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Given two matrices A, B, respectively, and an equation XA=B; how can we obtain X and discuss the system?

begin{pmatrix}
1 & a\
a & 1 \
end{pmatrix} begin{pmatrix}
a & 1 \
a^2 & 0 \
1 & 1 \
end{pmatrix}

I am not familiar with this structure and I can't even picture what the dimensions of X are. Any general matrix I write down looks uncomplete.

$X$ must be a $3 times 2$ matrix; we may write

$X = begin{pmatrix} x_{11} & x_{12} \ x_{21} & x_{22} \ x_{31} & x_{32} end{pmatrix}; tag 1$

we may thus write out the system

$XA = B tag 2$

as

$begin{pmatrix} x_{11} & x_{12} \ x_{21} & x_{22} \ x_{31} & x_{32} end{pmatrix} begin{pmatrix} 1 & a \ a & 1 end{pmatrix} = begin{pmatrix} a & 1 \ a^2 & 0 \ 1 & 1 end{pmatrix}; tag 3$

here

$A = begin{pmatrix} 1 & a \ a & 1 end{pmatrix}, tag 4$

and

$B = begin{pmatrix} a & 1 \ a^2 & 0 \ 1 & 1 end{pmatrix}; tag 5$

we may write out the system (3) as a set of $6$ linear equations in two variables each:

$x_{11} + ax_{12} = a, tag 6$

$ax_{11} + x_{12} = 1; tag 7$

$x_{21} + ax_{22} = a^2, tag 8$

$ax_{21} + x_{22} = 0, tag 9$

$x_{31} + ax_{32} = 1, tag{10}$

$ax_{31} + x_{32} = 1; tag{11}$

the system of equations (6)-(11) is in fact easy to solve, one pair $(x_{i1}, x_{i2})$ at a time; for example, we multiply (6) by $a$:

$ax_{11} + a^2x_{12} = a^2, tag{12}$

and subtract (7) from (12):

$(a^2 - 1) x_{12} = a^2 - 1, tag{13}$

whence

$x_{12} = 1, tag{14}$

provided of course that

$a ne pm 1; tag{15}$

having $x_{12}$ we find via (6) that

$x_{11} = a(1 - x_{12}) = 0; tag{16}$

the remaining equations (8)-(11) may be similarly solved.

The system (2), (3) may also be addressed in a somewhat more matrix-oriented manner by setting

$Y = begin{pmatrix} x_{11} & x_{12} \ x_{21} & x_{22} end{pmatrix}, ; mathbf y = (x_{31}, x_{32}), ; C = begin{pmatrix} a & 1 \ a^2 & 0 end{pmatrix}; mathbf d = (1, 1); tag {17}$

then (3) may be written in the from

$begin{pmatrix} Y \ mathbf y end{pmatrix} A = begin{pmatrix} C \ mathbf d end{pmatrix}, tag{18}$

leading to

$YA = C, ; mathbf y A = mathbf d, tag{19}$

whence

$Y = CA^{-1}, ; mathbf y = mathbf d A^{-1}; tag{20}$

where it is easily seen that

$A^{-1} = dfrac{1}{1 - a^2} begin{pmatrix} 1 & -a \ -a & 1 end{pmatrix}. tag{21}$

We assume $a = pm 1$ in deriving (17)-(21); in the event that $a = pm 1$ we see that (8)-(9) become

$x_{21} pm x_{22} = 1, ; pm x_{21} + x_{22} = 0, tag{22}$

which lead to the contradicion

$pm 1 = 0; tag{23}$

thus we see there is no solution to (2) when $a = pm 1$.