Linear Algebra - finding matrix A

[SmellyGoomba]

Homework Statement

Find A if (2A^-1 - 3I)^T =

2*$$\begin{pmatrix}<br /> -1 & 2\\<br /> 5 & 4<br /> \end{pmatrix}$$

Homework Equations

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
$$\begin{pmatrix}<br /> -2 & 4\\<br /> 10 & 8<br /> \end{pmatrix}$$

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

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

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

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

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

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

+
$$\begin{pmatrix}<br /> 3 & 0\\<br /> 0 & 3<br /> \end{pmatrix}$$
)

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

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

=

A *
$$\begin{pmatrix}<br /> 1 & 10\\<br /> 4 & 11<br /> \end{pmatrix}$$

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...

 

[Mark44]

Homework Statement

Find A if (2A^-1 - 3I)^T =

2*$$\begin{pmatrix}<br /> -1 & 2\\<br /> 5 & 4<br /> \end{pmatrix}$$

Homework Equations

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
$$\begin{pmatrix}<br /> -2 & 4\\<br /> 10 & 8<br /> \end{pmatrix}$$

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

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.

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

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

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

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

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

+
$$\begin{pmatrix}<br /> 3 & 0\\<br /> 0 & 3<br /> \end{pmatrix}$$
)

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

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

=

A *
$$\begin{pmatrix}<br /> 1 & 10\\<br /> 4 & 11<br /> \end{pmatrix}$$

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...