Matrix problems

Printable View

July 18th 2013, 10:19 AM
Paze

1 Attachment(s)

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.
July 18th 2013, 10:42 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 ${\left( {\begin{array}{rrr} 1&{ - 3}&0 \\ 0&1&1 \\ 2&{ - 1}& 4 \end{array}} \right)^{ - 1}}\left( {\begin{array}{r} 2 \\ { - 2} \\ { 1} \end{array}} \right)$
July 18th 2013, 10:52 AM
Paze

Re: Matrix problems

Quote:

Originally Posted by Plato

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

Oh, I see. Thanks!
July 18th 2013, 11:26 AM
HallsofIvy

Re: Matrix problems

For anyone who was wondering, the original problem, $x_1- 3x_2= 2$ , $x_2+ x_3= -2$ , $2x_1- x_2+ 4x_3= 1$ is equivalent to the matrix equation
$\begin{pmatrix}1 & -3 & 0 \\ 0 & 2 & 1 \\ 2 & -1 & 4 \end{pmatrix}\begin{pmatrix}x_1\\ x_2\\ x_3\end{pmatrix}= \begin{pmatrix}2 \\ -2 \\ 1\end{pmatrix}$

Thinking of that as "Ax= b", multiply both sides by $A^{-1}$ to get $x= A^{-1}b$ .
July 18th 2013, 06:51 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.

Set up your augmented matrix

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

R3 - 2R1 -> R3

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

R3 - 5R2 -> R3

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

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

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