Find the value of alpha and beta (matrix equation)

[Richie Smash]

Homework Statement

Hello, I have three matrices
A =
[3 -4
2 1]

B =
[α
β]

C=
[α
5]

The question now states AB = C

Find the values of (alpha) and (beta).

Homework Equations

The Attempt at a Solution

I found the inverse of A, then I multiplied each side of the equation AB=C by the inverse, and I was left with

alpha = alpha
Beta = 5

But apparently this answer is wrong.

 

[Dick]

Homework Statement

Hello, I have three matrices View attachment 218962 , View attachment 218963 and View attachment 218964

The question now states AB = C

Find the values of (alpha) and (beta).

Homework Equations

The Attempt at a Solution

I found the inverse of A, then I multiplied each side of the equation AB=C by the inverse, and I was left with

alpha = alpha
Beta = 5

But apparently this answer is wrong.

Yes, it's wrong. No way to tell why unless you show some of the details of how you got that.

 

[Richie Smash]

You cannot see the matrices pictures I posted?

 

[Dick]

You cannot see the matrices pictures I posted?

No, I can't, just the problem statement. BTW computing the inverse is really unnecessary. Just solve the linear equations you get by multiplying the original matrix equation out.

 

[Dick]

A =
3 -4
2 1

B =
α
β

c=
α
5

It looks like you are trying to do things the hard way to layout the response. I'd encourage you to learn TeX. But in the meantime, just skip to showing the equations you get by multiplying out the matrix problem.

 

[Ray Vickson]

A =
3 -4
2 1

B =
α
β

c=
α
5

Use Latex:
$$A = \pmatrix{3&-4\\2&1}, \; B = \pmatrix{\alpha\\ \beta}, \; C = \pmatrix{\alpha \\ 5}$$
You can type this as quickly as the hard-to-read version you just typed avove.

 

[Dick]

right so I found the inverse of A which is
1/11 , 4 /11
-2/11 , 3/11

Multiplied that on both sides of AB=C

and I got
α = (1/11)α, 20/11
β (-2/11)α , 15/11

You can split that into two equations, right? The first one only has $\alpha$ in it. What's $\alpha$?

 

[Richie Smash]

\begin{pmatrix}α \\ β \end{pmatrix}= \begin{bmatrix}
\frac{1}{11}a &\frac{20}{11} \\
\frac{-2}{11}a& \frac{15}{11}
\end{bmatrix}

here is the proper end result of what I got when I multiplied AB = C by the inverse of A.

Um I do not know how I can split that into two equations.

 

[Dick]

\\binom{\alpha }{\beta }= \begin{bmatrix}
\frac{1}{11}a &\frac{20}{11} \\
\frac{-2}{11}a& \frac{15}{11}
\end{bmatrix}

here is the proper end result of what I got when I multiplied AB = C by the inverse of A.

Um I do not know how I can split that into two equations.

The result of $A^{-1}C$ is not a matrix. It's a column vector. Maybe review how matrix multiplication works. You are missing two '+' signs.

 

[Mark44]

A =
[3 -4
2 1]

The above, using LaTeX:
$\begin{bmatrix}3 & -4 \\ 2 & 1 \end{bmatrix}$

Here's the unrendered script: \begin{bmatrix}3 & -4 \\ 2 & 1 \end{bmatrix}
I prefer to write matrices with brackets (using bmatrix, as above), rather than with parentheses (using pmatrix). I think that most linear algebra textbooks use the brackets.

 

[Richie Smash]

OK finally I've done it, Hi guys, yes I multiplied both sides of the equation after finding the inverse of A, and the result I got was in post number 8.

But now I see that Dick is right it should be a column vector and I got \begin{pmatrix}\frac{21}{11}a \\ \frac{13}{11}a \end{pmatrix}

 

[Dick]

OK finally I've done it, Hi guys, yes I multiplied both sides of the equation after finding the inverse of A, and the result I got was in post number 8.

But now I see that Dick is right it should be a column vector and I got \begin{pmatrix}\frac{21}{11}a \\ \frac{13}{11}a \end{pmatrix}

Shouldn't that be:

\begin{bmatrix}
\frac{1}{11}a + \frac{20}{11} \\
\frac{-2}{11}a + \frac{15}{11}
\end{bmatrix}

Not the same. Now equate that to $\begin{bmatrix} a \\ b \end{bmatrix}$.

 

[Richie Smash]

Sorry I don't know where I went wrong, I was following a website's instructions and reading up on this and they said to solve a matrix equation, first find the inverse of the coefficient matrix, so in this case it is the inverse of A.

Then they said multiply each side of the equation by the inverse, and then you should be left with the variables on the left and the constant on the right.

So that's what i did here and I ended up with this, is there no easier way??

 

[Dick]

Sorry I don't know where I went wrong, I was following a website's instructions and reading up on this and they said to solve a matrix equation, first find the inverse of the coefficient matrix, so in this case it is the inverse of A.

Then they said multiply each side of the equation by the inverse, and then you should be left with the variables on the left and the constant on the right.

So that's what i did here and I ended up with this, is there no easier way??

That advice only applies if you have all of the variables in the column vector on the side of the equation with the matrix. You have variables on both sides. The inverse matrix isn't helping. Just multiply out the original matrix equation getting column vectors on both sides. Then equate the components of the column vectors. Giving you two equations in two unknowns.

 

[Richie Smash]

WOW that really worked!

It was like a simultaneous equation thanks Dick!

 

[Dick]

WOW that really worked!

It was like a simultaneous equation thanks Dick!

You're welcome and, yes, it IS a simultaneous equation problem. You could also have pushed on and solved $\frac{a}{11}+\frac{20}{11}=a$ and $\frac{-2a}{11}+\frac{15}{11}=b$ which you'd get after multiplying by the inverse. Multiplying by the inverse didn't hurt. It just didn't help.