I am really new to matrices and these concepts, so any help would be much appreciated.
I have the following square matrix $$A = \begin{bmatrix}1&-3&0\\1&4&-3\\0&1&5\end{bmatrix}$$
We're asked to let each row = $a_1, a_2, a_3$ going down and then calculate the cross products and triple scalar products which I was able to do.
I calculated the cross products of $a_1 \times a_3$, and then $a_2 \times a_3$: which are:
$$a_1 \times a_3 = \begin{bmatrix}-15\\-5\\1\end{bmatrix}$$ and,
$$a_2 \times a_3 = \begin{bmatrix}23\\-5\\1\end{bmatrix}$$
If $Au = e_1$ and $Av = e_2$, and I need to determine vectors $u$ and $v$, could someone explain how to find u and v. I know that if I multiply A by the cross product vectors I get $[38, 0, 0]$ and $[0, -38, 0]$ but this confuses me because I thought to satisfy the equation the answer should be $[1, 0, 0]$ and $[0,1,0]$. I have also tried finding the inverse of A and multiply this by $e_1$ and $e_2$, but the results were not even close to the result of the cross products?? I am really trying to learn, would appreciate some advice.
Is there another method to solve this??
Best Answer
You are very close. Your cross product strategy gives you a vector that produces multiples of $e_1$ and $e_2$. But remember that since $A$ is a linear transformation, $A(\lambda x) = \lambda (A x)$ for any scalar lambda and any vector $x$. So how can you choose scalars $\lambda, \mu$ so that $A(\lambda u) = e_1$ and $A(\mu v) = e_2$? (Here $u$ and $v$ are the vectors you calculated).
You were also close with the strategy of finding the inverse! Note that if the inverse has columns $x_1$, $x_2$, and $x_3$ (in that order from left to right), then $$A A^{-1} = I = [e_1, e_2, e_3] = A [x_1, x_2, x_3] = [A x_1, A x_2, A x_3]$$ So in fact that first column of the inverse is the $u$ you're looking for, and the second column of the inverse is the $v$ you're looking for.
EDIT: OP asked a good question, namely whether the constants $\lambda$ and $\mu$ can be deduced without having to calculate $A u$ and $Av$ and then reverse-engineering them. And the answer is yes. We'll show it for $\lambda$, but the process is identical for $\mu$. Plugging in our expression for $u$, we notice that $$A u = A (a_2 \times a_3) = \begin{bmatrix} a_1 \cdot (a_2 \times a_3) \\ a_2 \cdot (a_2 \times a_3) \\ a_3 \cdot (a_2 \times a_3) \end{bmatrix} = \begin{bmatrix} a_1 \cdot (a_2 \times a_3) \\ 0 \\ 0\end{bmatrix}$$ the last equality following from the properties of the triple product. So we can deduce that $$e_1 = \begin{bmatrix} 1 \\ 0 \\ 0\end{bmatrix} = \frac{(a_2 \times a_3)}{a_1 \cdot (a_2 \times a_3)}$$ And you can further note that this expression is only ill-defined (i.e. the denominator is zero) when $a_1$ is a linear combination of $a_2$ and $a_3$, which would mean that $A$ is not invertible, in which case we would have to adopt a different method to solve this problem.