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If possible, using elementary row transformations, find the inverse of the following matrices
\begin{bmatrix} 2 &3 &-3 \\-1 &-2 &2 \\1 &1 &-1 \end{bmatrix}

Answers (1)

Let A =  \begin{bmatrix} 2 &3 &-3 \\-1 &-2 &2 \\1 &1 &-1 \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{array}{l} {\left[\begin{array}{ccc} 2 & 3 & -3 \\ -1 & -2 & 2 \\ 1 & 1 & -1 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \mathrm{A}} \\\\ \text { Applying } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}+\mathrm{R}_{3} \\ {\left[\begin{array}{ccc} 2 & 3 & -3 \\ 0 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array}\right] \mathrm{A}}\\ \\ \text { Applying } \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-2 \mathrm{R}_{3} \\ {\left[\begin{array}{ccc} 0 & 1 & -1 \\ 0 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & -2 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array}\right] \mathrm{A}} \\\\ \text { Applying } \mathrm{R}_{2} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{2} \\ {\left[\begin{array}{ccc} 0 & 1 & -1 \\ 0 & 0 & 0 \\ 1 & 1 & -1 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & -2 \\ -1 & 1 & -1 \\ 0 & 0 & 1 \end{array}\right] \mathrm{A}} \end{array}

As second row of LHS contains all zeros, so we aren’t going to get any matrix in LHS.

∴ Inverse of A does not exist.

Hence, A-1 does not exist.

 

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