Animated Solution for Mathematics - Matrices and Determinants: Let $A=\{X=(x,y,z{)}^T:PX=0\text{ and }{x}^2+y^2+z^2=1\}$, where $P=\left(\begin{array}{ccc}1& 2& 1\\ -2& 3& -4\\ 1& 9& -1\end{array}\right)$ then the set $A$ :

Imagine you are standing in a three-dimensional coordinate system. You have a target set $A$, defined by two distinct, powerful constraints.

The first is a homogeneous system of linear equations, $PX=0$, where $P$ is a $3\times 3$ matrix. The second is the classic equation of a unit sphere, $x^2+y^2+z^2=1$.

To find the set $A$, we are essentially looking for the intersection of these two geometric entities. It is a beautiful dance between linear algebra and spatial geometry.

The Gatekeeper

Before we dive into the algebra, we must ask: what kind of object is $PX=0$? In linear algebra, the determinant of the matrix $P$ acts as a gatekeeper.

If $|P| eq 0$, the system has a unique solution: the trivial solution, $X=(0,0,0{)}^T$. But as we noted, the origin does not satisfy the sphere equation. Thus, if $|P| eq 0$, the set $A$ would be empty.

Let us calculate the determinant of $P=\left(\begin{array}{ccc}1& 2& 1\\ -2& 3& -4\\ 1& 9& -1\end{array}\right)$ to see what we are dealing with.

Expanding along the first row, we calculate:

Simplifying this, we get:

This leads to $|P|=33-12-21=0$. The gatekeeper has spoken! Because the determinant is zero, our system $PX=0$ does not collapse to a single point. It expands into a line passing through the origin.

Now that we know we are dealing with a line, we need its equation. We write out the system:

By subtracting the third equation from the first, we eliminate $x$ and find a relationship between $y$ and $z$: $-7y+2z=0$, or $z=\frac{7}{2}y$.

Substituting this back into the first equation, $x+2y+\frac{7}{2}y=0$, we find $x=-\frac{11}{2}y$. To make our lives easier, let $y=2k$.

Then our parametric coordinates become $x=-11k$, $y=2k$, and $z=7k$. This is the line of solutions.

We are almost there. We need to find where this line pierces the unit sphere. We substitute our parametric expressions into $x^2+y^2+z^2=1$:

This simplifies to $121k^2+4k^2+49k^2=1$, or $174k^2=1$. Solving for $k$, we get:

Because we have two distinct values for $k$, we have two distinct points of intersection. Geometrically, this is exactly what we expect: a line passing through the center of a sphere will always intersect the surface at exactly two diametrically opposite points.

The set $A$ contains exactly two elements. We have successfully navigated the constraints and uncovered the elegant truth hidden within the matrix.