Solution of an overdetermined system of linear equations

Question

According to my textbook "Matrix Operations for Engineers and Scientists - An Essential Guide in Linear Algebra" by the late Alan Jeffrey the following system of equations

is impossible. To quote the author: System (a) has no solution. This can be shown in more than one way. The most elementary way being to solve the first three equations for $x_1$, $x_2$ and $x_3$, and then to substitute these values into the last equation to show that they do not satisfy it. Thus the last equation contradicts the other three, so there can be no solution set.

Nevertheless, according to my understanding of the Kronecker-Capelli theorem the system under question has a unique solution. The rank of the matrix of the coefficients of unknowns $A=\begin{bmatrix}1&-2&2\\1&1&-1\\1&3&-3\\1&1&1\end{bmatrix}$ is 3. The rank of the augmented matrix of the system $\begin{bmatrix}1&-2&2&6\\1&1&-1&0\\1&3&-3&-4\\1&1&1&3\end{bmatrix}$ is also 3. Finally the number of unknowns is 3 as well. The reduced row echelon form of the matrix that I found is $\begin{bmatrix}1&0&0&2\\0&1&0&-0.5\\0&0&1&1.5\\0&0&0&0\end{bmatrix}$. Thus, according to my understanding $$\begin{matrix}x&=&2\\y&=&-0.5\\y&=&1.5\end{matrix}$$ The following statements Matlab verifies my findings.

>> A = [1 -2 2; 1 1 -1; 1 3 -3; 1 1 1]; b = [6; 0; -4; 3];
>> rank(A), rank([A b])
ans =
     3
ans =
     3
>> A\b
ans =
    2.0000
   -0.5000
    1.5000
>> rref([A b])
ans =
    1.0000         0         0    2.0000
         0    1.0000         0   -0.5000
         0         0    1.0000    1.5000
         0         0         0         0

What am I missing here?

Answer 1

Let's see if you have reduced correctly. You start with augmented matrix $\begin{bmatrix} 1 & -2 & 2 & 6 \\ 1 & 1 & -1 & 0 \\ 1 & 3 & -3 & -4 \\ 1 & 1 & 1 & 3 \end{bmatrix}$.

Subtract the first row from each of the other rows to get $\begin{bmatrix}1 & -2 & 2 & 6 \\ 0 & 3 & -3 & -6 \\ 0 & 5 & -5 & -10 \\ 0 & 3 & -1 & -3 \end{bmatrix}$.

Divide the second row by 3: $\begin{bmatrix}1 & -2 & 2 & 6 \\ 0 & 1 & -1 & -2 \\ 0 & 5 & -5 & -10 \\ 0 & 3 & -1 & -3 \end{bmatrix}$.

Subtract 5 times the second row from the third row and three times the second row from the fourth row: $\begin{bmatrix}1 & -2 & 2 & 6 \\ 0 & 1 & -1 & -2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 2 & 3 \end{bmatrix}$.

Swap the third and fourth rows: $\begin{bmatrix}1 & -2 & 2 & 6 \\ 0 & 1 & -1 & -2 \\ 0 & 0 & 2 & 3 \\ 0 & 0 & 0 & 0 \end{bmatrix}$.

The third row is equivalent to 2z= 3 so, yes, z= 3/2. Then the second row is equivalent to y- z= y- 3/2= -2 so y= 3/2- 2= -1/2. An the first row is equivalent to x- 2y+ 2z= x+ 1+ 3= x+ 4= 6 so x= 2.

Yes, I get exactly what you did!