Matrix problem: No inverse exists

Solution 1:

$$A+B+C=3 \tag{1}$$ $$2B+C=2 \tag{2}$$ $$2A+C=4\tag{3}$$

As @K.defaoite pointed out, if you add equations (2) and (3) together, you get $$2A+2B+2C=6\tag{4}$$ $$A+B+C=3\tag{5}$$

So the third equation is the first equation in disguise. The system of equations is underdetermined - you have 3 variables but only two equations to solve. As you correctly point out, these means that the determinant is zero and there is no unique solution (there are in fact an infinite number of solutions).

Solution 2:

When the determinant is zero it means the matrix either has infinitely many solutions or no solutions. To check if it has infinitely many solutions or no solutions RREF can be used.

$$ \left[ \begin{array}{ccc|c} 1&1&1&3\\ 2&0&1&4\\ 0&2&1&2 \end{array} \right] $$

Add -2 times row1 to row2 and replace row2.

$$ \left[ \begin{array}{ccc|c} 1&1&1&3\\ 0&-2&-1&-2\\ 0&2&1&2 \end{array} \right] $$

Divide row2 by -2 and replace row2.

$$ \left[ \begin{array}{ccc|c} 1&1&1&3\\ 0&1&1/2&1\\ 0&2&1&2 \end{array} \right] $$

Add -1 times row2 to row1 and replace row1. Add -2 times row2 to row3 and replace row3.

$$ \left[ \begin{array}{ccc|c} 1&0&1/2&2\\ 0&1&1/2&1\\ 0&0&0&0 \end{array} \right] $$

The left and right of the last row of augmented matrix is all zeros. Then it will have infinitely many solutions. If left has all zeros and right had a value then it won't have any solution.

A+1/2C=2

B+1/2C=1

for any value of C, A and B can be computed.

eg. if C=2 then A=1, B=0