For what values of $p$ and $b$ is the vector $(b,8,b+7)$ a solution of this system? Tried to solve this - didn't work.

According to the exercise, the answer is $p = 1 \ , b = -5$. I am getting something way more different.



$$\begin{bmatrix}
1 &amp; 3p &amp; -1\\
-3 &amp; p &amp; -7
\end{bmatrix}\begin{bmatrix}
17 \\ 9
\end{bmatrix}$$


  For what values of $p$ and $b$ is the vector $(b,8,b+7)$ a solution of
  the system?


Well, I don't know - but it always helps to have the system in reduced row echelon form:

$$3r_1+r_2$$
$$\begin{bmatrix}
1 &amp; 3p &amp; -1\\
0 &amp; 10p &amp; -10
\end{bmatrix}\begin{bmatrix}
17 \\ 60
\end{bmatrix}$$

$$\frac{1}{10}r_2$$

$$3r_1+r_2$$
$$\begin{bmatrix}
1 &amp; 3p &amp; -1\\
0 &amp; p &amp; -1
\end{bmatrix}\begin{bmatrix}
17 \\ 6
\end{bmatrix}$$

Clearly, this system will have infinite solutions depending on one variable. It can be either the second or the third column:

If $p = 0$ then the second column is the one that is free. If $p \not = 0$ then the free one is the third.



For $p = 0$

$$\begin{bmatrix}
1 &amp; 0 &amp; -1\\
0 &amp; 0 &amp; -1
\end{bmatrix}\begin{bmatrix}
17 \\ 6
\end{bmatrix}$$

$$-r_2 \ \ , \ \ r_2 + r_1$$

$$\begin{bmatrix}
1 &amp; 0 &amp; 0\\
0 &amp; 0 &amp; 1
\end{bmatrix}\begin{bmatrix}
11 \\ -6
\end{bmatrix}$$

The solution would be

$$S = \{(x,y,z) = (11, t, -6)\} \ \forall t \in \mathbb{R}$$

$(b,8,b+7)$ cannot be a solution in this case, because if $p = 0$, we would need $b = 11$, but clearly $11 + 7 \not = -6$.



For $p \not = 0$

$$\begin{bmatrix}
1 &amp; 3p &amp; -1\\
0 &amp; p &amp; -1
\end{bmatrix}\begin{bmatrix}
17 \\ 6
\end{bmatrix}$$

$$\frac{1}{p}r_2$$

$$\begin{bmatrix}
1 &amp; 3p &amp; -1\\
0 &amp; 1 &amp; -1/p
\end{bmatrix}\begin{bmatrix}
17 \\ 6/p
\end{bmatrix}$$

$$(-3p)r_2+r_1$$

$$\begin{bmatrix}
1 &amp; 0 &amp; 2\\
0 &amp; 1 &amp; -1/p
\end{bmatrix}\begin{bmatrix}
-1 \\ 6/p
\end{bmatrix}$$

So

$$S = \{(x,y,z) = (-1-2t,(6+t)/p, t) \} \ \ \forall t \in \mathbb{R}$$

If you want $(b,8,b+7)$ to be a solution here we have to solve the system:

$$-1 - 2t = b\\
(6+t)/p = 8\\
t = b+7$$

The solutions:

$$b = 13/3\\
p = 13/6$$



So my answer is

$$b = 13/3\\
p = 13/6$$

Which is wrong.



What did I do wrong? What should I have done?
Is the system $Mv=u$ where $M$ is the matrix and $v$ is your solution vector (it isn't clear from what you have written, which presents as an undefined matrix product). i.e. the system is $b+24p-(b+7)=17$ and $-3b+8p-7(b+7)=9$

The first of these cancels nicely to give $p=1$ and the second then gives $b=-5$

(corrected for comment)
The solution of 

$$\left\{ \begin{align} -1-2t &amp; = b;\\
\frac{6+t}{p} &amp; =8; \\
t &amp; =b+7 \end{align} \right.$$

is

$$t=2,
b=-5,
p=1$$