The following matrix is obtained from a augmented matrix, system of equations.
1 0 3 20
0 1 2 -2
0 0 1 4 my answer is 20,-2,4 is this right, thanks for checking my answer.
You have:
x_1 + 3x_3 = 20
x_2 + 2x_3 = -2
x_3 = 4;
You can check yourself to see if this works.
20 + 4(3) = 24? No.
Get the matrix in row reduced echelon form. First, -2*Row 3 + Row 2
= Matrix([[1, 0, 3, 20],[0, 1, 0, -10],[0, 0, 1, 4]]
-3*Row 3 + Row 1
= Matrix([[1, 0, 0, 8],[0, 1, 0, -10],[0, 0, 1, 4]]
Now:
x_1 = 8
x_2 = -10
x_3 = 4
8 + 3(4) = 20? Yes.
-10 + 2(4) = -2? Yes.
x_3 = 4? Yes.
Hello, kwtolley!
You're into matrices and you still don't know how to check your answers ??
. . Shame! .Go to your room!
The following matrix is obtained from a augmented matrix, system of equations:
. . $\displaystyle \begin{vmatrix}1 & 0 &3 &|& 20 \\
0 & 1& 2 &|& \text{-}2 \\ 0& 0& 1 &|& 4\end{vmatrix}$
Most of it already reduced; we need to clear the third column only.
$\displaystyle \begin{array}{cccc}R_1-3R_3 \\ R_2-2R_3 \\ \\ \end{array}
\begin{vmatrix}1 & 0 & 0 & | & 8 \\
0 & 1 & 0 & | &\text{-}10 \\
0 & 0 & 1 & | & 4 \end{vmatrix}$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
The original system of equations is ridiculously simple:
. . . .$\displaystyle x \quad + 3z \:=\:20$
. . . . . $\displaystyle y + 2z\:=\:\text{-}2$
. . . . . . . . .$\displaystyle z\:=\:4$
We already have one answer: .$\displaystyle \boxed{z \,= \,4}$
Substitute into the second equation: .$\displaystyle y + 2(4) \:=\:\text{-}2\quad\Rightarrow\quad\boxed{y \,= \,\text{-}10}$
Substitute into the first equation: .$\displaystyle x + 3(4)\:=\:20\quad\Rightarrow\quad\boxed{ x \,= \,8}$