Matrix Problem of form Ax=B The matrix $A$ is given by
$$\left(\begin{array}{ccc}
1 & 2 & 3 & 4\\
3 & 8 & 11 & 8\\
1 & 3 & 4 & \lambda\\
\lambda & 5 & 7 & 6\end{array} \right)$$
Given that $\lambda$=$2$, $B$=$\left(\begin{array}{ccc}
2 \\
4 \\
\mu \\
3 \end{array} \right)$ and $X$=$\left(\begin{array}{ccc}
x \\
y \\
z \\
t \end{array} \right)$
Find the value of $\mu$ for which the equations defined by $AX=B$ are consistent and solve the equations in this case. State the rank of A.
So I began by reducing matrix $A$ to reduced row echelon form (kind of like taking the null space, except I'm dealing with $Ax=B$ instead of $Ax=0$) but since I have 5 variables and only 4 equations, I'm not sure how to continue onward.
 A: Here's an easier way to solve the problem, without using determinants.
The system of linear equations $Ax=b$ is solvable exactly when $b$ is a vector
in the column space of $A \equiv \text{col}(A) = \{ x \in \mathbb{R^4}: x=Ay, \text{ for some } y\in \mathbb{R^4} \}$.
Looking at $b = \begin{bmatrix} 2 & 4 & \mu & 3 \end{bmatrix}^{T}$ and at the fourth column of $A$, $A_{\bullet4} = \begin{bmatrix} 4 & 8 & 2 & 6 \end{bmatrix}^{T}$, we easily observe that if $\mu = 1$, then $b = \frac{1}{2}A_{\bullet4} = 0 \cdot A_{\bullet1} + 0 \cdot A_{\bullet2} + 0 \cdot A_{\bullet3} + \frac{1}{2}A_{\bullet4}$, hence $b$ is a linear combination of the columns of $A$, so, in that case, $b \in \text{col}(A)$, and the system of equations $Ax=b$ is consistent.
To find the rank of $A$, $\text{rk}(A)$, which is equal to the number of pivots in any echelon form of $A$, you just need to row reduce your matrix $A$, arriving at
$$\begin{bmatrix} 1 & 2 & 3 & 4 \\
3 & 8 & 11 & 8 \\
1 & 3 & 4 & 2\\
2 & 5 & 7 & 6
\end{bmatrix}
\underbrace{\rightarrow}_{\text{row reducing}}
\begin{bmatrix} 1 & 0 & 1 & 8 \\
0 & 1 & 1 & -2 \\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0
\end{bmatrix}.$$
Since there are only two pivots in the echelon form of $A$, $\text{rk}(A) = 2$. 
A: First, note that 
$$
\det A=2\,(\lambda-2)^2
$$
so $A$ is invertible if and only if $\lambda\neq 2$. In this case we can use Cramer's rule to solve for $X$. For example, we have
$$
x=
\frac{
\begin{vmatrix}
2   & 2 & 3  & 4 \\
4   & 8 & 11 & 8 \\
\mu & 3 & 4  & \lambda \\
3   & 5 & 7  & 6
\end{vmatrix}
}
{
\begin{vmatrix}
1 & 2 & 3 & 4\\
3 & 8 & 11 & 8\\
1 & 3 & 4 & \lambda\\
\lambda & 5 & 7 & 6
\end{vmatrix}
}
=
\frac{0}{2\,(\lambda-2)^2}
=
0
$$
and solving for $y$, $z$, and $t$ is similar.
Now, if $\lambda=2$, then row-reducing shows that
$$
\DeclareMathOperator{rref}{rref}\rref
\begin{bmatrix}
2 & 2 & 3 & 4 & 2 \\
3 & 8 & 11 & 8 & 4 \\
1 & 3 & 4 & 2 & \mu \\
2 & 5 & 7 & 6 & 3
\end{bmatrix}
=
\begin{bmatrix}
1 & 0 & 1 & 8 & 0 \\
0 & 1 & 1 & -2 & 0 \\
0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & 0
\end{bmatrix}
$$
which implies the system $AX=B$ is inconsistent (do you see why?).
Finally, the above also shows that 
$$
\rref A
=
\begin{bmatrix}
1 & 0 & 1 & 8  \\
0 & 1 & 1 & -2  \\
0 & 0 & 0 & 0  \\
0 & 0 & 0 & 0 
\end{bmatrix}
$$
What does this say about the rank of $A$?
A: Gauss-Jordan elimination is one of standard methods for solving linear systems.
$
\left(\begin{array}{cccc|c}
  1 & 2 & 3 & 4 & 2 \\
  3 & 8 &11 & 8 & 4 \\
  1 & 3 & 4 & 2 & \mu \\
  2 & 5 & 7 & 6 & 3
\end{array}\right)\sim
\left(\begin{array}{cccc|c}
  1 & 2 & 3 & 4 & 2 \\
  2 & 5 & 7 & 6 & 3 \\
  3 & 8 &11 & 8 & 4 \\    
  1 & 3 & 4 & 2 & \mu
\end{array}\right)\sim
\left(\begin{array}{cccc|c}
  1 & 2 & 3 & 4 & 2 \\
  0 & 1 & 1 &-2 &-1 \\
  0 & 2 & 2 &-4 & 0 \\
  1 & 3 & 4 & 2 & \mu
\end{array}\right)\sim
\left(\begin{array}{cccc|c}
  1 & 0 & 1 & 8 & 4 \\
  0 & 1 & 1 &-2 &-1 \\
  0 & 0 & 0 & 0 & 0 \\
  1 & 3 & 4 & 2 & \mu
\end{array}\right)\sim
\left(\begin{array}{cccc|c}
  1 & 0 & 1 & 8 & 4 \\
  0 & 1 & 1 &-2 &-1 \\
  0 & 3 & 3 &-6 & \mu-4 \\
  0 & 0 & 0 & 0 & 0
\end{array}\right)\sim
\left(\begin{array}{cccc|c}
  1 & 0 & 1 & 8 & 4 \\
  0 & 1 & 1 &-2 &-1 \\
  0 & 0 & 0 & 0 & \mu-1 \\
  0 & 0 & 0 & 0 & 0
\end{array}\right)
$
In this case you should discuss what happens depending on the value of $\mu$. The third row corresponds to the equation $0x+0y+0z+0t=\mu-1$, or - to put it simpler - $0=\mu-1$. What can you say about solutions of this system if $\mu-1=0$? What can you say if $\mu-1\ne0$?
You can also check your work in WA like this or this
