Solve the Augmented Matrix: Where did I go wrong?

[Lime]

1 0 1 = 1
0 1 -1 = -1
2 1 1 = 2

1 0 1 = 1 ---> R1
0 1 -1 = -1 --> R2
0 1 -1 = -2 --> R3*R2

1 0 1 = 1 ---> R1
0 1 -1 = -1 --> R2
0 0 -1 = -2 --> R3*R1

1 0 0 = -1 ---> R1 + R3
0 1 0 = 1 ---> R2 - R3
0 0 1 = 2 ---> -R3

X1 = -1
X2 = 1
X3 = 2

Turns out this matrix does not have a solution, so I need to know what I did incorrectly.[/url]

 

[skeeter]

1 0 1 = 1 ... R1
0 1 -1 = -1 ... R2
2 1 1 = 2 ... R3

1 0 1 = 1 ---> R1
0 1 -1 = -1 --> R2
0 1 -1 = 0--> R3 - 2(R1)

1 0 1 = 1 ---> R1
0 1 -1 = -1 --> R2
0 0 0 = 1 --> R3 - R2

stop here ... what does the last R3 tell you?

 

[soroban]

Hello, Lime!

\(\displaystyle \begin{pmatrix}1 & 0 & 1 & | & 1 \\
0 & 1 &-1 & | &-1 \\
2 & 1 & 1 & | & 2 \end{pmatrix}\)

\(\displaystyle \begin{pmatrix}1 & 0 & 1 & | & 1 \\
0 & 1 & -1 & | & -1 \\
0 & 1 &-1 & | & -2 \end{pmatrix}
\begin{array}{c}\;\;\rightarrow\;\; \\ \\ \\ \\ \rightarrow \\ \\ \\ \\ \rightarrow\end{array}
\begin{array}{c}R1 \\ \\ \\ R2 \\ \\ \\ R3\cdot R2\end{array}\)
. . . . . . . . . . . . . ↑
. . . . . . . . . . . Here!

You can't "multiply two rows" like that . . .

 

[Lime]

Hello, Lime!

\(\displaystyle \begin{pmatrix}1 & 0 & 1 & | & 1 \\
0 & 1 &-1 & | &-1 \\
2 & 1 & 1 & | & 2 \end{pmatrix}\)

\(\displaystyle \begin{pmatrix}1 & 0 & 1 & | & 1 \\
0 & 1 & -1 & | & -1 \\
0 & 1 &-1 & | & -2 \end{pmatrix}
\begin{array}{c}\;\;\rightarrow\;\; \\ \\ \\ \\ \rightarrow \\ \\ \\ \\ \rightarrow\end{array}
\begin{array}{c}R1 \\ \\ \\ R2 \\ \\ \\ R3\cdot R2\end{array}\)
. . . . . . . . . . . . . ↑
. . . . . . . . . . . Here!

You can't "multiply two rows" like that . . .

Why not?

What is the specific math law on that?

Any other laws I should know?

 

[stapel]

Why not? What is the specific math law on that?

What "law" says that you can multiply rows (that is, linear expression) to get another row (that is, another linear expression)? What rule says that the product (ax + b)(cx + d) of two linear factors will somehow be another linear expression, rather than a quadratic?

Eliz.