Find matrices X given an equation.

[candymountain]

Homework Statement

find all matrices x that satisfy the given matrix equation
[ 1 2 3
4 5 6] * X = I_2

I_2 is the identity matrix 2x2

Homework Equations

The Attempt at a Solution

I just inverted the square matrix
[ 1 2
4 5]
so it becomes
[5 -2
-4 1 ]
so X should be
[ 5 -2
-4 1
0 0]

but my book solution introduces 2 variables S and T to capture all the solutions, how do i do this?
*ps, how do i make my matrices neater?

 

[hunt_mat]

The way you went about it is a little odd. I personally would have written:
<br /> \left(<br /> \begin{array}{ccc}<br /> 1 &amp; 2 &amp; 3 \\<br /> 4 &amp; 5 &amp; 6<br /> \end{array}<br /> \right) \left(<br /> \begin{array}{cc}<br /> a_{1} &amp; a_{2} \\<br /> a_{3} &amp; a_{4} \\<br /> a_{5} &amp; a_{6}<br /> \end{array}<br /> \right) =\left(<br /> \begin{array}{cc}<br /> 1 &amp; 0 \\<br /> 0 &amp; 1<br /> \end{array}<br /> \right)<br />
Multiplied them out and solved the linear equations. You will get four equations to solve for 6 unknowns, this is where the variables t & s come into it.

 

[Raskolnikov]

I would just treat it as two simple linear algebra problems of finding the solution space for Ax=b:<br /> <br /> \left(<br /> \begin{array}{ccc}<br /> 1 &amp; 2 &amp; 3 \\<br /> 4 &amp; 5 &amp; 6<br /> \end{array}<br /> \right)

simplifies to the reduced row-echelon form matrix

<br /> \left(<br /> \begin{array}{ccc}<br /> 1 &amp; 0 &amp; -1 \\<br /> 0 &amp; 1 &amp; 2<br /> \end{array}<br /> \right)

So we have
<br /> <br /> \left(<br /> \begin{array}{ccc}<br /> 1 &amp; 0 &amp; -1 \\<br /> 0 &amp; 1 &amp; 2<br /> \end{array}<br /> \right) \left(<br /> \begin{array}{c}<br /> a_{1} \\<br /> a_{3} \\<br /> a_{5}<br /> \end{array}<br /> \right) =\left(<br /> \begin{array}{c}<br /> 1 \\<br /> 0 <br /> \end{array}<br /> \right)<br /> <br />

and
\left(<br /> \begin{array}{ccc}<br /> 1 &amp; 0 &amp; -1 \\<br /> 0 &amp; 1 &amp; 2<br /> \end{array}<br /> \right) \left(<br /> \begin{array}{c}<br /> a_{2} \\<br /> a_{4} \\<br /> a_{6}<br /> \end{array}<br /> \right) =\left(<br /> \begin{array}{c}<br /> 0 \\<br /> 1 <br /> \end{array}<br /> \right)<br /> <br />

 

[candymountain]

thanks for the replies.
I see how it's more intuitive to do it the algebraic way, so I'll just discard the inverse trick.
Using the rref makes it pretty simple, but when i reduce the one on the left side, do I reduce it on the right as well, if say we're given a matrix other than an I_n ?

 

[Raskolnikov]

thanks for the replies.
I see how it's more intuitive to do it the algebraic way, so I'll just discard the inverse trick.
Using the rref makes it pretty simple, but when i reduce the one on the left side, do I reduce it on the right as well, if say we're given a matrix other than an I_n ?

Yea, sorry I forgot to change the right side. But yep, just apply Gauss-Jordan elimination to the augmented matrix (which includes the right-hand side).

 

[candymountain]

can anyone verify this?

my 1st column came out to be
a = t
b = -2 -2t
c = 5/3 +t

and 2nd column
d= s
e= 1-2s
f= -2/3+s