$A=\begin{bmatrix} a&1&0\\ 1&b&d\\ 1&b&c \end{bmatrix}, B=\begin{bmatrix} a&1&1\\ 0&d&c\\ f&g&h\end{bmatrix}, U=\begin{bmatrix} f\\ g\\ h \end{bmatrix} U=\begin{bmatrix} a^2\\ 0\\ 0 \end{bmatrix}, X=\begin{bmatrix} x\\ y\\ z \end{bmatrix}$

If $AX = U$ has infinitely many solutions, then prove that $BX=V$ cannot have a unique solution. If further $afd\ne0$, then prove that $BX=V$ has no solution.

**My attempt is as follows:-**

If $AX=U$ has infinitely many solutions, then $|A|=0$

$$\begin{vmatrix} a&1&0\\ 1&b&d\\ 1&b&c \end{vmatrix}=0$$

$$a(bc-bd)-(c-d)=0$$ $$ab(c-d)-(c-d)=0$$ $$(c-d)(ab-1)=0$$

So either $c=d$ or $ab=1$

**Case $1$: $c=d$**

$$AX=U$$

Taking $adj(A)$ on both sides

$$adj(A)AX=adj(A)U$$ $$OX=adj(A)U$$

$$adj(A)=\begin{bmatrix} bc-bd&-c&d\\ -(c-d)&ac&-ad\\ 0&-(ab-1)&ab-1 \end{bmatrix}$$

$$adj(A)U=0$$ $$\begin{bmatrix} (bc-bd)f-cg+dh\\ -(c-d)f+acg-adh\\ -(ab-1)g+(ab-1)h \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$$

As $c-d=0$

$$\begin{bmatrix} -cg+ch\\ acg-ach\\ -(ab-1)g+(ab-1)h \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$$ $$g=h$$

Now let's prove $BX=V$ cannot have a unique solution

$$|B|=a(dh-cg)-(0-cf)+(0-df)$$ $$|B|=0$$

$$BX=V$$

Taking $adj(B)$ on both sides

$$OX=adj(B)V$$

$$adj(B)=\begin{bmatrix}\ dh-gc&g-h&c-d\\ cf&ah-f&-ac\\ -df&-(ag-f)&ad \end{bmatrix}$$

$$adj(B)V=\begin{bmatrix} a^2(dh-gc)\\ a^2cf\\ -a^2df \end{bmatrix}$$

$$adj(B)V=a\begin{bmatrix} a(dh-gc)\\ acf\\ -adf \end{bmatrix}$$

So if $afd$ is not equal to zero, then there will be no solution otherwise it will have infinitely many solutions.

**Case $2$: $ab-1=0$**

$$adj(A)U=0$$ $$\begin{bmatrix} (bc-bd)f-cg+dh\\ -(c-d)f+acg-adh\\ -(ab-1)g+(ab-1)h \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$$ $$b(c-d)f-cg+dh=0\tag{1}$$ $$-(c-d)f+acg-adh=0\tag{2}$$

Multiplying first equation by $-a$

$$-(c-d)f+acg-adh=0$$

So first and second equation are same.

Now let's prove $BX=V$ cannot have a unique solution

$$|B|=a(dh-cg)-(0-cf)+(0-df)$$ $$|B|=a(dh-cg)+cf-df$$

$$|B|=-(c-d)f+cf-df$$ $$|B|=0$$

Taking $adj(B)$ on both sides

$$OX=adj(B)V$$

$$adj(B)V=a\begin{bmatrix} a(dh-gc)\\ acf\\ -adf \end{bmatrix}$$

So yes here also if $afd\ne0$, then $BX=V$ will have no consistent solution.

My solution went very long, any easy to solve this?