linear system: solving using matrices

[mathstresser]

Find an equation relating a, b, and c so that the linear system

2x + 2y + 3z = a
3x - y + 5 = b
x – 3y + 2 = c

is consistent for any values of a, b, and c that satisfy that equation.

The array is (I,II,III)

. . .\(\displaystyle \L A\, =\, \left[\, \begin{array}{rrrr}2&2&3&a\\ 3&-1&5&b\\ 1&-3&2&c \end{array}\, \right]\)

I combine I and 2II and get 8x +13z = a + 2b.
I combine 3I + 2III and get 8x + 13z = 3a + 2c.

So, a + 2b = 3a + 2c.

2b = 2a + 2c
b = a + c

This may be a long and drawn out way, but is this how to do it- and is this the answer they're looking for?

 

[stapel]

What you've done is fine. Now notice that, if a + 2b and 3a + 2c are NOT equal, then you can subtract the one row from the other, and get an x,y,z-row of zeroes being equal to something that is NOT zero. That is, if a + 2b does NOT equal 3a + 2c, then the system will be inconsistent.

So what condition do you have on "a", "b", and "c"?

Eliz.

Edit: Ne'mind; answer is below.

 

[soroban]

Hello, mathstresser!

I assume you left out some \(\displaystyle z\)'s in the last two equations.

Find an equation relating \(\displaystyle a,\,b,\,c\) so that the linear system

. . \(\displaystyle \begin{array}{ccc}2x\,+\,2y\,+\,3z & \:=\: & a \\
3x\,-\,y\,+\,5z &\:=\:& b\\
x\,-\,3y\,+\,2z & \:=\:& c \end{array}\)

is consistent for any values of \(\displaystyle a,\,b,\,c\) that satisfy that equation.

Your work and your conclusion are correct.
Here's how I approached it . . .

We have:\(\displaystyle \;\begin{vmatrix}2&2&3&| &a\\ 3&-1&5&|&b\\ 1&-3&2&|&c \end{vmatrix}\)

\(\displaystyle \begin{array}{r}\qquad \\ \quad\\ \quad\\ \quad\\ \quad \\ \quad\\ \quad \\ \quad \\ \quad \\ R3+R1\end{array}\;\;
\begin{vmatrix}2 & 2 & 3 & | & a \\
3 & -1 & 5 & | & b \\
3 & -1 & 5 &| & a+c \end{vmatrix}\)

\(\displaystyle \begin{array}{r}\qquad\\ \quad\\ \quad\\ \quad\\ \quad\\ \quad\\ \quad \\ \quad \\ \quad \\ R3-R2\end{array}\;\;
\begin{vmatrix}2 & 2 & 3 & | & a \\
3 & -1 & 5 & | & b\\
0 & 0 & 0 & | & a-b+c\end{vmatrix}\)


We have a row of zeros.
The system is consistent if the rightmost element is also zero.

. . Therefore: \(\displaystyle \:\fbox{a\,-\,b\,+\,c\:=\:0}\)