Question

Inverse matrices.

Answer 1

find the values of a and b given that matrix\[A= \left[\begin{matrix}a & -4& -6 \\ -8 & 5 & 7\\ -5& 3& 4\end{matrix}\right]\]
is the inverse of the matrix
\[B= \left[\begin{matrix}1 & 2& -2 \\ 3 & b & 1\\ -1 & 1& -3 \end{matrix}\right]\]

Answer 2

A= 8
B= 10
i did this last week in math

Answer 3

For A to be the inverse matrix of B, it must satisfy the equation:
AB = I (I being the identity matrix)

Try multiplying them together and see if you can find a suitable a and b value that change it into the identity matrix?

Answer 4

lover123 how did you get that?

Answer 5

one sec i will atach a file

Answer 6

Thank you

Answer 7

B | I augmented might work

or simply work out the matrix multiplications and solve for the unknowns

Answer 8

lol, 3x3; i counted wrong :)

Answer 9

\[ \left[\begin{matrix}1 & 2& -2 &|&1&0&0\\ 3 & b & 1&|&0&1&0\\ -1 & 1& -3&|&0&0&1 \end{matrix}\right]\]
and augment to your hearts content maybe

Answer 10

multiply both by the identity matrix?

Answer 11

http://www.wolframalpha.com/input/?i=rref%7B%7B1%2C2%2C-2%2C1%2C0%2C0%7D%2C%7B3%2Cb%2C1%2C0%2C1%2C0%7D%2C%7B-1%2C1%2C-3%2C0%2C0%2C1%7D%7D

ewww, must be a better way :)

Answer 12

I suppose it'd be easier to focus just on the key parts of that multiplication of AB = BA = I
Since the corresponding parts should be equal...
-4 * 3 + 5*b + 3*1 = 1 Solve for b
a * 1 + (-8) * 2 + (-5)*(-2) = 1 Solve for a

Answer 13

definantly

Answer 14

:/ that looks confusing.
I'll try to decipher the process.

Answer 15

[a -4 -6] [1]
[3] = a -12 + 6 = 1 ; a = 7 with any luck
[-1]

Answer 16

r2.c2 would have to equal 1 as well

-8 5 7
2 b 1
----------
-16+5b+7 = 1 ; b = 2

Answer 17

http://www.wolframalpha.com/input/?i=rref%7B%7B1%2C2%2C-2%2C1%2C0%2C0%7D%2C%7B3%2C2%2C1%2C0%2C1%2C0%7D%2C%7B-1%2C1%2C-3%2C0%2C0%2C1%7D%7D

lol, much nicer

Answer 18

Thanks amistre64.

Answer 19

Oh, it looks like he already got to what I was going to say. lol

If you need, I was illustrating the process a bit better on paint and can show you what I was doing. :P

Answer 20

http://i44.tinypic.com/107ozf6.png

Answer 21

thanks