 & 3 \\ - 1 & 1 & 2 \\ 3 & 1 & 1 \end{array}] = [\begin{array}{lll} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array}] A

Applying

R_{2} \Rightarrow R_{2} + R_{1}

and

R_{3} \Rightarrow R_{3} - 3 R_{1}

we get

\Rightarrow [\begin{array}{ccc} 1 & 2 & 3 \\ 0 & 3 & 5 \\ 0 & - 5 & - 8 \end{array}] = [\begin{array}{ccc} 0 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & - 3 & 1 \end{array}] A

Applying

R_{1} \Rightarrow R_{1} - 2 / 3 R_{2}

\Rightarrow [\begin{array}{ccc} 1 & 0 & - 1 / 3 \\ 0 & 3 & 5 \\ 0 & - 5 & - 8 \end{array}] = [\begin{array}{ccc} - 2 / 3 & 1 / 3 & 0 \\ 1 & 1 & 0 \\ 0 & - 3 & 1 \end{array}] A

Applying

R_{2} \Rightarrow 1 / 3 R_{2}

[\begin{array}{ccc} 1 & 0 & - 1 / 3 \\ 0 & 1 & 5 / 3 \\ 0 & - 5 & - 8 \end{array}] = [\begin{array}{ccc} - 2 / 3 & 1 / 3 & 0 \\ 1 / 3 & 1 / 3 & 0 \\ 0 & - 3 & 1 \end{array}] A

Applying

R_{3} \Rightarrow R_{3} + 5 R_{2}

[\begin{array}{ccc} 1 & 0 & - 1 / 3 \\ 0 & 1 & 5 / 3 \\ 0 & 0 & 1 / 3 \end{array}] = [\begin{array}{ccc} - 2 / 3 & 1 / 3 & 0 \\ 1 / 3 & 1 / 3 & 0 \\ 5 / 3 & - 4 / 3 & 1 \end{array}] A

Applying

R_{3} \Rightarrow 3 R_{3}

[\begin{array}{ccc} 1 & 0 & - 1 / 3 \\ 0 & 1 & 5 / 3 \\ 0 & 0 & 1 \end{array}] = [\begin{array}{ccc} - 2 / 3 & 1 / 3 & 0 \\ 1 / 3 & 1 / 3 & 0 \\ 5 & - 4 & 3 \end{array}] A

Applying

R_{1} \Rightarrow R_{1} + 1 / 3 R_{3}

and

R_{2} \Rightarrow R_{2} - 5 / 3 R_{3}

[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] = [\begin{array}{ccc} 1 & - 1 & 1 \\ - 8 & 7 & - 5 \\ 5 & - 4 & 3 \end{array}] A

Hence

A^{- 1} = [\begin{array}{ccc} 1 & - 1 & 1 \\ - 8 & 7 & - 5 \\ 5 & - 4 & 3 \end{array}]

Adjoint and Inverse of Matrix excercise 6.2 question 17

Answer:

[\begin{array}{ccc} 3 & - 2 & 1 \\ - 4 & 1 & - 2 \\ 2 & 0 & 1 \end{array}]

Hint: Here, we use the concept of matrix inverse using elementary row operation
Given:

A = [\begin{array}{ccc} 1 & 2 & 3 \\ 2 & 5 & 7 \\ - 2 & - 4 & - 5 \end{array}]

Solution:

A = [\begin{array}{ccc} 1 & 2 & 3 \\ 2 & 5 & 7 \\ - 2 & - 4 & - 5 \end{array}]

\begin{aligned} A = I A \\ A = [\begin{array}{ccc} 1 & 2 & 3 \\ 2 & 5 & 7 \\ - 2 & - 4 & - 5 \end{array}], I = [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] \end{aligned}

\Rightarrow [\begin{array}{ccc} 1 & 2 & 3 \\ 2 & 5 & 7 \\ - 2 & - 4 & - 5 \end{array}] = [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] A

Applying

R_{2} \to R_{2} + R_{3}

\Rightarrow [\begin{array}{ccc} 1 & 2 & 3 \\ 0 & 1 & 2 \\ - 2 & - 4 & - 5 \end{array}] = [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array}] A

Applying

R_{3} \to 2 R_{1} + R_{3}

\Rightarrow [\begin{array}{lll} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{array}] = [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 2 & 0 & 1 \end{array}] A

Applying

R_{1} \to R_{1} - 3 R_{3}

\Rightarrow [\begin{array}{lll} 1 & 2 & 0 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{array}] = [\begin{array}{ccc} - 5 & 0 & - 3 \\ 0 & 1 & 0 \\ 2 & 0 & 1 \end{array}] A

