$AX=B$ solve for $X$ ....... in MATRIX

Question

$$ 2x - 3y + 4z = -19\\ 6x + 4y - 2z =8 \\ x + 5y + 4z = 23 $$ what I have done so far is I put the nubmer and $x, y $ and $ z$ in matrix form: $$ \begin{bmatrix} 2 & -3 & 4\\ 6 & 4 &-2\\ 1 & 5 & 4 \end{bmatrix} \begin{bmatrix} x\\ y\\ z \end{bmatrix}= \begin{bmatrix} -19\\ 8\\ 23 \end{bmatrix} $$ step 2: I don't know where to go from here

Answer 1

You are trying to solve a system of equation in the form $$ A\cdot x = B $$ where A is a $3$x$3$ matrix, $x$ is your $3$ elements vector and $B$ is your constant vector. You get your $x$ doing $$ x=A^{-1}\cdot B $$ Since I am lazy I used the computer to solve it. Your result is $$ 20/9, 7/9, 38/9 $$

Answer 2

You can use Gaussian Elimination on the augmented coefficient matrix to solve for $x, y, z$ by expressing the matrix in reduced row echelon form.

$$\begin{bmatrix} 2 & -3 & 4 &\mid&19\\ 6 & 4 &-2&\mid & 8\\ 1 & 5 & 4& \mid &23 \end{bmatrix}$$

If you do this correctly, you should obtain the following: $$\begin{bmatrix} 1 & 0 & 0 &\mid&20/9\\ 0& 1 &0&\mid & 7/9\\ 0 & 0 & 1& \mid &38/9 \end{bmatrix}$$ This means that $$\begin{bmatrix} x\\y\\z\end{bmatrix} = \begin{bmatrix} 20/9 \\ 7/9\\38/9\end{bmatrix} $$

Answer 3

You can also use Cramer's Rule: $$ x=\frac{\left|\begin{array}{r}\color{#C00000}{-19}&-3&4\\\color{#C00000}{8}&4&-2\\\color{#C00000}{23}&5&4\end{array}\right|}{\left|\begin{array}{r}2&-3&4\\6&4&-2\\1&5&4\end{array}\right|}=-2 $$ $$ y=\frac{\left|\begin{array}{r}2&\color{#C00000}{-19}&4\\6&\color{#C00000}{8}&-2\\1&\color{#C00000}{23}&4\end{array}\right|}{\left|\begin{array}{r}2&-3&4\\6&4&-2\\1&5&4\end{array}\right|}=5 $$ $$ z=\frac{\left|\begin{array}{r}2&-3&\color{#C00000}{-19}\\6&4&\color{#C00000}{8}\\1&5&\color{#C00000}{23}\end{array}\right|}{\left|\begin{array}{r}2&-3&4\\6&4&-2\\1&5&4\end{array}\right|}=0 $$ In the numerator, replace the column in the matrix corresponding to the given variable by the column of results. Note that the bars denote the determinant of the matrix.