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find the values of $a$ and $b$ for which the following system is consistent

$x+ y-z+w=1$



Note: I have been using gauss elimination, but I got confused when completed "power-1" in the 3rd line

how to move in column 4?

I feel the value that can be entered is a $2$ and $a=1$, but I do not have proof for that

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im sorry i mean i feel the value is a isn't 1 or 2... im sorry for my bad english – Zuhair Oct 27 '12 at 15:35
What do you mean by "power-1$? – Kevin Carlson Oct 27 '12 at 16:18

We have the augmented matrix

$$\left[\begin{array}{cccc|c} 1&1&-1&1&1\\ a&1&1&1&b\\ 3&2&0&a&1+a \end{array}\right]\;.$$

Subtracting $a$ times the first row from the second row and $3$ times the first row from the third row gives us this matrix:

$$\left[\begin{array}{cccc|c} 1&1&-1&1&1\\ 0&1-a&1+a&1-a&b-a\\ 0&-1&3&a-3&a-2 \end{array}\right]\;.$$

Interchange the last two rows and multiply the new middle row by $-1$:

$$\left[\begin{array}{cccc|c} 1&1&-1&1&1\\ 0&1&-3&3-a&2-a\\ 0&1-a&1+a&1-a&b-a\\ \end{array}\right]\;.$$

Subtract $1-a$ times the second row from the third row:

$$\left[\begin{array}{cccc|c} 1&1&-1&1&1\\ 0&1&-3&3-a&2-a\\ 0&0&4-2a&-a^2+3a-2&b-a^2+2a-2\\ \end{array}\right]\;.$$

If $a\ne 2$, then $4-2a\ne 0$, and we can pivot on $4-2a$ to complete the Gaussian elimination, and the system will be consistent. If $a=2$, the matrix is

$$\left[\begin{array}{cccc|c} 1&1&-1&1&1\\ 0&1&-3&1&0\\ 0&0&0&0&b-2\\ \end{array}\right]\;,$$

which is consistent if and only if $b=2$.

Thus, the system is consistent when $a\ne 2$, and it is consistent when $a=b=2$.

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thank you for your solution... :) – Zuhair Oct 27 '12 at 22:22

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