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# If the product of two non-null square matrices is a null matrix, show that both of them must be singular matrices.

Last updated date: 13th Jun 2024
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Hint:
Here we use the properties of non-null matrices, null matrices, singular matrices, and non-singular matrices. In the given question the product of two non-null square matrices is a null matrix. In the solution part we assume one of the matrices as a nonsingular matrix and calculate the property of another matrix.

Complete step by step solution:
Here we know from the problem that the product of two non-null square matrices is a null matrix.
Now. We suppose A and B be two non-null matrices of the same order matrix $n \times n$.
Here we write the product of two matrices is a null matrix.
${\rm{AB}} = 0$
Now, we assume the matrix A is a non-singular matrix then the inverse of matrix A exists that is ${{\rm{A}}^{ - 1}}$. Multiply ${{\rm{A}}^{ - 1}}$ in the expression ${\rm{AB}} = 0$.
${{\rm{A}}^{ - 1}}\left( {{\rm{AB}}} \right) = {{\rm{A}}^{ - 1}}0$
We substitute the expressions ${\rm{AB}} = 0$ and ${{\rm{A}}^{ - 1}}{\rm{A}} = I$ (where $I$ is the identity matrix) in the expression ${{\rm{A}}^{ - 1}}\left( {{\rm{AB}}} \right) = {{\rm{A}}^{ - 1}}0$.
$\Rightarrow {{\rm{A}}^{ - 1}}\left( {{\rm{AB}}} \right) = {{\rm{A}}^{ - 1}}0\\ \Rightarrow {{\rm{A}}^{ - 1}}{\rm{A}}\left( {\rm{B}} \right) = 0\\ \Rightarrow {\rm{IB}} = 0\\ \Rightarrow {\rm{B}} = 0$
Hence, matrix B is the null matrix. This means matrix A is a singular matrix.

Now, we assume B is a non-singular matrix then the inverse of matrix B exists that is ${{\rm{B}}^{ - 1}}$. Multiply ${{\rm{B}}^{ - 1}}$ in the expression ${\rm{AB}} = 0$.
$\left( {{\rm{AB}}} \right){{\rm{B}}^{ - 1}} = {{\rm{B}}^{ - 1}}0$
We substitute the expressions ${\rm{AB}} = 0$ and ${\rm{B}}{{\rm{B}}^{ - 1}} = I$ (where I is the identity matrix) in the expression $\left( {{\rm{AB}}} \right){{\rm{B}}^{ - 1}} = {{\rm{B}}^{ - 1}}0$.
$\Rightarrow \left( {{\rm{AB}}} \right){{\rm{B}}^{ - 1}} = {{\rm{B}}^{ - 1}}0\\ \Rightarrow {\rm{A}}\left( {{\rm{B}}{{\rm{B}}^{ - 1}}} \right) = 0\\ \Rightarrow {\rm{AI}} = 0\\ \Rightarrow {\rm{A}} = 0$

Hence, matrix A is the null matrix. This means matrix B is a singular matrix.
And, the conclusion is that both of them must be singular matrices.

Note:
We can solve this problem with determinant methods. The determinant for the product ${\rm{AB}} = 0$ , use the property to show the zero or null matrix. Here, $\left| {\rm{A}} \right| = 0$ or $\left| {\rm{B}} \right| = 0$ or both. But the determinant method may not give an absolute solution.