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I have a problem that I need to solve. I am a newbie to the linear algebra so explain the solution as clearly as you can.

Find a so that the system is compatible and solve it. I guess I could handle the solving part, I don't know how to find the a. Please explain me how you did it. Here it is:

$x_1+x_2+2x_3=2$

$2x_1+3x_2-x_3=5$

$3x_1+4x_2+x_3=a$

Thanks a lot.

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3 Answers

up vote 4 down vote accepted

Note that for compatibility condition is:

Rank of $A$ = Rank of the augmented matrix $\bar{A}$.

In your case $A=\left[\begin{matrix}1 & 1 & 2 \\ 2 & 3 & -1\\ 3 & 4 &1\end{matrix}\right] $ and $\bar{A}=\left[\begin{matrix}1 & 1 & 2&2 \\ 2 & 3 & -1&5\\ 3 & 4 &1&a\end{matrix}\right] $. So start with the augmented matrix $\bar{A}=\left[\begin{array}{ccc|c}1 & 1 & 2&2 \\ 2 & 3 & -1&5\\ 3 & 4 &1&a\end{array}\right] $. Apply row operation ($R_3-(R_1+R_2)$) to get $\bar{A}\equiv\left[\begin{array}{ccc|c}1 & 1 & 2&2 \\ 2 & 3 & -1&5\\ 0 & 0 &0&a-7\end{array}\right] $. Hence for the compatibility condition, we must have $a-7=0\Rightarrow a=7$.

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Sorry, please read the first sentence as "The compatibility condition is: " –  Tapu Oct 24 '11 at 18:15
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You can see that the coefficients of the third equation are the sums of the coefficients of the first and second equations. So for the system to be compatible, $a$ must be the sum of 2 and 5.

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I recommend you to keep the matrix-vector product style for these problems. Here is the problem again

$$ \pmatrix{1&1&2\\2&3&-1\\3&4&1}\pmatrix{x_1\\x_2\\x_3} = \pmatrix{2\\5\\a} $$ This is the common $Ax=b$ type or set of linear equations. Now if the matrix is invertible then we can directly multiply from the left by $A^{-1}$ and obtain the solution $x=A^{-1}b$. But we can not since, without the details, the matrix is not invertible. That tells us that some of the rows/columns are linearly dependent to the others.

As AMPerrine pointed out, we are lucky to see that the sum of first two rows equals the third one. This means there are in fact 3 unknowns but only 2 independent equations. Let us show what we have concluded by using a matrix multiplying from the left as follows: $$ \pmatrix{1&0&0\\0&1&0\\-1&-1&1}\pmatrix{1&1&2\\2&3&-1\\3&4&1}\pmatrix{x_1\\x_2\\x_3} = \pmatrix{1&0&0\\0&1&0\\-1&-1&1}\pmatrix{2\\5\\a} $$ This leads to $$ \pmatrix{1&1&2\\2&3&-1\\0&0&0}\pmatrix{x_1\\x_2\\x_3} = \pmatrix{2\\5\\a-7} $$ Now the last equation gives $0=a-7$ and this is the requirement for consistency.

After obtaining a consistent set of equations, we can discard the last equation and continue with the remaining ones. Because as we showed it introduces no extra information. $$ \pmatrix{1&1&2\\2&3&-1}\pmatrix{x_1\\x_2\\x_3} = \pmatrix{2\\5} $$ This leads to the famous Gauss elimination method where we try to simplify things as much as possible via obtaining zero entries in the matrix: $$ \pmatrix{1&0\\-2&1}\pmatrix{1&1&2\\2&3&-1}\pmatrix{x_1\\x_2\\x_3} = \pmatrix{1&0\\-2&1}\pmatrix{2\\5} $$ This gives, $$ \pmatrix{1&1&2\\0&1&-5}\pmatrix{x_1\\x_2\\x_3} = \pmatrix{2\\1} $$ This makes it possible to obtain the solution family : from the second equation we obtain $x_2 = 1+5x_3$ and plugging this into the first gives $x_1+7x_3=1$

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I think it will be helpful (at least at the OP's site) if you kindly mention that those pre-multiplier matrices are not caught arbitrarily, rather those are elementary matrices (obtained by suitable row operations on the identity). –  Tapu Oct 24 '11 at 18:36
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