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  • If possible, using elementary row transformations, find the inverse of the following matrices A =[2 -1 3]

If possible, using elementary row transformations, find the inverse of the following matrices

\begin{bmatrix} 2 &-1 & 3\\-5 &3 &1\\-3 &2 &3 \end{bmatrix}

Answers (1)

Let A = \begin{bmatrix} 2 &-1 & 3\\-5 &3 &1\\-3 &2 &3 \end{bmatrix}

To apply elementary row transformations we write:

A = IA where I is the identity matrix

We proceed with solving the problem in such a way that LHS becomes I and the transformations in I give us a new matrix such that

I = XA

And this X is called inverse of A = A^{-1}

Note: Never apply row and column transformations simultaneously over a matrix.

So we get,

\begin{aligned} &\left[\begin{array}{ccc} 2 & -1 & 3 \\ -5 & 3 & 1 \\ -3 & 2 & 3 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \mathrm{A}\\ &\text { Applying } R_{2} \rightarrow R_{2}+R_{1}\\ &\Rightarrow\left[\begin{array}{ccc} 2 & -1 & 3 \\ -3 & 2 & 4 \\ -3 & 2 & 3 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \mathrm{A}\\ &\text { Applying } \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{2}\\ &\Rightarrow\left[\begin{array}{ccc} 2 & -1 & 3 \\ -3 & 2 & 4 \\ 0 & 0 & -1 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 1 & 1 & 0 \\ -1 & -1 & 1 \end{array}\right] \mathrm{A}\\ &\text { Applying } R_{1} \rightarrow R_{1}+R_{2}\\ &\Rightarrow\left[\begin{array}{ccc} -1 & 1 & 7 \\ -3 & 2 & 4 \\ 0 & 0 & -1 \end{array}\right]=\left[\begin{array}{ccc} 2 & 1 & 0 \\ 1 & 1 & 0 \\ -1 & -1 & 1 \end{array}\right] \mathrm{A} \end{aligned}

Applying R2→ R2 - 3R1

\begin{aligned} &\left[\begin{array}{ccc} -1 & 1 & 7 \\ 0 & -1 & -17 \\ 0 & 0 & -1 \end{array}\right]=\left[\begin{array}{ccc} 2 & 1 & 0 \\ -5 & -2 & 0 \\ -1 & -1 & 1 \end{array}\right] \mathrm{A}\\ &\text { Applying } \mathrm{R}_{3} \rightarrow(-1) \mathrm{R}_{3}\\ &\Rightarrow\left[\begin{array}{ccc} -1 & 1 & 7 \\ 0 & -1 & -17 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 2 & 1 & 0 \\ -5 & -2 & 0 \\ 1 & 1 & -1 \end{array}\right] \mathrm{A}\\ \end{aligned}

\text { Applying } R_{1} \rightarrow R_{1}+R_{2}

\left[\begin{array}{ccc} -1 & 0 & -10 \\ 0 & -1 & -17 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} -3 & -1 & 0 \\ -5 & -2 & 0 \\ 1 & 1 & -1 \end{array}\right] A

\\\text { Applying } \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+10 \mathrm{R}_{3} \text { and } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}+17 \mathrm{R}_{3}\\ \Rightarrow\left[\begin{array}{ccc} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 7 & 9 & -10 \\ 12 & 15 & -17 \\ 1 & 1 & -1 \end{array}\right] \mathrm{A}\\ \text { Applying } \mathrm{R}_{1} \rightarrow(-1) \mathrm{R}_{1} \text { and } \mathrm{R}_{2} \rightarrow(-1) \mathrm{R}_{2}\\

\Rightarrow\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} -7 & -9 & 10 \\ -12 & -15 & 17 \\ 1 & 1 & -1 \end{array}\right] A

\text { As we have an Identity Matrix in LHS, }\\ \\\therefore \mathrm{A}^{-1}=\left[\begin{array}{ccc} -7 & -9 & 10 \\ -12 & -15 & 17 \\ 1 & 1 & -1 \end{array}\right]

 

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