Thread: Matrice - Ax=b A^-1b =x

1. Matrice - Ax=b A^-1b =x

Hi

I am hoping I am in the right spot - it is a linear algebra class but 1st year so take pity on me.....

A is 1 1 1 1
2 3 4 4
3 4 4 4
4 5 4 5

B is 4
5
6
3

So let's pretend that I did the calculation of A^-1 correctly....

I got

1 0 0 0 -1 0 -2 2
0 1 0 0 -4 3 -2 8
0 0 1 0 1 0 1 -1
0 0 0 1 0 -1 0 5

Now how do I tie A^-1 back to B do I really just multiply them and how does this tell me what is x1 and x2 etc.

Thanks for any insight you can give me....

2. Originally Posted by calcbeg
Hi

I am hoping I am in the right spot - it is a linear algebra class but 1st year so take pity on me.....

A is 1 1 1 1
2 3 4 4
3 4 4 4
4 5 4 5

B is 4
5
6
3

So let's pretend that I did the calculation of A^-1 correctly....

I got

1 0 0 0 -1 0 -2 2
0 1 0 0 -4 3 -2 8
0 0 1 0 1 0 1 -1
0 0 0 1 0 -1 0 5

Now how do I tie A^-1 back to B do I really just multiply them and how does this tell me what is x1 and x2 etc.

Thanks for any insight you can give me....

So apparently you have $\displaystyle Ax=b$ ,and in full form, according to what you wrote, $\displaystyle \begin{pmatrix}1&1&1&1\\2&3&4&4\\3&4&4&4\\4&5&4&5\ end{pmatrix}\,\begin{pmatrix}x_1\\x_2\\x_3\\x_4\en d{pmatrix}=\begin{pmatrix}4\\5\\6\\3\end{pmatrix}$

Now, assuming A is invertible, we get $\displaystyle Ax=b\Longleftrightarrow x=A^{-1}b$ , so assuming you did calculate the inverse of A correctly , you

get $\displaystyle \begin{pmatrix}-1&0&-2&2\\-4&3&-2&8\\\;\;\;1&0&\;\;1&\!\!\!-1\\\;\;\;0&\!\!\!-1&\;\;0&5\end{pmatrix}\begin{pmatrix}4\\5\\6\\6\en d{pmatrix}=\begin{pmatrix}x_1\\x_2\\x_3\\x_4\end{p matrix}$ , and the you only do the matrix product on the left to find out what the vector x (i.e., x_1,x_2,...) is.

Unfortunately, your calculation of $\displaystyle A^{-1}$ is incorrect, as multiplying it by A we do NOT get 0 in the entry 1-2, as we should...so check this and then do as above.

Tonio