## anonymous 3 years ago Express the inverse of the following matrix (assuming it exists) as a matrix containing expressions in terms of k. -3 0 k 12 8 16 4 2 4

1. zzr0ck3r

um you gave a list of numbers that in no way resembles a matrix

2. anonymous

$\left[\begin{matrix}3 & 0 & k \\ 12 & 8 & 16\\ 4 & 2 & 4\end{matrix}\right]^{-1}$

3. zzr0ck3r

can you reduce it to I?

4. anonymous

i think so

5. anonymous

i write where i am write now

6. zzr0ck3r

k brb wife just got home.

7. anonymous

this is just the coefficient matrix$\left[\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 2\\ 0 & 0 & k\end{matrix}\right]$

8. anonymous

Use Gauss-Jordan elimination on $\left[\begin{matrix}3 & 0 & k & 1 & 0 & 0\\ 12 & 8 & 16 & 0 & 1 & 0\\ 4 & 2 & 4 & 0 & 0 & 1\end{matrix}\right]$ That's how I do it. (There's probably an easier way..)

9. anonymous

yes i started w/ that

10. anonymous

Hmm, is that 3 in a_11 positive or negative?

11. anonymous

this is where im at now$\left[\begin{matrix}1 & 0 & 0 & 0 & \frac{ -1 }{ 4} & 1 \\ 0 & 1 & 2 & 0 & \frac{ 1 }{ 2 } & \frac{ -3 }{ 2 } \\ 0 & 0 & k & 1 & \frac{ -3 }{ 4 } & 3\end{matrix}\right]$

12. anonymous

postive

13. anonymous

how would i express k on the inverse matrix

14. anonymous

Continue to reduce it using row operations until you have only the identity matrix on the left.

15. anonymous

could i multiplay R3 by 1/k?

16. anonymous

multiply*

17. anonymous

Looks like you have to.

18. anonymous

thanks

19. anonymous

The last step ought to be subtracting your new R3 from R2.

20. anonymous

*Sorry, twice R3 from R2..