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Find inverse, by elementary row operations (if possible), of the following matrices.

\begin{bmatrix} 1 &-3 \\-2 & 6 \end{bmatrix}

Answers (1)

Let B=\begin{bmatrix} 1 &-3 \\-2 & 6 \end{bmatrix}

To apply elementary row transformations we write:

B = IB 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 = XB

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

So we get,

\begin{array}{l} {\left[\begin{array}{cc} 1 & -3 \\ -2 & 6 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \mathrm{B}} \end{array}

 By Applying R2→ R2 + 2R1

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

We have got all zeroes in one of the row of matrix in LHS.

So by any means we can't make identity matrix in LHS.

∴ inverse of B does not exist.

B^{-1} does not exist.

 

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infoexpert22

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