Starting with (A7), and using only elementary row operations, find the inverse of A = 12 1 1) 1 2 1 1 1 2) Again, you'll be grad

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3 years ago

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iragen [17] · Answer 1 · 2022-08-01T13:22:11+03:00

Answer:

The inverse of the matrix A is $A^{-1}=\left[\begin{array}{ccc}3/4&-1/4&-1/4\\-1/4&3/4&-1/4\\-1/4&-1/4&3/4\end{array}\right]$

Step-by-step explanation:

We have the following matrix $A=\left[\begin{array}{ccc}2&1&1\\1&2&1\\1&1&2\end{array}\right]$

Step 1 - Adjoin the identity matrix to the given matrix

$\left[\begin{array}{ccc|ccc}2&1&1&1&0&0\\1&2&1&0&1&0\\1&1&2&0&0&1\end{array}\right]$

Step 2 - Transform the matrix to the reduced row echelon form

$\left[\begin{array}{ccc|ccc}1&1/2&1/2&1/2&0&0\\1&2&1&0&1&0\\1&1&2&0&0&1\end{array}\right]$

$\left[\begin{array}{ccc|ccc}1&1/2&1/2&1/2&0&0\\0&3/2&1/2&-1/2&1&0\\1&1&2&0&0&1\end{array}\right]$

$\left[\begin{array}{ccc|ccc}1&1/2&1/2&1/2&0&0\\0&3/2&1/2&-1/2&1&0\\0&1/2&3/2&-1/2&0&1\end{array}\right]$

$\left[\begin{array}{ccc|ccc}1&1/2&1/2&1/2&0&0\\0&1&1/3&-1/3&2/3&0\\0&1/2&3/2&-1/2&0&1\end{array}\right]$

$\left[\begin{array}{ccc|ccc}1&1/2&1/2&1/2&0&0\\0&1&1/3&-1/3&2/3&0\\0&0&4/3&-1/3&-1/3&1\end{array}\right]$

$\left[\begin{array}{ccc|ccc}1&1/2&1/2&1/2&0&0\\0&1&1/3&-1/3&2/3&0\\0&0&1&-1/4&-1/4&3/4\end{array}\right]$

$\left[\begin{array}{ccc|ccc}1&1/2&1/2&1/2&0&0\\0&1&0&-1/4&3/4&-1/4\\0&0&1&-1/4&-1/4&3/4\end{array}\right]$

$\left[\begin{array}{ccc|ccc}1&1/2&0&5/8&1/8&-3/8\\0&1&0&-1/4&3/4&-1/4\\0&0&1&-1/4&-1/4&3/4\end{array}\right]$

$\left[\begin{array}{ccc|ccc}1&0&0&3/4&-1/4&-1/4\\0&1&0&-1/4&3/4&-1/4\\0&0&1&-1/4&-1/4&3/4\end{array}\right]$

As can be seen, we have obtained the identity matrix to the left. So, we are done.

The inverse of the matrix A is $A^{-1}=\left[\begin{array}{ccc}3/4&-1/4&-1/4\\-1/4&3/4&-1/4\\-1/4&-1/4&3/4\end{array}\right]$