Matrix equation: Given matrices A, B, use eqn to solve for X, Y

richiesmasher

Junior Member
Joined
Dec 15, 2017
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111
Hello, I have two matrices here, A
{2 3}
-4 5

AND B,
{-5 -7}
::2 1
I have determined the inverse of B to be
{1/9 7/9}
- 2/9 - 5/9

and the Identity matrix of B to be
{1 0}
0 1

Now the question says: Solve the following equation for the unknown values of X and Y
[(AB) B^-1] {x} = { 20}
:::::::::::::: 8 ::::::6y


After multiplying the entire left side I have obtained: { 2x + 24} = {20}
::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: -3x + 40 ::::::6y

Now I've been reading about matrix equations, and apparently I need the coefficient matrix, the variable matrix, and the constant matrix.
My problem with this is that I don't have a value for Y on the left side, so when I try to get the coefficient matrix, I'm simply getting x, and its all downhill from there I'm not sure what to do at this point, can someone guide me please?

Please ignore the colons, it's just so I could space the matrices
 
Last edited:
Hello, I have two matrices here, A
{2 3}
-4 5

AND B,
{-5 -7}
::2 1
I have determined the inverse of B to be
{1/9 7/9}
- 2/9 - 5/9

and the Identity matrix of B to be
{1 0}
0 1

Now the question says: Solve the following equation for the unknown values of X and Y
[(AB) B^-1] {x} = { 20}
:::::::::::::: 8 ::::::6y


After multiplying the entire left side I have obtained: { 2x + 24} = {20}
::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: -3x 40 ::::::6y

Now I've been reading about matrix equations, and apparently I need the coefficient matrix, the variable matrix, and the constant matrix.
My problem with this is that I don't have a value for Y on the left side, so when I try to get the coefficient matrix, I'm simply getting x, and its all downhill from there I'm not sure what to do at this point, can someone guide me please?

Please ignore the colons, it's just so I could space the matrices

What you have now is the matrix equation

\(\displaystyle \begin{pmatrix}2x+24\\ -3x+40\end{pmatrix} = \begin{pmatrix}20\\ 6y\end{pmatrix}\).

Two matrices are equal when their corresponding elements are equal; so this matrix equation means

\(\displaystyle \left\{\begin{matrix}2x+24=20\\ -3x+40=6y\end{matrix}\right. \).

Solve that system of equations (by any means you want -- you don't have to use matrices).
 
What you have now is the matrix equation

\(\displaystyle \begin{pmatrix}2x+24\\ -3x+40\end{pmatrix} = \begin{pmatrix}20\\ 6y\end{pmatrix}\).

Two matrices are equal when their corresponding elements are equal; so this matrix equation means

\(\displaystyle \left\{\begin{matrix}2x+24=20\\ -3x+40=6y\end{matrix}\right. \).

Solve that system of equations (by any means you want -- you don't have to use matrices).

Ah thank you Dr, I figured it out, and in doing so made an error, it shouldnt be -3x +40, it's actually -4x +40, I made one tiny error and it messed the whole thing up, But i've solved the equation I didn't know that rule for matrices existed.
 
Ah thank you Dr, I figured it out, and in doing so made an error, it shouldnt be -3x +40, it's actually -4x +40, I made one tiny error and it messed the whole thing up, But i've solved the equation I didn't know that rule for matrices existed.

I thought I'd checked that you had it right, but I was most interested in the method, and I guess it's good practice for you to have found the error yourself!

Knowing what it means for two things to be equal is, in principle, one of the first things you learn; but it is easily overlooked in all the complex things you learn just after that!

By the way, I assume you recognized that you don't have to do the multiplication (AB)B^-1, but can just know that it is A.
 
I thought I'd checked that you had it right, but I was most interested in the method, and I guess it's good practice for you to have found the error yourself!

Knowing what it means for two things to be equal is, in principle, one of the first things you learn; but it is easily overlooked in all the complex things you learn just after that!

By the way, I assume you recognized that you don't have to do the multiplication (AB)B^-1, but can just know that it is A.

Only now that I see it I realize it's A, wow I did not see that, is there some sort of rule?
 
Only now that I see it I realize it's A, wow I did not see that, is there some sort of rule?

Have you heard of the associative property?

(AB)B^-1 = A(BB^-1) = AI = A

I imagine someone wanted you to discover this and be impressed ...
 
