Solving $3\times 3$ matrix equations:

Question

I am familiar with finding the inverse of matrices, but struggle to formulate matrix equations.

In this particular question, one is asked to find the elementary matrix E where $E*A = B$. $A$ is given as matrix

$$\begin{bmatrix}2 & 2 & -1\\1 & 1 & 1\\ -1 & -2 & 0\end{bmatrix}$$

and $B$ is given as matrix $$\begin{bmatrix}2 & 2 & -1\\-1 & -2 & 0\\ 1 & 1 & 1\end{bmatrix}$$

I have been able to find $A^{-1}$ as

$$\begin{bmatrix}\frac 23 & \frac 23 & 1\\\frac{-1}{3} & \frac{-1}{3} & -1\\\frac{-1}{3}& \frac 23 &0\end{bmatrix}$$ However I am unsure on how to use this to find matrix $E$.

Thanks in advance.

Answer 1

$I_3=\begin{pmatrix} 1&0&0\\ 0&1&0\\ 0&0&1\end{pmatrix}$

By definition, $I$ is the identity matrix such that $I_nM=MI_n=M$ for all $n\times n$ matrices $M$.

So you know $EA=B$, if we multiply $A^{-1}$ to the right of both sides:

$$EAA^{-1}=BA^{-1}$$

$$EI=E=BA^{-1}$$

Answer 2

You needn't compute $A^{-1}$: Form the $3 \times 6$ "block" matrix $\left[\begin{array}{@{}c|c@{}}A & I_{3}\end{array}\right]$ and perform a single elementary row operation to convert the left-hand block from $A$ to $B$. The "new" right-hand block is the elementary matrix $E$ you seek.