2x +2y+3z=10
4x-y+z=-5
5x-2y+6z=1
This is a problem using cramer's rule and I am lost trying to get this one to work out.
Could you specify what the problem is? It's just applying Cramer's rule and thus filling in the formula. It comes down to computing 4 determinants, perhaps a lot of work but not difficult.
Do you have trouble in understanding the rule or with the arithmetic of the computations of the determinants?
Understanding the rule is not a problem? The determinant of the coefficient matrix is used 3 times, but you only have to compute it once of course. In the nominator, to solve for the i-th unknown, the i-th column is replaced by the column of constants. Its determinant divided by the determinant of the coefficient matrix gives the solution. For x, this is:
$\displaystyle
x = \frac{{\left| {\begin{array}{*{20}c}
{10} & 2 & 3 \\
{ - 5} & { - 1} & 1 \\
1 & { - 2} & 6 \\
\end{array}} \right|}}{{\left| {\begin{array}{*{20}c}
2 & 2 & 3 \\
4 & { - 1} & 1 \\
5 & { - 2} & 6 \\
\end{array}} \right|}} = \frac{{55}}{{ - 55}} = - 1
$
I'll show how to compute a determinant, I'll do the bottom one as an example.
$\displaystyle
\left| {\begin{array}{*{20}c}
2 & 2 & 3 \\
4 & { - 1} & 1 \\
5 & { - 2} & 6 \\
\end{array}} \right| = \left| {\begin{array}{*{20}c}
{10} & 0 & 5 \\
4 & { - 1} & 1 \\
7 & 0 & 9 \\
\end{array}} \right| = - \left| {\begin{array}{*{20}c}
{10} & 5 \\
7 & 9 \\
\end{array}} \right|
$
I added the second row to the last one and I added it twice to the first one, creating two zeroes in column 2 and leaving only a -1. You then expand into the (only one left) minor. We can now easily find:
$\displaystyle
- \left| {\begin{array}{*{20}c}
{10} & 5 \\
7 & 9 \\
\end{array}} \right| = 5 \cdot 7 - 10 \cdot 9 = 35 - 90 = - 55
$
Re write the equations in matrix form:Originally Posted by kbryant05
$\displaystyle
\left[\begin{array}{ccc}
2&+2&3\\
4&-1&1\\
5&-2&6
\end{array}\right]
\left[\begin{array}{c}
x\\
y\\
z
\end{array}\right]=
\left[\begin{array}{c}
10\\
-5\\
1
\end{array}\right]
$
Then use Cramers rule to invert the matrix, and then use:
$\displaystyle \left[ \begin{array}{c}
x\\
y\\
z
\end{array}\right]=
\left[\begin{array}{ccc}
2&+2&3\\
4&-1&1\\
5&-2&6
\end{array}\right]^{-1}
\left[\begin{array}{c}
10\\
-5\\
1
\end{array}\right]
$
to solve for $\displaystyle x,\ y$ and $\displaystyle z$.
RonL
Hello,Originally Posted by kbryant05
with 3X3-determinants there exist another method to calculate their values. Maybe this is of some help:
Because I don't know to write determinants in Latex I've attached an image to sho you what you have to calculate:
Copy the first two columns of the determinant and add them to the determinant. Then multiply always three numbers as I've indicated by black and red lines. You have to add the products which are produced by the black lines and then you have to subtract the products which are produced by the red lines.
Greetings
EB
Hello,Originally Posted by ThePerfectHacker
you are right. It is a kind of condensed Sarrus' rule.
But in school determinants are not very often bigger than 3X3. So this rule (if you are used to it and trained to use it) is very helpful. If you use integers only (like here) you can even solve such a problem much faster than those poor kids who only learned to hit keys on their calculators.
Greetings
EB