Problem on linear equations

$\newcommand{\bfA}{\mathbf{A}}$ $\newcommand{\bfB}{\mathbf{B}}$ $\newcommand{\bfC}{\mathbf{C}}$ $\newcommand{\bfF}{\mathbf{F}}$ $\newcommand{\bfI}{\mathbf{I}}$ $\newcommand{\bfa}{\mathbf{a}}$ $\newcommand{\bfb}{\mathbf{b}}$ $\newcommand{\bfc}{\mathbf{c}}$ $\newcommand{\bfd}{\mathbf{d}}$ $\newcommand{\bfe}{\mathbf{e}}$ $\newcommand{\bfi}{\mathbf{i}}$ $\newcommand{\bfj}{\mathbf{j}}$ $\newcommand{\bfk}{\mathbf{k}}$ $\newcommand{\bfn}{\mathbf{n}}$ $\newcommand{\bfr}{\mathbf{r}}$ $\newcommand{\bfu}{\mathbf{u}}$ $\newcommand{\bfv}{\mathbf{v}}$ $\newcommand{\bfw}{\mathbf{w}}$ $\newcommand{\bfx}{\mathbf{x}}$ $\newcommand{\bfy}{\mathbf{y}}$ $\newcommand{\bfz}{\mathbf{z}}$
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
    The matrix product of an $n \times m$ matrix with an $m \times \ell$ matrix is an $n \times \ell$ matrix.

    The $(i,j)$ entry of the matrix product $\bfA \mathbf{B}$ is the dot product of the $i$th row of $\bfA$ with the $j$th column of $\mathbf{B}$.
    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 $\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}1\\2\\1 \end{pmatrix}.$

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