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Question and answer from an exam paper, how on earth are you supposed to work this out from the beginning. I can see how the answer works, but for a similar question how do I figure out what to times/add by what to make the two equal.

Question:

Find the value of a which allows
solutions to the system of equations
4x − y + 2z = 7
x + y − 3z = −3
2x − 3y + 8z = a

Solution:

4x − y + 2z = 7 [1]
x + y − 3z = −3 [2]
2x − 3y + 8z = a [3]
[1] + [2], 5x − z = 4
3 × [2] + [3], 5x − z = −9 + a
Equations are consistent if 4 = −9 + a i.e. a = 13

Another way of asking the question; is there a logical method for working these types of questions out.

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A very similar question was asked recently: math.stackexchange.com/questions/39844/linear-algebra-question/… –  lhf May 25 '11 at 22:55
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4 Answers

up vote 3 down vote accepted

Yes there is. It is called the Row Echelon form. Row echelon form is sometimes called reduced row echelon form or Hermite normal form. A matrix is in row echelon form if:

  • the first nonzero element in every row is 1,
  • the first nonzero element in every row occurs to the right of the first nonzero element in the row above it, and
  • all the elements above the first nonzero element in a row are 0.

For your case you construct the following from the coefficients and the right hand side:

$$ \begin{bmatrix}\begin{array}{ccc|c} 4 & -1 & 2 & 7\\ 1 & 1 & -3 & -3\\ 2 & -3 & 8 & a\end{array}\end{bmatrix} $$

and try to eliminate the lower triangular elements. To do so make the first column equal to 4 in all the rows by multiplying by 4 and 2 rows 2 and 3 to get

$$ \begin{bmatrix}\begin{array}{ccc|c} 4 & -1 & 2 & 7\\ 4 & 4 & -12 & -12\\ 4 & -6 & 16 & 2a\end{array}\end{bmatrix} $$

and then subtract the 1st row from all the other rows to get

$$ \begin{bmatrix}\begin{array}{ccc|c} 4 & -1 & 2 & 7\\ 0 & 5 & -14 & -19\\ 0 & -5 & 14 & 2a-7\end{array}\end{bmatrix} $$

The repeat for the 2nd column starting from the 2nd row and down. This gets us to

$$ \begin{bmatrix}\begin{array}{ccc|c} 4 & -1 & 2 & 7\\ 0 & 5 & -14 & -19\\ 0 & 0 & 0 & 26-2a\end{array}\end{bmatrix} $$

which obviously cannot be solved for $x$, $y$, $z$ at the same time because all their coefficients are zero in the last equation.

The above is expanded to

$$ \begin{aligned}4\, x-y+2\, z & =7\\ 5\, y-14\, z & =-19\end{aligned} $$

which is solved for $x$ and $y$ in terms of $z$ for example, and

$$ 2\,a = 26 $$.

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Question: Find the value of a which allows solutions to the system of equations:

$4x −\; y + 2z = \;\; 7 \qquad\; [1]$
$\;\; x + \;y − 3z = −3 \qquad [2]$
$2x − 3y + 8z = \; a \qquad\; [3]$

Solution: First of all, take a look at equations $[1],[2],[3]$ and note any similarities between the coefficients of the $x$'s, $y$'s, and $z$'s. The goal is to reduce the system of equations in such a way that we can solve for a single variable, or, in this case, we need only reduce the system to two equations in $x$ and $z$.

I. Note that in the 1st and 2nd equation, the coefficients of the $y$-terms are, respectively, $-1$ and $1$. So simply by adding equations $[1]$ and $[2]$, we can eliminate $y$.

II. That said, it would then make the most sense to eliminate the $y$-variable in the third equation, as well; we can do so my multiplying equation $[2]$ by $3$, then adding the modified equation $[2]$ to $[3]$.

We thus end up with two equations which resolve the problem nicely. More details:

$[1] + [2] \to 5x − z = 4\qquad$ Add eq. $[1]$; the $y$-terms cancel out, leaving $5x - z = 4$:

$\; 4x-y+2z = 7$
$\underline{+ x + y - 3z = -3}$
$\;5x+0y-z =4 \implies 5x - z =4\qquad$ [I]

$3 \times [2] + [3] \to 5x-z=-9\quad$ Multiply $[2]$ by $3$: $\quad 3(x+y-3z=-3) = 3x +3y - 9z=-9$

*Note: Take this last equation [$3x +3y - 9z=-9$], and add it to equation $[3]$:

$\quad 3x +3y-9z=-9$
$\underline{+ 2x -3y +8z = a}$
$\;\;5x +0y-z=-9 +a \implies 5x -z = -9 +a\qquad$ [II]

So we're left with [I] $5x - z = 4$ and [II] $5x - z = -9 + a$.

For [I] and [II] to both hold, we must have that $4 = -9 + a$, from which we can solve for $a$ by adding 9 to both sides of the equation, giving us: $a = 13$.

The goal, using this approach, is to eliminate a variable by adding/subtracting equations (as with equation $[1]$ and $[2]$), or adding/subtracting a multiple of an equation (as we did when we multiplied equation $[2]$ by $3$, and then added to equation $[3]$).

This is essentially what row reduction will be like, except you won't have the variables cluttering everything up!

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Another way of thinking about it: you really have 3 equations in 4 unknowns $x$,$y$,$z$,$a$ (except that you only care about the value of $a$ in the solution). So do row reduction on the matrix $\left[ \matrix{4 & -1 & 2 & 0 & 7\cr 1 & 1 & -3 & 0 & -3\cr 2 & -3 & 8 & -1 & 0\cr} \right]$.

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If you look at the augmented matrix, and start to row reduce, it only takes two reductions (depending how you count) before you can see that $6+a=19$.

$\begin{bmatrix} 4 & -1 & 2 & 7\\ 1 & 1 & -3 & -3\\ 2 & -3 & 8 & a\end{bmatrix}\rightarrow^{-4\textrm{R}_2+\textrm{R}_1\rightarrow\textrm{R}_1}_{-2\textrm{R}_2+\textrm{R}_3\rightarrow\textrm{R}_3}\rightarrow\begin{bmatrix} 0 & -5 & 14 & 19\\ 1 & 1 & -3 & -3\\ 0 & -5 & 14 & 6+a\end{bmatrix}$

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Given the tag chosen by the OP, I don't know that the OP has encountered matrices and row reduction (i.e. linear algebra)...This seems more at the algebra/precalculus level, where students are taught to solve n number of equations in n variables, usually n = 2 or 3... –  amWhy May 25 '11 at 22:28
    
That 9 should be an 8 –  Robert Israel May 25 '11 at 22:31
    
We have been taught matrices, but not row reduction, sounds of interest though –  Pez Cuckow May 25 '11 at 22:31
    
@Pez See the non-matrix answer below: it is essentially using "row reduction", as you will learn, but may be more familiar to you now. –  amWhy May 25 '11 at 23:20
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