# Matrix problems

• Jul 18th 2013, 09:19 AM
Paze
Matrix problems
Hi, I have a few matrix problems I am to solve.

It says: Solve the following equations using matrices:

Attachment 28857

I'm unfamiliar with this so I would like a hint on where to learn this. I understand what a matrix is but this seems different than normal matrices...E.g. boxes with numbers in them, lol.
• Jul 18th 2013, 09:42 AM
Plato
Re: Matrix problems
Quote:

Originally Posted by Paze
Hi, I have a few matrix problems I am to solve.
It says: Solve the following equations using matrices:
Attachment 28857

Solve $\displaystyle {\left( {\begin{array}{rrr} 1&{ - 3}&0 \\ 0&1&1 \\ 2&{ - 1}& 4 \end{array}} \right)^{ - 1}}\left( {\begin{array}{r} 2 \\ { - 2} \\ { 1} \end{array}} \right)$
• Jul 18th 2013, 09:52 AM
Paze
Re: Matrix problems
Quote:

Originally Posted by Plato
Solve $\displaystyle {\left( {\begin{array}{rrr} 1&{ - 3}&0 \\ 0&1&1 \\ 2&{ - 1}& 4 \end{array}} \right)^{ - 1}}\left( {\begin{array}{r} 2 \\ { - 2} \\ { 1} \end{array}} \right)$

Oh, I see. Thanks!
• Jul 18th 2013, 10:26 AM
HallsofIvy
Re: Matrix problems
For anyone who was wondering, the original problem, $\displaystyle x_1- 3x_2= 2$, $\displaystyle x_2+ x_3= -2$, $\displaystyle 2x_1- x_2+ 4x_3= 1$ is equivalent to the matrix equation
$\displaystyle \begin{pmatrix}1 & -3 & 0 \\ 0 & 2 & 1 \\ 2 & -1 & 4 \end{pmatrix}\begin{pmatrix}x_1\\ x_2\\ x_3\end{pmatrix}= \begin{pmatrix}2 \\ -2 \\ 1\end{pmatrix}$

Thinking of that as "Ax= b", multiply both sides by $\displaystyle A^{-1}$ to get $\displaystyle x= A^{-1}b$.
• Jul 18th 2013, 05:51 PM
Prove It
Re: Matrix problems
Quote:

Originally Posted by Paze
Hi, I have a few matrix problems I am to solve.

It says: Solve the following equations using matrices:

Attachment 28857

I'm unfamiliar with this so I would like a hint on where to learn this. I understand what a matrix is but this seems different than normal matrices...E.g. boxes with numbers in them, lol.

\displaystyle \displaystyle \begin{align*} \left[ \begin{matrix} 1 & -3 & 0 & \phantom{-}2 \\ 0 & \phantom{-}1 & 1 & -2 \\ 2 & -1 & 4 & \phantom{-}1 \end{matrix} \right] \end{align*}

R3 - 2R1 -> R3

\displaystyle \displaystyle \begin{align*} \left[ \begin{matrix} 1 & -3 & 0 & \phantom{-}2 \\ 0 & \phantom{-}1 & 1 & -2 \\ 0 & \phantom{-}5 & 4 & -3 \end{matrix} \right] \end{align*}

R3 - 5R2 -> R3

\displaystyle \displaystyle \begin{align*} \left[ \begin{matrix} 1 & -3 & \phantom{-}0 & \phantom{-}2 \\ 0 & \phantom{-}1 & \phantom{-}1 & -2 \\ 0 & \phantom{-}0 & -1 & \phantom{-}7 \end{matrix} \right] \end{align*}

So it can be seen that \displaystyle \displaystyle \begin{align*} -x_3 = 7 \implies x_3 = -7 \end{align*}. Then

\displaystyle \displaystyle \begin{align*} x_2 + x_3 &= -2 \\ x_2 - 7 &= -2 \\ x_2 &= 5 \\ \\ x_1 - 3x_2 &= 2 \\ x_1 - 3(5) &= 2 \\ x_1 - 15 &= 2 \\ x_1 &= 17 \end{align*}