# finding the values of a such that the system $AB\cdot x=0$ has infinitely many solutions

the following matrix are given: $$A= \begin{bmatrix} a & 1 & 1 \\ a & a & 2 \\ a & a & a \\ \end{bmatrix}$$ $$B= \begin{bmatrix} a-3 & 1 & 1 \\ a-3 & a-3 & 2 \\ a-3 & a-3 & a-3 \\ \end{bmatrix}$$ find the values of a such that the system $AB\cdot x=0$ has infinitely many solutions. I know that a matix has infinitely many solutions $\iff det(AB)=0$, so this is what I did : $$A'= \begin{bmatrix} a & 1 & 1 \\ 0 & a-1 & 2 \\ 0 & 0 & a-2 \\ \end{bmatrix}$$ and $det(A) =0$ if $a\cdot(a-1)\cdot(a-2)=0$ , so if $a=0,1,2.$ A has infinitely many solutions. $$B= \begin{bmatrix} a-3 & 1 & 1 \\ 0 & a-4 & 2 \\ 0 & 0 & a-5 \\ \end{bmatrix}$$ and $det(B) =0$ if $(a-3)\cdot(a-4)\cdot(a-5)=0$ , so if $a=3,4,5.$ B has infinitely many solutions. So $AB\cdot x=0$ , if $a=0,1,2,3,4,5$, BUT the textbook answer is $a=1,2,3$. What am I doing wrong?

• The curse of the wrong textbook strikes. Mar 7, 2018 at 12:20
• You can even see at once that $a=0$ must be a solution as the matrix $A$ has a column with only zeroes, so the determinant of this and any product must be zero as well. So yes, text book has to be wrong. Mar 7, 2018 at 12:33

## 1 Answer

Your answer is right (I checked it) and therefore your textbook is wrong.