## Problem on linear equations

Write this matrix equation as a system of 3 equations. Solve for $x,y,z$: $$\begin{pmatrix}1 & 1& 1\\0 & 1 & 1\\ 0 &0 & 1 \end{pmatrix} \begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}4\\3\\1\end{pmatrix}$$
• ## Solution

Recall that
We identify $\bfA = \begin{pmatrix}1&1&1\\0&1&1\\0&0&1 \end{pmatrix}$ and $\bfB = \begin{pmatrix}x\\y\\z \end{pmatrix}.$
The left hand side of the matrix equaion is the matrix product $\bfA \bfB$, which is a $3 \times 1$ matrix.
The first entry of $\bfA \bfB$ is the product of the first row of $\bfA$ with the first (and only) column of $\bfB$: $$(\bfA \bfB)_{1} = 1\cdot x + 1 \cdot y + 1 \cdot z.$$
Similarly the second entry of $\bfA \bfB$ is the product of the second row of $\bfA$ with the first column of $\bfB$: $$(\bfA \bfB)_{2} = 0\cdot x + 1 \cdot y + 1 \cdot z.$$
Finally, the third entry of $\bfA \bfB$ is the product of the third row of $A$ with the first column of $\bfB$: $$(\bfA \bfB)_{3} = 0\cdot x + 0 \cdot y + 1 \cdot z.$$
Hence, we rewrite the matrix equation as the system \begin{align}
x+y+z &= 4\\y+z &=3 \\ z &=1.
\end{align}

#### Solving for $x$,$y$, $z$

To solve for $x,y,z$ we observe that the third equation gives us $z$ explicitly. From that, we could plug into the second equation to get $y$. Then we could plug into the first equation to get $x$.
Using $z=1$, the second equation gives $y = 2$.
Plugging into the first equation, we get $x+2+1 = 4$. Hence, $x=1$.
The solution to the matrix equation is thus $\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}1\\2\\1 \end{pmatrix}.$