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Let
$$A=beginpmatrix 1 & 5 & -1\ -2 & -10 & 5\ -2 & -10 & -1endpmatrix$$

and
$$B=beginpmatrix 0 \ 15 \ -15endpmatrix.$$

To find a vector $U$ such that $A$ maps $U$ to $B$ then you must solve the system $AU = B$.

Isolating for U yields $U = A^-1B$

However, when attempting to solve the inverse of $A$ produces all $0$'s in the bottom row during matrix inversion algorithm, otherwise known as a determinant of $0$.

After Gaussian elimination we get a system in echelon form,

$$beginpmatrix1&5&-1\0&0&3\0&0&0endpmatrixbeginpmatrixx\y\zendpmatrix=beginpmatrix0\15\0endpmatrix.$$

So by the second equation, $z=5$ and by the first $x+5y=5$. This is all you can say.

Yes, it is true that $det(A)=0$. So $A^-1$ is meaningless. Nevertheless we can solve $Ax=b$ with $b=(0,15,-15)^T$ by, say, $x=(1,4/5,5)$. For this, we write $x=(x_1,x_2,x_3)$ and obtain three linear equations, namely
$$
x_1+5x_2-x_3= - 2x_1 - 10x_2 + 5x_3 - 15= - 2x_1 - 10x_2 - x_3 + 15=0.
$$
If we substitute $x_3$ from the last equation, we obtain $- x_1 - 5x_2 + 5=0$. With $x_1=5-5x_2$ all equations are satisfied.

The most general solution of a square linear system $AX=B$ can be written

$$X=X_0+X_1$$

where :

• 1) $X_0$ is a particular solution of system $AX=B$.




• 2) $X_1$ is the general solution of the associated homogeneous system $AX=0$ (in more abstract terms : the generic element of the kernel of $A$).

In the case at hand,

1) has solution (see explanations below):
$$X_0=beginpmatrix5\0\5endpmatrix.$$

2) has solution
$$X_1=kbeginpmatrix-5\1\0endpmatrix textfor any real k.$$

Thus the general solution is

$$X_0+X_1=beginpmatrix5-5k\k\5endpmatrix textfor any real k.$$

Remark : this formulation "particular solution of... + general solution of ..." is the same for linear differential equations.

Explanation for (1) : There are in fact many solutions ; here is a way to obtain one :

In system

$$begincasesx&+&5y&-&z&=&0& (a)\-2x&-&10y&+&5z&=&15& (b)\-2x&-&10y&-&z&=&-15& (c)endcases$$

Substracting (c) from (b), one gets $z=5$. Plugging this value in (a),(b) or (c) gives the same equation $x+5y=5$. It suffices then to take $x=5$ and $y=0$.

Explanation for (2) :

In homogeneous system

$$begincasesx&+&5y&-&z&=&0& (a)\-2x&-&10y&+&5z&=&0& (b)\-2x&-&10y&-&z&=&0& (c)endcases$$

Substracting (c) from (b), one gets this time $z=0$. Plugging this value in (a),(b) or (c) gives the same equation $x+5y=0$. It suffices then to take $x=-5y$.

Thus the general solution of the homogeneous equation is :

$$beginpmatrixx\y\zendpmatrix=beginpmatrix-5y\y\0endpmatrix=ybeginpmatrix-5\1\0endpmatrix.$$

Please note that, in fact, arbitrary constant $k$ is ordinate $y$.