Find matrices X given an equation.

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Homework Help Overview

The discussion revolves around finding matrices X that satisfy a given matrix equation involving a 2x3 matrix and the 2x2 identity matrix. The problem is situated within the context of linear algebra and matrix operations.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore different methods for solving the matrix equation, including inverting a matrix and using reduced row-echelon form. There are discussions about the implications of introducing variables to capture the solution space and how to handle the right side of the equation when performing row reductions.

Discussion Status

Several participants have provided insights into alternative approaches to the problem, with some expressing a preference for algebraic methods over matrix inversion. Questions remain about the application of row reduction techniques and the handling of augmented matrices.

Contextual Notes

Participants are considering the implications of using variables to express multiple solutions and are discussing the process of applying Gauss-Jordan elimination in various contexts.

candymountain
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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?
 
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The way you went about it is a little odd. I personally would have written:
[tex] \left(<br /> \begin{array}{ccc}<br /> 1 & 2 & 3 \\<br /> 4 & 5 & 6<br /> \end{array}<br /> \right) \left(<br /> \begin{array}{cc}<br /> a_{1} & a_{2} \\<br /> a_{3} & a_{4} \\<br /> a_{5} & a_{6}<br /> \end{array}<br /> \right) =\left(<br /> \begin{array}{cc}<br /> 1 & 0 \\<br /> 0 & 1<br /> \end{array}<br /> \right)[/tex]
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.
 
I would just treat it as two simple linear algebra problems of finding the solution space for Ax=b:[tex] <br /> \left(<br /> \begin{array}{ccc}<br /> 1 & 2 & 3 \\<br /> 4 & 5 & 6<br /> \end{array}<br /> \right)[/tex]

simplifies to the reduced row-echelon form matrix

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

So we have
[tex] <br /> \left(<br /> \begin{array}{ccc}<br /> 1 & 0 & -1 \\<br /> 0 & 1 & 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 /> [/tex]

and
[tex]\left(<br /> \begin{array}{ccc}<br /> 1 & 0 & -1 \\<br /> 0 & 1 & 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 /> [/tex]
 
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 ?
 
candymountain said:
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).
 
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
 

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