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OpenStudy (across):
So?
OpenStudy (anonymous):
OpenStudy (anonymous):
can you help in solving it further
OpenStudy (across):
You now have three equations and three unknowns, which is solvable.
\[\begin{matrix}
2c_1 & + & 4c_2 & + & 8c_3 & = & 0\\
c_1 & + & 4c_2 & + & 12c_3 & = & 2\\
& & 2c_2 & + & 12c_3 & = & 6
\end{matrix}\]
OpenStudy (across):
This is equivalent to
\[\begin{bmatrix}
2 & 4 & 8\\
1 & 4 & 12\\
0 & 2 & 12
\end{bmatrix}\begin{bmatrix}
c_1\\
c_2\\
c_3
\end{bmatrix}=\begin{bmatrix}
0\\
2\\
6
\end{bmatrix}\]
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OpenStudy (across):
All that you are left to do is find the inverse of the \(3\times3\) matrix and left-multiply it with the RHS.
OpenStudy (anonymous):
thanks now we just need to apply crammers rule
OpenStudy (across):
Indeed. This is what I obtained. I will leave it here as a sanity check.
\[\begin{bmatrix}
c_1\\
c_2\\
c_3
\end{bmatrix}=\begin{bmatrix}
2\\
-3\\
1
\end{bmatrix}\]
OpenStudy (anonymous):
for C1 i am getting 4 why?
OpenStudy (across):
Show me your reasoning.
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OpenStudy (anonymous):
for C determinant im getting 16
and for C1 im getting
OpenStudy (anonymous):
OpenStudy (anonymous):
for C1 iam getting 2
OpenStudy (anonymous):
@across
OpenStudy (across):
\(c_1\) is \(2\) indeed.
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