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Can you please tell how to solve this: Ax=b

$$A=\begin{pmatrix} 2 & 4 & -8 & 4 & 1\\ 4 & 10 &-16 & 8 &-8\\ -4 & -12 & 17 & -8 & 20\\ -2 & -10 & 10 & -3 & 34 \end{pmatrix}$$

$$b=\begin{pmatrix}-33\\-36\\4\\-62\end{pmatrix}$$

I have to give a fundamental and particular solutions.

I have tried so: enter image description here

But I do not know whether it is right and what to do next ... Please, help me! Thanks.

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I'm not sure I am familiar with your notation (is the comma representative of a decimal point?) but just based on the fact that you have 5 unknowns and 4 equations, you will either get no solutions or infinitely many solutions. As a check there is an excellent resource here (see "Solving a linear system of equations"). –  process91 Nov 16 '11 at 19:27
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The 2nd element of the 2nd row of the 2nd matrix should be 2 (not -6). I hope you can do it now yourself. –  Tapu Nov 16 '11 at 19:31

2 Answers 2

up vote 4 down vote accepted

The first thing to notice is that you have four equations in five variables -- this means you will have an infinite number of solutions or zero solutions. In the case of an infinite number, you can solve for the set of solutions in terms of one parameter.

The easiest way to do this by hand is to reduce the augmented matrix $\left( A | b\right)$ to reduced row-echelon form. This means proceed with Gaussian elimination until it reaches the form $$ \left( \begin{array}{cccccc} 1 && 0 && 0 && 0 && \star && \star\\ 0 && 1 && 0 && 0 && \star && \star \\ 0 && 0 && 1 && 0 && \star && \star\\ 0 && 0 && 0 && 1 && \star && \star \end{array} \right) $$

where $\star$ is some unknown value.

For your matrix, this results in $$ \left( \begin{array}{cccccc} 1 && 0 && 0 && 0 && 16.5 && -52.5\\ 0 && 1 && 0 && 0 && -5 && 15\\ 0 && 0 && 1 && 0 && 2 && -2\\ 0 && 0 && 0 && 1 && 1 && -1 \end{array} \right) $$

Thus, we see that $$ \begin{align*} x_1 + 16.5x_5 &= -52.5 \\ x_2 - 5 x_5 &= 15 \\ x_3 + 2 x_5 &= -2 \\ x_4 + x_5 &= -1 \end{align*} $$

which I assume is the "general" solution. For a particular solution you can just pick any $x_5$ and solve for the remaining variables.

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$$\left(\begin{array}{ccccc|c}2&4&-8&4&1&-33\\ 4&10&-16&8&-8&-36\\ -4&-12&17&-8&20&4\\-2&-10&10&-3&34&-62\end{array}\right)\to\left(\begin{array}{ccccc|c}2&4&-8&4&1&-33\\0&2&0&0&-10&30\\ 0&-2&1&0&12&-32\\0&-6&2&1&35&-95\end{array}\right)\to$$

$$\left(\begin{array}{ccccc|c}2&4&-8&4&1&-33\\0&1&0&0&-5&15\\ 0&0&1&0&2&-2\\0&0&2&1&5&-5\end{array}\right)\to\left(\begin{array}{ccccc|c}2&4&-8&4&1&-33\\0&1&0&0&-5&15\\ 0&0&1&0&2&-2\\0&0&0&1&1&-9\end{array}\right)\to$$

$$\left(\begin{array}{ccccc|c}2&0&0&0&33&-73\\0&1&0&0&-5&15\\ 0&0&1&0&2&-2\\0&0&0&1&1&-9\end{array}\right)$$ So your general solution is $(-\frac{73}{2},15,-2,-9,0)+x(-\frac{33}{2},5,-2,-1,1)$. Choosing a particular value of $x$ (e.g., $x=0$) you will get a particular solution.

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