# Linear Algebra - finding matrix A

1. Sep 17, 2012

### SmellyGoomba

1. The problem statement, all variables and given/known data
Find A if (2A-1 - 3I)T =

2*$$\begin{pmatrix} -1 & 2\\ 5 & 4 \end{pmatrix}$$

2. Relevant equations

3. The attempt at a solution

I have no idea if I'm even on the right track of solving this question...

I simplified the right hand side down to
$$\begin{pmatrix} -2 & 4\\ 10 & 8 \end{pmatrix}$$

I took the transpose of both sides so the left hand side is just (2A-1 - 3I)
The right hand side is now
$$\begin{pmatrix} -2& 10\\ 4 & 8 \end{pmatrix}$$

I multiplied both sides by A so
LH: A(2A-1 - 3I)
RH: A*$$\begin{pmatrix} -2& 10\\ 4 & 8 \end{pmatrix}$$

Distribute the A so
LH: (A2A-1 - A3I) => (2I - A3I)

Bring the A3I part to the right side
LH = 2I
RH = A3I + A*$$\begin{pmatrix} -2& 10\\ 4 & 8 \end{pmatrix}$$

Simplify the A3I part on the right hand side to get
A * $$\begin{pmatrix} 3 & 0\\ 0 & 3 \end{pmatrix}$$

Factor out the A on the right hand side to get
A * ($$\begin{pmatrix} -2 & 10\\ 4 & 8 \end{pmatrix}$$

+
$$\begin{pmatrix} 3 & 0\\ 0 & 3 \end{pmatrix}$$
)

Add the two matrices together in the brackets to get
A * $$\begin{pmatrix} 1 & 10\\ 4 & 11 \end{pmatrix}$$

So I'm left with
$$\begin{pmatrix} 2 & 0\\ 0 & 2 \end{pmatrix}$$

=

A *
$$\begin{pmatrix} 1 & 10\\ 4 & 11 \end{pmatrix}$$

Now there's no way I can somehow get A like that so I screwed up somewhere in there.. probably from the start lol

This looks so messy... am I allowed to upload my handwritten work on a site then post it here? Seems like a better alternative to the mess I have up there >_> Anyways I'm not really looking for a step by step on how to do this. Just a push in the right direction is all...

2. Sep 17, 2012

### Staff: Mentor

It looks like you did a lot more work than you needed to.

You have
$$2A^{-1} - 3I = \begin{pmatrix} -2& 10\\ 4 & 8 \end{pmatrix}$$

Just add 3I to both sides, and then you'll have an expression for 2A-1, from which you can fairly easily find A.