Finding reverse matrix from equation

Question

we have the following matrices:

     ( a''      b''     c'' )              ( a   b   c )
B =  ( 3a-2a'   3b-2b'  3c-2c')  and A =   ( a'  b'  c')
     ( 5a'      5b'     5c')               ( a'' b'' c'')

and an equation: B = PA, i need to find P.

this is my attemp to solve it: we need to find P and B=PA.

we duplicate with A^-1 and get

BA^-1 = (PA)A^-1 --> BA^-1 = P(AA^-1) --> BA^-1 = P

so we need to find A^-1. in order to find P. I thought to try to solve it like (A|I) and to reach I on the left side in order to get the reverse matrix. but I'm not sure how I can get from matrix A to I. since all of the elements are different.. a/a''/a'.

Answer 1

If $B=PA$, then consider the lines of $B$. The first line of $B$ is the linear combination of lines of $A$ with coefficients at this linear combination being the coefficients in the first line of $P$.

In other words, if $B$ consists of lines $L_1$, $L_2$, $L_3$, the first line of $P$ writes $p_{11},p_{12},p_{13}$ and $A$ consists of lines $l_1$, $l_2$, and $l_3$, then you have $$L_1 = p_{11}l_1+p_{12}l_2+p_{13}l_3.$$

By looking at the first line of $B$ we can easily see that we can take $p_{11}=p_{12}=0$ and $p_{13}=1$.

Apply the same thing for the second and the third lines of $B$ to obtain $$P=\begin{pmatrix}0&0&1\\ 3&-2&0\\0&5&0\end{pmatrix}.$$

If $A$ is invertible, then this $P$ is unique; if $\det A=0$, then you have an infinity of possible $P$.

Answer 2

Finding $A^{-1}$ is going to be more than difficult. Instead, you should develop a slightly deeper understanding of things. Whether you've done row reduction or not, this might allow you to recontextualise some of that work, when you do it or now.

Firstly, recall that all matrix products are row by column, and the simplest rows are the standard basis. Note that $$ (1, 0, 0)A = (a, b, c),\ \ (0, 1, 0)A = (a', b', c')\ \text{ and }\ (0, 0, 1)A = (a'', b'', c'').$$

If we make $v$ the first row $P$ then $vA$ will be the first row of $PA = B$. So, immediately, you should be able to see what your first row of $P$ should be. Next, have a look at the third row of $B$ and again, using the results from multiplication by the standard basis, it should be easy to see what the third row is: all you really need to remember is that scalar multiplication works such that $(\lambda v) A = \lambda (vA)$.

Finally, the second row of $B$ is again a simple linear combination of the rows of A, so follow your nose and you should be able to work out what the second row of $P$ needs to be.