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Mathematics 18 Online
OpenStudy (anonymous):

olve equation with boundary equations

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 \[\begin{bmatrix} 2 & 4 & 8\\ 1 & 4 & 12\\ 0 & 2 & 12 \end{bmatrix}\begin{bmatrix} c_1\\ c_2\\ c_3 \end{bmatrix}=\begin{bmatrix} 0\\ 2\\ 6 \end{bmatrix}\]

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.

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.

OpenStudy (anonymous):

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