Question

SoLvE the matrix equation: 2x-5[2 -8
-4 2] = [4 -6
2 -8]

Answer 1

so the 2x-5 is getting multiplied into the first matrix? and then we solve for x?

Answer 2

Is it this

\[\Large 2X + 5\begin{bmatrix}2 & -8\\-4 & 2\end{bmatrix} = \begin{bmatrix}4 & -6\\2 & -8\end{bmatrix}\]

???

Answer 3

yeah im trying to see if thats it..

Answer 4

yes it is. exactly like tht lol

Answer 5

or if it's (2x+5) time each of the elements as a scalar

Answer 6

which would be painful..

Answer 7

Start by multiplying the 5 through to each element in the matrix


\[\Large 2X - 5\begin{bmatrix}2 & -8\\-4 & 2\end{bmatrix} = \begin{bmatrix}4 & -6\\2 & -8\end{bmatrix}\]


\[\Large 2X - \begin{bmatrix}2*5 & -8*5\\-4*5 & 2*5\end{bmatrix} = \begin{bmatrix}4 & -6\\2 & -8\end{bmatrix}\]


\[\Large 2X - \begin{bmatrix}10 & -40\\-20 & 10\end{bmatrix} = \begin{bmatrix}4 & -6\\2 & -8\end{bmatrix}\]


What's next?

Answer 8

do i multiply everything by 2 next...

Answer 9

sorry idk

Answer 10

that will be one of your last steps, the next step is to add that matrix you see on the left side to both sides like this

\[\Large 2X - \begin{bmatrix}10 & -40\\-20 & 10\end{bmatrix} = \begin{bmatrix}4 & -6\\2 & -8\end{bmatrix}\]


\[\Large 2X - \begin{bmatrix}10 & -40\\-20 & 10\end{bmatrix} + \begin{bmatrix}10 & -40\\-20 & 10\end{bmatrix} = \begin{bmatrix}4 & -6\\2 & -8\end{bmatrix}+ \begin{bmatrix}10 & -40\\-20 & 10\end{bmatrix} \]


\[\Large 2X + 0 = \begin{bmatrix}4 & -6\\2 & -8\end{bmatrix}+ \begin{bmatrix}10 & -40\\-20 & 10\end{bmatrix} \]


\[\Large 2X = \begin{bmatrix}4 & -6\\2 & -8\end{bmatrix}+ \begin{bmatrix}10 & -40\\-20 & 10\end{bmatrix} \]

Answer 11

Now add the two matrices on the right side

Answer 12

what do you get when you do this?

Answer 13

[14 -46
-18 2]

Answer 14

good, so

\[\Large 2X = \begin{bmatrix}14& -46\\-18 & 2\end{bmatrix}\]

the last step is to multiply both sides by 1/2 to completely isolate X

\[\Large X = \frac{1}{2}\begin{bmatrix}14& -46\\-18 & 2\end{bmatrix}\]

\[\Large X = \begin{bmatrix}\frac{1}{2}*14& \frac{1}{2}(-46)\\\frac{1}{2}(-18) & \frac{1}{2}*2\end{bmatrix}\]

\[\Large X = \begin{bmatrix}7& -23\\-9 & 1\end{bmatrix}\]