# Given three equations, calculate the value of $a,b$ such that the matrix has no solutions, one solutions, and infinitely many

The original equations were:

$$x+y+z = 1$$ $$y-z=1$$ $$2x+3y+z(a+2)=b+3 \text{ where a,b \in \mathbb{R}}$$ \begin{array}{ccc|c} 1 & 1 & 1 & 1 \\ 0 &1 & -1 & 1 \\ 2 & 3 & a+2 &b+2 \\ \end{array}

Performing a series of row operations, I get down to:

\begin{array}{ccc|c} 1 & 0 & 0 & \frac{-2b}{a+1} \\ 0 &1 & 0 & \frac{a+b+1}{a+1} \\ 0 & 0 & 1 &\frac{b}{a+1} \\ \end{array}

I am asked to find the solutions of $a,b$, such that the matrix has no solutions, one solution, and infinitely many.

A matrix has $no$ solutions if there is a row of $0$ before the partition, and a non-zero number after the partition.

A matrix has one solution (the trivial solution), if all the pivots equal $0$.

A matrix has infintely many solutions if there exists a column without a pivot.

Knowing this, how can I find values of $a,b$? Since all the columns have pivots, can there even be a solution for infinity? or no solutions?

I would consider the row-echelon form of the system rather than the reduced row-echelon form:

$$\begin{array}{ccc|c} 1 & 1 & 1 & 1 \\ 0 &1 & -1 & 1 \\ 0 & 0 & a+1 &b \\ \end{array}$$

Can you choose $a$ and $b$ so that

$$a+1=0\ \mathrm{and}\ b\neq 0\ ? \mathrm{\ \ \ \ (no\ solution)}$$ $$a+1=1\ ?\mathrm{\ \ \ \ (one\ solution)}$$ $$a+1=b=0\ ?\mathrm{\ \ \ \ (infinitely\ many\ solutions)}$$

One solution doesn't mean that all pivots equal zero, it just means that there is a pivot in each column (i.e. every column has a leading $1$).

• How do you know that $a+1=0$ and $b \neq 0$ mean no solution? – K Split X Feb 21 '17 at 16:06
• If $a=-1$ and $b\neq 0$ then our system looks like this: $$\begin{array}{ccc|c} 1 & 1 & 1 & 1 \\ 0 &1 & -1 & 1 \\ 0 & 0 & 0 &b \\ \end{array}$$ – yung_Pabs Feb 21 '17 at 16:20
• Yes sorry, I got confused for a second since it was in REF , thanks for the answer – K Split X Feb 21 '17 at 17:51