System with a parameter

Solve the system with augmented matrix

$$\left(\begin{array}{cccc|c} a&1&1&1&1\\ 1&a&1&1&1\\ 1&1&a&1&1\\ 1&1&1&a&1\\ \end{array}\right)$$

depending on the parameter $$a$$.

• Hint

Use the eliminatin, but avoid division by zero. This might appear for some choices of $$a$$.

• Resolution

We first switch the first and the last rows, and then reduce the first column. In the next reduction we add the second and the third row to the last one.

$$\left(\begin{array}{cccc|c} a&1&1&1&1\\ 1 & a & 1 & 1 & 1\\ 1 & 1 & a & 1 & 1\\ 1 & 1 & 1 & a & 1\\ \end{array}\right)\sim \left(\begin{array}{cccc|c} 1 & 1 & 1 & a & 1\\ 0 & a-1 & 0 & 1-a & 0\\ 0 & 0 & a-1 & 1-a & 0\\ 0 & 1-a & 1-a & 1-a^2 & 1-a\\ \end{array}\right)\sim\\ \left(\begin{array}{cccc|c} 1 & a & 1 & 1 & 1\\ 0 & a-1 & 0 & 1-a & 0\\ 0 & 0 & a-1 & 1-a & 0\\ 0 & 0 & 0 & (a+3)(1-a) & 1-a\\ \end{array}\right)$$

Now it suffices to ditinguish cases $$a=-3$$ and $$a=1$$.

• Result

When $$a=1$$ the system has solution set $$\{(1-p-q-r,p,q,r)^T\mid p,q,r\in\mathbb R\}$$.

For $$a=-3$$ the system has no solution.

Otherwise the solution is $$(\frac{1}{a+3},\frac{1}{a+3},\frac{1}{a+3},\frac{1}{a+3})^T$$.