If $B=\begin{bmatrix} 1& 3&4\\ 3&-1&5\\ -2&4&-3\end{bmatrix}$ and $X=\begin{bmatrix} 3& -1&5\\ 1&3&4\\ 4&-8&6\end{bmatrix}$, Solve $AB=X$.
My attempt:
We have $AB=X$.
Multiplying By $B^{-1}$ on both sides, we get $A=X.B^{-1}$. But this method is very lengthy and I'm supposed to do this question in 2 minutes. Is there any other short way?
On
This problem is meant to be solved by inspecting the matrices involved. $X$ is almost the same as $B$ but with a few changes. Swapping the first two rows of $B$ and multiplying the last row by $-2$ you arrive at the matrix $X$. Since these are row operations they can be expressed using elementary matrices and so we calculate $$E_1=\pmatrix{0 &1& 0 \\ 1& 0& 0 \\ 0& 0 &1}, E_2=\pmatrix{1 &0& 0 \\ 0& 1& 0 \\ 0& 0 &-2}, A=E_1E_2=\pmatrix{0 &1& 0 \\ 1& 0& 0 \\ 0& 0 &-2}$$
You must make observations to do this question faster.
Indeed, note that the first row of $X$ is the second row of $B$, and the second row of $X$ is the first row of $B$. Finally, the third row of $X$ is $-2$ times the third row of $B$.
Focus on the first row of $A$ : multiplying this row matrix with $B$ would give the first row of $X$. But the first row of $X$ is the second row of $B$, so the first row of $A$ must somehow neglect the first and third entries of each column of $B$ while leaving the second intact. This suggests $A = [0,1,0]$. Multiplying $[0,1,0]$ with $B$ , the answer is seen to be $[3,-1,5]$, the first row of $X$.
In similar fashion, the second row of $A$ must be $[1,0,0]$ , and the third row must be $[0,0,-2]$. Hence the answer is just these put together which is : $$ A = \begin{pmatrix} 0&1&0\\ 1&0&0\\ 0&0&-2 \end{pmatrix} $$
EDIT : Note that were it not for these observations, this question could not be classed as one doable in two minutes. Indeed, this question would involve determinant computation of both $B$ and all it's $2 \times 2$ submatrices, which would (on average) take at least four to five minutes, and that's even before computing $XB^{-1}$.