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

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

Explanation

Solution

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×nn \times n.
Here we write the product of two matrices is a null matrix.
AB=0{\rm{AB}} = 0
Now, we assume the matrix A is a non-singular matrix then the inverse of matrix A exists that is A−1{{\rm{A}}^{ - 1}}. Multiply A−1{{\rm{A}}^{ - 1}} in the expression AB=0{\rm{AB}} = 0.
A−1(AB)=A−10{{\rm{A}}^{ - 1}}\left( {{\rm{AB}}} \right) = {{\rm{A}}^{ - 1}}0
We substitute the expressions AB=0{\rm{AB}} = 0 and A−1A=I{{\rm{A}}^{ - 1}}{\rm{A}} = I (where II is the identity matrix) in the expression A−1(AB)=A−10{{\rm{A}}^{ - 1}}\left( {{\rm{AB}}} \right) = {{\rm{A}}^{ - 1}}0.
⇒A−1(AB)=A−10 ⇒A−1A(B)=0 ⇒IB=0 ⇒B=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 B−1{{\rm{B}}^{ - 1}}. Multiply B−1{{\rm{B}}^{ - 1}} in the expression AB=0{\rm{AB}} = 0.
(AB)B−1=B−10\left( {{\rm{AB}}} \right){{\rm{B}}^{ - 1}} = {{\rm{B}}^{ - 1}}0
We substitute the expressions AB=0{\rm{AB}} = 0 and BB−1=I{\rm{B}}{{\rm{B}}^{ - 1}} = I (where I is the identity matrix) in the expression (AB)B−1=B−10\left( {{\rm{AB}}} \right){{\rm{B}}^{ - 1}} = {{\rm{B}}^{ - 1}}0.
⇒(AB)B−1=B−10 ⇒A(BB−1)=0 ⇒AI=0 ⇒A=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 AB=0{\rm{AB}} = 0 , use the property to show the zero or null matrix. Here, ∣A∣=0\left| {\rm{A}} \right| = 0 or ∣B∣=0\left| {\rm{B}} \right| = 0 or both. But the determinant method may not give an absolute solution.