Question

Solve AX=B. Find X

Answer 1

x=b/a

Answer 2

Equation is:
\[AX=B\]Where:
\[A=\begin{bmatrix}
2& 1\\
6& 4\\
\end{bmatrix}\]
\[A=\begin{bmatrix}
6& 4\\
2& 9\\
\end{bmatrix}\]

Answer 3

Sorry, second one is \(B\)

Answer 4

ok then i'll help

Answer 5

Thank you @ttb123456789 :). Please help @iambatman

Answer 6

@ttb123456789?

Answer 7

yeah

Answer 8

@iambatman, I can give a hint if that helps?

Answer 9

\[AX=B\] is the same as \[X=A^{-1}B\]

Answer 10

According to my teacher

Answer 11

ok i'll help just a minue

Answer 12

k

Answer 13

x=0 1
-23 8

Answer 14

How is
\[X=\begin{bmatrix}
0& 1 \\
-23& 8\\
\end{bmatrix}\]
@ttb123456789?

Answer 15

@Abhisar, please help

Answer 16

Yes, because it's a matrix inverse, it's like saying 2*2^(-1) = 1 which is right because 2^(-1) = 1/2 and 1/2*2 = 1 right?

But now since we're dealing with matrix's so for example lets say we have:

\[A=\begin{bmatrix} 2\\ 1\\ \end{bmatrix}, ~~~ B=\begin{bmatrix} 4\\ 2\\ \end{bmatrix}\]
Then you can solve for AB

\[AB=\begin{bmatrix} 2\\ 1\\ \end{bmatrix}\begin{bmatrix} 4\\ 2\\ \end{bmatrix} = A=\begin{bmatrix} 12\\ 6\\ \end{bmatrix}\]

and now using the same method you solve for BA, which I'm sure won't = AB.

To do the matrix A

\[A=\begin{bmatrix} 2& 1\\ 6& 4\\ \end{bmatrix} = 2 \times 4 - 1 \times 6 \neq 0 \]

but since you want the inverse of the matrix, \[A^{-1} = \frac{ 1 }{ 2 \times 4 - 1 \times 6 } \begin{bmatrix} 4& -(-1)\\ -6& 2\\ \end{bmatrix}\] or really it's easier if I put it as:

\[A^{-1} \frac{ 1 }{ ad-bc }=\begin{bmatrix} d& -b\\ -c& a\\ \end{bmatrix}\] so you can see for yourself what its going.
If you've noticed something, the ad - bc = determinant.

So now you should know how to find the inverse of A and be able to compute...
\[X = A^{-1}B\]

Answer 17

Uh there's a mistake with equal signs, let me fix that quickly.

\[A^{-1}= \frac{ 1 }{ ad-bc }\begin{bmatrix} d& -b\\ -c& a\\ \end{bmatrix}\]

Answer 18

Oh, that makes sense @iambatman. Whats confusing is whether or not its \(A^{-1}B\) or \(BA^{-1}\) since in most algebraic equationa \(ax=b\) is the same as \(x=b\times a^{-1}\)

Answer 19

\(\Large A^{-1}B \neq BA^{-1}\)

matrix is commutative only in some cases

if

\(\Large AX=B\)
then
\(\Large X=BA^{-1}\neq A^{-1}B\)

Answer 20

So how do you know if to put the A in the front, or the back @mathmath333

Answer 21

Was i right?

Answer 22

oh wait here i did that wrong

\(\Large X=A^{-1}B\)
\(\Large X\neq BA^{-1}\)

Answer 23

if its inverse we keep it in front while
tranferring to RHS

Answer 24

Nah, sorry @ttb123456789

Answer 25

Its:
\[\begin{bmatrix}
11& 3.5 \\
-16& -3
\end{bmatrix}\]

Answer 26

@mathmath333, what happens its like
\[ABX=C\]
And we wan't to move the \[A\] and \[B\]?

Answer 27

it should be

\(\large ABX=C\\
\large BX=A^{-1}C\\
\large X=B^{-1}[A^{-1}C]\)

Answer 28

Oh, so you would one at a time, right @mathmath333?

Answer 29

yes its not like numbers

Answer 30

btw the way u asked rarely seen in questions

Answer 31

The question I asked at the begining came in a test, @mathmath333, thats why