Animated Solution for Mathematics - Matrices and Determinants: Let $A$ and $B$ be two $3\times 3$ non-zero real matrices such that $AB$ is a zero matrix. Then

We are given two $3\times 3$ non-zero real matrices, $A$ and $B$, such that their product $AB$ is the zero matrix $O$.

At first glance, one might assume that at least one of the matrices must be the zero matrix. However, in the elegant and counter-intuitive world of linear algebra, this is not necessarily true.

To understand the nature of these matrices, we utilize the property of determinants. We start with the given equation:

Taking the determinant of both sides, we have $|AB|=|O|$. Since the determinant of a zero matrix is $0$, we obtain:

Invoking the multiplicative property of determinants, $|AB|=|A||B|$, we arrive at the crucial realization:

This equation implies that at least one of the determinants must be zero.

Suppose, for the sake of argument, that $|A| eq 0$. If this were true, matrix $A$ would be non-singular and possess an inverse, $A^{-1}$.

We test this assumption by pre-multiplying the original equation $AB=O$ by $A^{-1}$:

Using the associative property of matrix multiplication, we get $(A^{-1}A)B=O$. Since $A^{-1}A=I$ (the identity matrix), this simplifies to:

However, the problem explicitly states that $B$ is a non-zero matrix. This creates a contradiction, forcing us to conclude that $|A|=0$. Thus, matrix $A$ must be a singular matrix.

With $|A|=0$ established, we examine the homogeneous system of linear equations $AX=0$.

In linear systems, the determinant acts as a gatekeeper. When $|A|=0$, the transformation $A$ collapses space, projecting 3D space onto a lower-dimensional subspace such as a plane or a line.

Because the transformation is not injective, there exists a non-zero null space (or kernel). Any vector $X$ in this null space satisfies $AX=0$. Since the null space contains infinitely many vectors, the system $AX=0$ must have infinitely many solutions.

We have proven that $|A|=0$, which directly implies that the system $AX=0$ has infinitely many solutions.

As a further observation, consider the adjoint matrix, $\text{adj}(A)$. We know the identity:

For a $3\times 3$ matrix, this becomes:

Since $|A|=0$, the determinant of the adjoint is also zero, meaning $\text{adj}(A)$ is not invertible. The beauty of this problem lies in how it forces us to abandon the simple rules of scalar arithmetic and embrace the structural properties of linear transformations.