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anonymous
 4 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
anonymous
 4 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

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zzr0ck3r
 4 years ago
Best ResponseYou've already chosen the best response.0um you gave a list of numbers that in no way resembles a matrix

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\left[\begin{matrix}3 & 0 & k \\ 12 & 8 & 16\\ 4 & 2 & 4\end{matrix}\right]^{1}\]

zzr0ck3r
 4 years ago
Best ResponseYou've already chosen the best response.0can you reduce it to I?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i write where i am write now

zzr0ck3r
 4 years ago
Best ResponseYou've already chosen the best response.0k brb wife just got home.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this is just the coefficient matrix\[\left[\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 2\\ 0 & 0 & k\end{matrix}\right]\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Use GaussJordan 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..)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes i started w/ that

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Hmm, is that 3 in a_11 positive or negative?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this is where im at now\[\left[\begin{matrix}1 & 0 & 0 & 0 & \frac{ 1 }{ 4} & 1 \\ 0 & 1 & 2 & 0 & \frac{ 1 }{ 2 } & \frac{ 3 }{ 2 } \\ 0 & 0 & k & 1 & \frac{ 3 }{ 4 } & 3\end{matrix}\right]\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0how would i express k on the inverse matrix

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Continue to reduce it using row operations until you have only the identity matrix on the left.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0could i multiplay R3 by 1/k?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Looks like you have to.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The last step ought to be subtracting your new R3 from R2.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0*Sorry, twice R3 from R2..
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