1. ## Matrix Arithmetic

Solve each of the following matrix equations for X:
a) AX=C
b) AX+D=C
Where A= ⎡2 1 2⎤, C=⎡2⎤and D= ⎡-1⎤
⎢0 1 6⎥ ⎢0⎥ ⎢0 ⎥
⎣2 1 0⎦ ⎣2⎦ ⎣3 ⎦

2. Originally Posted by mandarep
Solve each of the following matrix equations for X:
a) AX=C
b) AX+D=C
Where A= ⎡2 1 2⎤, C=⎡2⎤and D= ⎡-1⎤
⎢0 1 6⎥ ⎢0⎥ ⎢0 ⎥
⎣2 1 0⎦ ⎣2⎦ ⎣3 ⎦
a) $\displaystyle X = A^{-1} C$

b) $\displaystyle X = A^{-1} (C-D)$

To get the inverse matrix you either have to follow the procedure that will be outlined in your textbook or class notes, or use a CAS. Since A is 3x3 matrix the former approach will require some small amount of time and effort on your part.

3. Originally Posted by mandarep
Solve each of the following matrix equations for X:
a) AX=C
b) AX+D=C
Where A= ⎡2 1 2⎤, C=⎡2⎤and D= ⎡-1⎤
⎢0 1 6⎥ ⎢0⎥ ⎢0 ⎥
⎣2 1 0⎦ ⎣2⎦ ⎣3 ⎦
$\displaystyle A^{-1}AX=A^{-1}C\Rightarrow X=A^{-1}C$

$\displaystyle \displaystyle X=((x_{ij}))=\sum_{k=1}^n (a_{ik}c_{kj})$

$\displaystyle \displaystyle A^{-1}=\frac{1}{\text{det}(A)}A$

$\displaystyle \text{det}(A)=-4$

$\displaystyle \displaystyle -\frac{1}{4}\begin{bmatrix}2&1&2\\0&1&6\\2&1&0\end{ bmatrix}\begin{bmatrix}2\\0\\2\end{bmatrix}=C_{3x1 }$

$\displaystyle \displaystyle X=((x_{11}))=-\frac{1}{4}\sum_{k=1}^n (a_{i1}c_{1j})=4+0+4=-2$

$\displaystyle \displaystyle X=((x_{21}))=-\frac{1}{4}\sum_{k=1}^n (a_{i2}c_{2j})=0+0+12=-3$

$\displaystyle \displaystyle X=((x_{31}))=-\frac{1}{4}\sum_{k=1}^n (a_{i3}c_{13})=\cdots$

4. Thanks so much guys.
I thought that was the method I would have to use.
You're both brilliant.

5. Alternatively, set up your matrix equations

$\displaystyle \displaystyle \left[\begin{matrix}2&2&1\\0&1&6\\2&1&0\end{matrix}\righ t]\left[\begin{matrix}x_1\\x_2\\x_3\end{matrix}\right] = \left[\begin{matrix}2\\0\\2\end{matrix}\right]$

and go through row operations to create an upper triangular matrix, which you will then be able to solve the system with.

$\displaystyle \displaystyle R_3 - R_1 \to R_3$

$\displaystyle \displaystyle \left[\begin{matrix}2&\phantom{-}2&\phantom{-}1\\0&\phantom{-}1&\phantom{-}6\\0&-1&-1\end{matrix}\right]\left[\begin{matrix}x_1\\x_2\\x_3\end{matrix}\right] = \left[\begin{matrix}2\\0\\0\end{matrix}\right]$

$\displaystyle \displaystyle R_3 + R_2 \to R_2$

$\displaystyle \displaystyle \left[\begin{matrix}2&2&1\\0&1&6\\0&0&5\end{matrix}\righ t]\left[\begin{matrix}x_1\\x_2\\x_3\end{matrix}\right] = \left[\begin{matrix}2\\0\\0\end{matrix}\right]$.

So $\displaystyle \displaystyle 5x_3 = 0 \implies x_3 = 0$

$\displaystyle \displaystyle x_2 + 6x_3 = 0 \implies x_2 = 0$

$\displaystyle \displaystyle 2x_1 + 2x_2 + x_3 = 2 \implies x_1 = 1$.

So the solution is $\displaystyle \displaystyle \mathbf{x} = \left[\begin{matrix}1\\0\\0\end{matrix}\right]$

6. Thank you for all your help.