1.

Find the inverse using elementary row transformation: [1 3 -2] [-3 0 1] [2 1 0]\(\begin{bmatrix}1&3&-2\\-3&0&1\\2&1&0\end{bmatrix}\)

Answer»

Let A = \(\begin{bmatrix}1&3&-2\\-3&0&1\\2&1&0\end{bmatrix}\)

We have A = I A

⇒ \(\begin{bmatrix}1&3&-2\\-3&0&1\\2&1&0\end{bmatrix}\) = \(\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}A\) 

Applying R2 → R2 + 3R1

R3 → R3 - 2R1

\(\begin{bmatrix}1&3&-2\\0&9&-5\\0&-5&4\end{bmatrix}\)= \(\begin{bmatrix}1&0&0\\3&1&0\\-2&0&1\end{bmatrix}A\) 

Applying R3 → 9R3 + 5R2

\(\begin{bmatrix}1&3&-2\\0&9&-5\\0&0&11\end{bmatrix}\) = \(\begin{bmatrix}1&0&0\\3&1&0\\-3&5&9\end{bmatrix}A\)

Applying R3 → R3/11

\(\begin{bmatrix}1&3&-2\\0&9&-5\\0&0&1\end{bmatrix}\) = \(\begin{bmatrix}1&0&0\\3&1&0\\-3/11&5/11&9/11\end{bmatrix}A\) 

Applying R1 → R1 + 2R3

R2  → R2 + 5 R3

\(\begin{bmatrix}1&3&0\\0&9&0\\0&0&1\end{bmatrix}\) = \(\begin{bmatrix}5/11&10/11&18/11\\18/11&36/11&45/11\\-3/11&5/11&9/11\end{bmatrix}A\) 

Applying R2 → R2/9

\(\begin{bmatrix}1&3&0\\0&1&0\\0&0&1\end{bmatrix}\) = \(\begin{bmatrix}5/11&10/11&18/11\\2/11&4/11&5/11\\-3/11&5/11&9/11\end{bmatrix}A\)

Applying R1 → R1 - 3R2

\(\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\) = \(\begin{bmatrix}-1/11&-2/11&3/11\\2/11&4/11&5/11\\-3/11&5/11&9/11\end{bmatrix}A\)

We obtain I = \(\begin{bmatrix}-1/11&-2/11&3/11\\2/11&4/11&5/11\\-3/11&5/11&9/11\end{bmatrix}A\)

\(\therefore\) A-1 = \(\begin{bmatrix}-1/11&-2/11&3/11\\2/11&4/11&5/11\\-3/11&5/11&9/11\end{bmatrix}\)(\(\because\) A-1 A = I)


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