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OpenStudy (anonymous):
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
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OpenStudy (zzr0ck3r):
um you gave a list of numbers that in no way resembles a matrix
OpenStudy (anonymous):
\[\left[\begin{matrix}3 & 0 & k \\ 12 & 8 & 16\\ 4 & 2 & 4\end{matrix}\right]^{-1}\]
OpenStudy (zzr0ck3r):
can you reduce it to I?
OpenStudy (anonymous):
i think so
OpenStudy (anonymous):
i write where i am write now
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OpenStudy (zzr0ck3r):
k brb wife just got home.
OpenStudy (anonymous):
this is just the coefficient matrix\[\left[\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 2\\ 0 & 0 & k\end{matrix}\right]\]
OpenStudy (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..)
OpenStudy (anonymous):
yes i started w/ that
OpenStudy (anonymous):
Hmm, is that 3 in a_11 positive or negative?
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OpenStudy (anonymous):
this is where im at now\[\left[\begin{matrix}1 & 0 & 0 & 0 & \frac{ -1 }{ 4} & 1 \\ 0 & 1 & 2 & 0 & \frac{ 1 }{ 2 } & \frac{ -3 }{ 2 } \\ 0 & 0 & k & 1 & \frac{ -3 }{ 4 } & 3\end{matrix}\right]\]
OpenStudy (anonymous):
postive
OpenStudy (anonymous):
how would i express k on the inverse matrix
OpenStudy (anonymous):
Continue to reduce it using row operations until you have only the identity matrix on the left.
OpenStudy (anonymous):
could i multiplay R3 by 1/k?
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OpenStudy (anonymous):
multiply*
OpenStudy (anonymous):
Looks like you have to.
OpenStudy (anonymous):
thanks
OpenStudy (anonymous):
The last step ought to be subtracting your new R3 from R2.
OpenStudy (anonymous):
*Sorry, twice R3 from R2..
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