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

$x+ y-z+w=1$

$ax+y+z+w=b$

$3x+2y+aw=1+a$

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

  • 0
    What do you mean by "power-1$?2012-10-27

1 Answers 1

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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... :)2012-10-27