Matrix Multiplication and Solving for X?

Find X if

[5 -8]X + [0 9] = [-3 6]

[8 -1].... [0 -8] ....[-1 -9]

X =?

My steps:

[5 -8]X + [0 9] = [-3 6]

[8 -1] ......[0 -8] ..[-1 -9]

Ax + B = C

Subtract matrix B on both sides gives

[5 -8]X = [-3 6] - [0 9]

[8 -1] .......[-1 -9] [0 -8]

[5 -8]X = [-3 -3]

[8 -1] .......[-1 -1]

A^-1 = 1/(ad -bc) [d -b]

[-c a]

[5 -8] = 1/(5*-1 - -8*8) [-1 8]

[8 -1] .............................[-8 5]

=1/59[-1 8]

..........[-8 5]

X = [-3 -3] * 1/59[-1 8]

.....[-1 -1] .............[-8 5]

= 1/59 [(-3*-1 + -3*-8) (-3*8 + -3*5)]

............[(-1*-1 + -1*-8) (-1*8 + -1*5)]

X= 1/59 [27 -39]

...............[9 - 13]

But for some reason that is not the answer for X, can someone help me please? Thank you!

Re: Matrix Multiplication and Solving for X?

here is what you did wrong:

AX + B = C

AX = C - B <---you're fine up to here

A^{-1}(AX) = A^{-1}(C - B) <--you multiplied C - B on the wrong side.

X = A^{-1}(C - B).

if you multiply C - B on the LEFT by A^{-1}, you get:

$\displaystyle X = \frac{1}{59} \begin{bmatrix}-1&8\\-8&5 \end{bmatrix} \begin{bmatrix}-3&-3\\1&-1 \end{bmatrix} = \frac{1}{59}\begin{bmatrix} 11&-5\\29&19 \end{bmatrix}$

with matrices, AB is NOT the same as BA.