Applyinng

R_{2} \to R_{2} - 2 R_{3}

\Rightarrow [\begin{array}{lll} 1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] = [\begin{array}{ccc} - 5 & 0 & - 3 \\ - 4 & 1 & - 2 \\ 2 & 0 & 1 \end{array}] A

Applying

R_{1} \to R_{1} - 2 R_{2}

\Rightarrow [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] = [\begin{array}{ccc} 3 & - 2 & 1 \\ - 4 & 1 & - 2 \\ 2 & 0 & 1 \end{array}] A

So,

A^{- 1} = [\begin{array}{ccc} 3 & - 2 & 1 \\ - 4 & 1 & - 2 \\ 2 & 0 & 1 \end{array}]

Adjoint and Inverse of Matrix excercise 6.2 question 18

Answer: $[\begin{array}{ccc} 0 & 1 & - 3 \\ - 2 & 9 & - 23 \\ - 1 & 5 & - 13 \end{array}]$
Hint: Here, we use the concept of matrix inverse using elementary row operation
Given: $[\begin{array}{ccc} 2 & - 3 & 5 \\ 3 & 2 & - 4 \\ 1 & 1 & - 2 \end{array}]$
Solution : Let $A = [\begin{array}{ccc} 2 & - 3 & 5 \\ 3 & 2 & - 4 \\ 1 & 1 & - 2 \end{array}]$
$\begin{aligned} A = I A \\ A = [\begin{array}{ccc} 2 & - 3 & 5 \\ 3 & 2 & - 4 \\ 1 & 1 & - 2 \end{array}], I = [\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] \\ \Rightarrow [\begin{array}{ccc} 2 & - 3 & 5 \\ 3 & 2 & - 4 \\ 1 & 1 & - 2 \end{array}] = [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] A \end{aligned}$
Applying $R_{1} \leftrightarrow R_{3}$
$[\begin{array}{ccc} 1 & 1 & - 2 \\ 3 & 2 & - 4 \\ 2 & - 3 & 5 \end{array}] = [\begin{array}{lll} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{array}] A$
Applying $R_{2} \Rightarrow R_{2} - 3 R_{1}$ and $R_{3} \Rightarrow R_{3} - 2 R_{1}$
$[\begin{array}{ccc} 1 & 1 & - 2 \\ 0 & - 1 & 2 \\ 0 & - 5 & 9 \end{array}] = [\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 1 & - 3 \\ 1 & 0 & - 2 \end{array}] A$
Applying $R_{2} \Rightarrow - R_{2}$ and $R_{1} = R_{1 +} R_{2}$
$[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & - 2 \\ 0 & - 5 & 9 \end{array}] = [\begin{array}{ccc} 0 & 1 & - 3 \\ 0 & - 1 & 3 \\ 1 & 0 & - 2 \end{array}] A$
Applying $R_{3} \Rightarrow R_{3} + 5 R_{2}$
$[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & - 2 \\ 0 & 0 & - 1 \end{array}] = [\begin{array}{ccc} 0 & 1 & - 3 \\ 0 & - 1 & 3 \\ 1 & - 5 & 13 \end{array}] A$
Applying $R_{3} \Rightarrow - R_{3}$
$[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & - 2 \\ 0 & 0 & 1 \end{array}] = [\begin{array}{ccc} 0 & 1 & - 3 \\ 0 & - 1 & 3 \\ - 1 & 5 & - 13 \end{array}] A$
Applying $R_{2} \Rightarrow R_{2} + 2 R_{3}$
$[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] = [\begin{array}{ccc} 0 & 1 & - 3 \\ - 2 & 9 & - 23 \\ - 1 & 5 & - 13 \end{array}] A$
So, $A^{- 1} = [\begin{array}{ccc} 0 & 1 & - 3 \\ - 2 & 9 & - 23 \\ - 1 & 5 & - 13 \end{array}]$

RD Sharma Class 12th Exercise 6.2 deals with the chapter Adjoint and Inverse of Matrix. It has 18 Level 1 questions that are relatively easy and can be completed in one go. The questions from this chapter are divided into Level 1 and Level 2, depending on their complexity and weightage.

The level one questions are pretty straightforward and take very little time to complete. Although Level 2 questions for Matrices are also on the more accessible side, they can still be lengthy and time-consuming. However, this particular exercise only has sums.

Chapter-wise RD Sharma Class 12 Solutions

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