Matrix Problem of form Ax=B

Question

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.

Answer 1

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 $A$?

Answer 2

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 $A$, $\text{rk}(A) = 2$.

Answer 3

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