Question

Please help me!!! ASAP!!
1. what is the solution of the matrix equation ?
[9 4] x= [-9 -6]
[2 1] [-1 -8]

A. [5 -26]
[9 -60]

B. [-5 26]
[-1 -8]

C. [-5 26]
[9 -60]

D. [9 26]
[9 1]

Answer 1

If you left multiply both sides of the equation by the inverse of the left matrix you will have the identity matrix on the left multiplied by X leaving just X behind so that will mean the right side will be your answer!

Answer 2

oh okay so wait what do you multiply by?

Answer 3

\[\large \left[\begin{matrix}9 & 4 \\ 2 & 1\end{matrix}\right]x = \left[\begin{matrix}-9 & -6 \\ -1 & -8\end{matrix}\right]\]

So if you multiply the both sides of this by the inverse of the left hand side's matrix...you will have your answer...why? because when you multiply a matrix by its inverse...you get the identity matrix...or just 1...leaving you with 'x = something'

so what is the inverse of that matrix?

\[\large \left[\begin{matrix}9 & 4 \\ 2 & 1\end{matrix}\right]\]

The first step would be to find the determinant. For this 2x2 matrix it would be \( (9 \times 1) - (4 \times 2) = 9 - 8 = 1\) Great...our determinant is 1

So now that we have that...we can multiply both sides by the inverse of the matrix...

We take

\[\large \left[\begin{matrix}9 & 4 \\ 2 & 1\end{matrix}\right]\]
switch the positions of 9 and 1 so

\[\large \left[\begin{matrix}1 & 4 \\ 2 & 9\end{matrix}\right]\]
and then make 2 and 4 negative
\[\large \left[\begin{matrix}1 & -4 \\ -2 & 9\end{matrix}\right]\]

Now we multiply this by \(\large \frac{1}{\text{determinant}}\) which we found to be 1...so just 1

So now we multiply that matrix we have now...
\[\large \left[\begin{matrix}1 & -4 \\ -2 & 9\end{matrix}\right]\]
to both sides of the equation

\[\large \left[\begin{matrix}1 & -4 \\ -2 & 9\end{matrix}\right] \times \large \left[\begin{matrix}9 & 4 \\ 2 & 1\end{matrix}\right]x = x\]

That's the left side of the equation...and for the right side

\[\large \left[\begin{matrix}1 & -4 \\ -2 & 9\end{matrix}\right] \times \left[\begin{matrix}-9 & -6 \\ -1 & -8\end{matrix}\right] = \left[\begin{matrix}((1 \times -9) + (-4 \times -1)) & ((1 \times -6) + (-4 \times -8)) \\ ((-2 \times -9) + (9 \times -1)) & ((-2 \times -6) + (9 \times -8))\end{matrix}\right]\]

\[\large \left[\begin{matrix}(-9 + 4)) & (-6 + 32) \\ (18 + -9) & (12 + -72)\end{matrix}\right]\]

\[\large \left[\begin{matrix}-5 & 26 \\ 9 & -60\end{matrix}\right]\]

Answer 4

omg thank you so much that helped a lot :)

Answer 5

Lol anytime :)

Answer 6

could you by chance help me with one more cause this one is confusing me :P @johnweldon1993

Answer 7

Lol of course...same process as up there...literally gonna copy and paste this but change the numbers lol

Answer 8

solve the system
{ 2x +y - 4z = -30
{ 3x + 2y +2z =-8
{5x + 5y + z = -26

A. (-30,-8,-26)
B.(4,-2,4)
C. (-4,-2,4)
D.(-4,2,-4)

Answer 9

Ahh fun! lol....

Well...want to go through the whole process? or do you know how to do it just getting thrown off?

Answer 10

Yeah i would like to go through the process again because i completely forgot :P

Answer 11

Lol okay hang on 1 sec okay? :P

Answer 12

Alrighty...ready? lol

Answer 13

"completely forgot"

just completely scroll to the top.