Solve for x1 and x2

frctl

Full Member
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Jun 29, 2019
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Picture1.png

I require help with this system the way it is presented
Where does the x1 and x2 go or with which coefficient?
 
well just apply matrix subtraction and multiplication as you usually would.
First subtract [MATH]\begin{pmatrix}3\\4\end{pmatrix}[/MATH] from both sides then the first equation is

[MATH]1 \cdot x_1 + 2 \cdot x_2 = 6[/MATH]
I leave you to figure out the second equation
 
1 · x1 + 2 · x2 = 6
1 · x1 + 1 · x2 = 5

This is not the answer
 
View attachment 17443

I require help with this system the way it is presented
Where does the x1 and x2 go or with which coefficient?
View attachment 17443

I require help with this system the way it is presented
Where does the x1 and x2 go or with which coefficient?
\(\left[ {\begin{array}{*{20}{c}} 1&2 \\ 1&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\
y \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 3 \\ 4 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
9 \\ 9 \end{array}} \right]\) Subtract from both sides.
\(\left[ {\begin{array}{*{20}{c}} 1&2 \\ 1&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\
y \end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
6 \\ 5 \end{array}} \right]\) Find the inverse.
\(\left[ {\begin{array}{*{20}{c}} x \\ y \end{array}} \right]=\left[ {\begin{array}{*{20}{c}} 1&2 \\ 1&1 \end{array}} \right]^{-1} \left[ {\begin{array}{*{20}{c}} 6 \\ 5 \end{array}} \right]\)
 
I found the inverse is

-1 2
1 -1

How does this help me solve for x1 and x2
 
the matrix
x1 and x2
That's not correct, frctl. You already posted your solution: x1=4 and x2=1 (post #8).

The second equation in post #6 shows the matrix containing elements 6 and 5.

We need to be more careful with language.

?
 
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