You can just solve it like a system of equations (Hint: use elimination), or you can turn it into a matrix and solve it like this:

**Step one:**
$$ \left[
\begin{array}{cc|c}
1&-3&b_{1}\\
3&1&b_{2}\\
1&7&b_{3}\\
2&4&b_{4}
\end{array}
\right] $$

Leave R1 intact.
Replace R2 with: R2 - 3(R1).
Replace R3 with: R3 - R1.
Replace R4 with: R4 - 2(R1).
$$ \left[
\begin{array}{cc|c}
1&-3&b_{1}\\
0&10&b_{2} - 3b_{1}\\
0&10&b_{3} - b_{1}\\
0&10&b_{4} - 2b_{1}
\end{array}
\right] $$

**Step two:**
Leave R1 intact.
Leave R2 intact.
Replace R3 with: R3 - R2.
Replace R4 with: R4 - R2.
$$ \left[
\begin{array}{cc|c}
1&-3&b_{1}\\
0&10&b_{2} - 3b_{1}\\
0&0&b_{3} - b_{1} - (b_{2} - 3b_{1})\\
0&0&b_{4} - 2b_{1} - (b_{2} - 3b_{1})
\end{array}
\right] $$

**Step three (answer):** Now that you've got an echelon form of the matrix, you can figure out what your solutions might be. Since the question wants you to find out what must be true of the b's so that you can have at least one solution, and the same rules of math apply, your b's must be something that makes the matrix true. Your equations are:
$$
\begin{matrix}
x & -3y & = b_{1} \\
0x & +10y & = b_{2} - 3b_{1} \\
0x & +0y & = b_{3} - b_{1} - (b_{2} - 3b_{1})\\
0x & +0y & = b_{4} - 2b_{1} - (b_{2} - 3b_{1}) \\
\end{matrix}
$$
Realize that zero must equal zero! Since 0x = 0 and 0y = 0, the right hand side of those last two equations has to equal zero.

**Step four (answer):** Set the right hand side of the last two equations equal to zero and solve. What you get (left hand side in terms of one thing, right hand side in terms of another) is the answer the book gave you.

0 = b3 - b1 - (b2 - 3b1)

0 = b4 + b1 - b2

Gaussâ€™s Method shows that this system is consistent if and only if both
b3 = -2b1 + b2 and b4 = -b1 + b2.

Both are false. For example, consider $m = 3$ and $n = 4$.

For part A, consider:
$$
A = \begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 \\
\end{bmatrix} \qquad\text{and}\qquad
b = \begin{bmatrix}
0 \\ 0 \\ 7
\end{bmatrix}
$$
Since the third row of $A$ is all zeroes, any linear combinations of its columns will give us a $0$ in the third entry (and not $7$).

For part B, use the same matrix $A$ from part A and observe that (besides the trivial solution), another solution to the homogeneous system is:
$$
x = \begin{bmatrix}
0 \\ 0 \\ 8 \\ 9
\end{bmatrix}
$$

## Best Answer

If you want to solve a linear system $A(x)=b$, where $A$ is an $n\times n$ matrix. The fact that $A$ is non singular means that $A$ is invertible, thus $A^{-1}(b)$ is a solution of the equation.

When $A$ is not invertible, you need first to determine the space of homogeneous solutions which are the solution of $A(x)=0$. Denote by $Ker(A)$ this space of homogeneous solutions, and let $x_0$ be a particular solution, $A(x_0)=b$, the solutions are $x_0+a,a\in ker(A)$.