Answer to Question #118715 in Linear Algebra for Nii Laryea

Question

A and X are the matrices
a b
c d
and
x y
u v
respectively, where b is not equal to zero. Prove that if AX = XA then u = cy/b and v = x + (d − a)y/b. Hence prove
that if AX = XA then there are numbers p and q such that X = pA + qI, and find
p and q in terms of a, b, x, y.

Answer 1

A=$\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ and X=$\begin{bmatrix} x & y \\ u & v \end{bmatrix}$ and b is not equal to zero.

now given,

AX = XA

or $\begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} x & y \\ u & v \end{bmatrix}$ = $\begin{bmatrix} x & y \\ u & v \end{bmatrix}\begin{bmatrix} a & b \\ c & d \end{bmatrix}$

or $\begin{bmatrix} ax+bu & ay+bv \\ cx+du & cy+dv \end{bmatrix}$ = $\begin{bmatrix} ax+cy & bx+dy \\ au+cv & bu+dv \end{bmatrix}$

As the two matrices are equal we can compare the corresponding elements of the two matrices.

$\therefore$ $ax+bu = ax+cy$

or $bu=cy$

or $u = cy/b$ ( as b is not equal to zero)

(proved)

Again,

or $ay+bv = bx+dy$

or $bv = dy-ay+bx$

or $bv = (d-a)y+bx$

or $v= (d-a)y/b+x$ (proved)

Now

or $X=pA+qI$

or $\begin{bmatrix} x & y \\ u & v \end{bmatrix}$ = $p\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ + $q$ $\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$

or $\begin{bmatrix} x & y \\ u & v \end{bmatrix}$ = $\begin{bmatrix} p a &pb \\ pc & pd \end{bmatrix}$ + $\begin{bmatrix} q & 0 \\ 0 & q \end{bmatrix}$

or $\begin{bmatrix} x & y \\ u & v \end{bmatrix}$ = $\begin{bmatrix} pa+q &p b \\ pc & pd+q \end{bmatrix}$

As the two matrices are equal we can compare the corresponding elements of two matrices.

$\therefore$ $x=pa+q$ and $y= pb$

Now $y=pb$

or $p = y/b$ . Answer

again

$x=pa+q$

or $x= (ay)/b+q$

or $q= x-((ay)/b)$ Answer