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Answer: $\frac{1}{11}\left[\begin{array}{ccc} -2 & 5 & -1 \\ -1 & -3 & 5 \\ 7 & -1 & -2 \end{array}\right]$

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

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

Solution: Let $A = IA$

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

Applying

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

Applying

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

Applying $R_{2} \rightarrow \frac{-1}{2} R_{2}$

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

Applying

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

Applying $R_{3} \rightarrow \frac{-2}{11} R_{3}$

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

Applying , $R_{1} \rightarrow R_{1}+\frac{1}{2} R_{3}$,$R_{2} \rightarrow R_{2}-\frac{5}{2} R_{3}$

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

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

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Answer:

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

Given:

Solution: Let

Applying

Applying

Applying

⇒10-12015200-112=-1212032-120-72121A

Applying

Applying R1→R1+12R3,
R2→R2-52R3

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