031 Review Part 2, Problem 8

Let  $A={\begin{bmatrix}1&3&8\\2&4&11\\1&2&5\end{bmatrix}}.$  Find  $A^{-1}$  if possible.

Foundations:
To find the inverse of a matrix  $A,$  you augment the matrix  $A$
with the identity matrix and row reduce  $A$  to the identity matrix.

Solution:

Step 1:
We begin by augmenting the matrix  $A$  with the identity matrix. Hence, we get
$\left[{\begin{array}{ccc|ccc}1&3&8&1&0&0\\2&4&11&0&1&0\\1&2&5&0&0&1\end{array}}\right].$

Step 2:
Now, we row reduce the matrix  $A$  to obtain the identity matrix. Hence, we have
${\begin{array}{rcl}\displaystyle {\left[{\begin{array}{ccc|ccc}1&3&8&1&0&0\\2&4&11&0&1&0\\1&2&5&0&0&1\end{array}}\right]}&\sim &\displaystyle {\left[{\begin{array}{ccc|ccc}1&3&8&1&0&0\\0&-2&-5&-2&1&0\\0&-1&-3&-1&0&1\end{array}}\right]}\\&&\\&\sim &\displaystyle {\left[{\begin{array}{ccc|ccc}1&3&8&1&0&0\\0&1&3&1&0&-1\\0&-2&-5&-2&1&0\end{array}}\right]}\\&&\\&\sim &\displaystyle {\left[{\begin{array}{ccc|ccc}1&3&8&1&0&0\\0&1&3&1&0&-1\\0&0&1&0&1&-1\end{array}}\right]}\\&&\\&\sim &\displaystyle {\left[{\begin{array}{ccc|ccc}1&3&0&1&-8&8\\0&1&0&1&-3&2\\0&0&1&0&1&-1\end{array}}\right]}\\&&\\&\sim &\displaystyle {\left[{\begin{array}{ccc|ccc}1&0&0&-2&1&2\\0&1&0&1&-3&2\\0&0&1&0&1&-1\end{array}}\right].}\end{array}}$
Therefore, the inverse of  $A$  is
$\left[{\begin{array}{ccc}-2&1&2\\1&-3&2\\0&1&-1\end{array}}\right].$

Final Answer:
$A^{-1}=\left[{\begin{array}{ccc}-2&1&2\\1&-3&2\\0&1&-1\end{array}}\right]$

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