For what value of $b$ does the system $3x-2y+z=b,5x-8y+9z=3,2x+y-3z=-1$ have infinite solutions? 
For what value of $b$ does the system $3x-2y+z=b,5x-8y+9z=3,2x+y-3z=-1$ have infinite solutions?

My Attempt:
Let $A$ be the coefficient matrix. And $B$ be the column matrix with elements $b,3,-1$. Also, $det(A)=0$
In Matrix method, for infinite solutions, $(\text{adj}(A))B=0$. My $(\text{adj}(A))B$ is coming a column matrix with entries $15b-25,33b-55,21b-35$. For this to be zero, $b=5/3$.
But the answer given is $1/3$.
If I use Crammer's method, calculating $D_x$ (i.e. in $A$, replacing coefficient of $x$ with constants), and making it zero, I do get $b=1/3$.
Did I make some mistake in the matrix method?
For $\text{adj}(A)$, my first row has elements $15,-5,-10$, second row $33,-11,-22$, and third $21,-7,-14$.

We have the adjoint matrix (this agrees with your result)
$$\begin{pmatrix}
 15 & -5 & -10 \\ 33 & -11 & -22 \\ 21 & -7 & -14 \\
\end{pmatrix}$$
We then form
$$\begin{pmatrix}
 15 & -5 & -10 \\  33 & -11 & -22 \\  21 & -7 & -14 
\end{pmatrix}.\begin{pmatrix}
 b  \\  3  \\  -1 
\end{pmatrix} = \begin{pmatrix}
 15 b-5\\33 b-11\\21 b-7 \end{pmatrix}$$
It appears you had a simple sign error in $b$ with the last value, $1$ instead of $-1$, when doing the multiplications.