Hello, I have two matrices here, A
{2 3}
-4 5

AND B,
{-5 -7}
::2 1
I have determined the inverse of B to be
{1/9 7/9}
- 2/9 - 5/9

and the Identity matrix of B to be
{1 0}
0 1

Now the question says: Solve the following equation for the unknown values of X and Y
[(AB) B^-1] {x} = { 20}
:::::::::::::: 8 ::::::6y


After multiplying the entire left side I have obtained: { 2x + 24} = {20}
::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: -3x + 40 ::::::6y <======== should be -4x + 40 = 6y

Now I've been reading about matrix equations, and apparently I need the coefficient matrix, the variable matrix, and the constant matrix.
My problem with this is that I don't have a value for Y on the left side, so when I try to get the coefficient matrix, I'm simply getting x, and its all downhill from there I'm not sure what to do at this point, can someone guide me please?

Please ignore the colons, it's just so I could space the matrices

As I understand it you want to solve the equation
(A B) B-1 \(\displaystyle \left(\begin{matrix}x\\8\end{matrix}\right)\, =\,
A (B B^{-1}) \left(\begin{matrix}x\\8\end{matrix}\right)\, =\,
A \left(\begin{matrix}x\\8\end{matrix}\right)\, =\,
\left(\begin{matrix}20\\6y\end{matrix}\right)\)
Which you have reduced to (after a small correction, see red above)
\(\displaystyle \left(\begin{matrix}2x+24\\-4x+40\end{matrix}\right)\, =\, \left(\begin{matrix}20\\6y\end{matrix}\right)\)

To get to the nomenclature which you have indicated you know, we first note that the 'nominal' coefficient matrix is A or, as initially written, (A B) B-1. Normally the column matrix on the left hand side would be the variable matrix, and the one on the right would be constant matrix. However, In this case, you can see that variable and constant matrices has been mixed which also has the effect of changing that 'nominal' coefficient matrix. To see this write the matrix equation as two equivalent equations:
+2x + 24 = 20
- 4x + 40 = 6y
which we could rewrite as
+2x + 0y = -4
- 4x - 6y = -40
and then back into matrix form
\(\displaystyle \left(\begin{matrix}2 & 0 \\ -4 & -6 \end{matrix}\right)\, \left(\begin{matrix}x\\y\end{matrix}\right)\, =\, C\, \left(\begin{matrix}x\\y\end{matrix}\right)\, =\,\left(\begin{matrix}-4\\-40\end{matrix}\right)\)
It is then much easier to see that the coefficient matrix is C, the variable matrix is \(\displaystyle \left(\begin{matrix}x\\y\end{matrix}\right)\) and the constant matrix is \(\displaystyle \left(\begin{matrix}-4\\-40\end{matrix}\right)\)

Thus we see that the problems are not always presented as the initial (sometimes simplified) nomenclature indicates.
 
As I understand it you want to solve the equation
(A B) B-1 \(\displaystyle \left(\begin{matrix}x\\8\end{matrix}\right)\, =\,
A (B B^{-1}) \left(\begin{matrix}x\\8\end{matrix}\right)\, =\,
A \left(\begin{matrix}x\\8\end{matrix}\right)\, =\,
\left(\begin{matrix}20\\6y\end{matrix}\right)\)
Which you have reduced to (after a small correction, see red above)
\(\displaystyle \left(\begin{matrix}2x+24\\-4x+40\end{matrix}\right)\, =\, \left(\begin{matrix}20\\6y\end{matrix}\right)\)

To get to the nomenclature which you have indicated you know, we first note that the 'nominal' coefficient matrix is A or, as initially written, (A B) B-1. Normally the column matrix on the left hand side would be the variable matrix, and the one on the right would be constant matrix. However, In this case, you can see that variable and constant matrices has been mixed which also has the effect of changing that 'nominal' coefficient matrix. To see this write the matrix equation as two equivalent equations:
+2x + 24 = 20
- 4x + 40 = 6y
which we could rewrite as
+2x + 0y = -4
- 4x - 6y = -40
and then back into matrix form
\(\displaystyle \left(\begin{matrix}2 & 0 \\ -4 & -6 \end{matrix}\right)\, \left(\begin{matrix}x\\y\end{matrix}\right)\, =\, C\, \left(\begin{matrix}x\\y\end{matrix}\right)\, =\,\left(\begin{matrix}-4\\-40\end{matrix}\right)\)
It is then much easier to see that the coefficient matrix is C, the variable matrix is \(\displaystyle \left(\begin{matrix}x\\y\end{matrix}\right)\) and the constant matrix is \(\displaystyle \left(\begin{matrix}-4\\-40\end{matrix}\right)\)

Thus we see that the problems are not always presented as the initial (sometimes simplified) nomenclature indicates.

Wow, thanks alot to both you and Dr Peterson for these great explanations, I really understand that much better now especially the three matrices, I see that the reason why I had trouble was because they were in a different order, and now i know the correspondance rule as well.
 
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