
If the product of two non-null square matrices is a null matrix, show that both of them must be singular matrices.
Answer
497.7k+ views
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.
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.
Recently Updated Pages
Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Trending doubts
Which one is a true fish A Jellyfish B Starfish C Dogfish class 11 biology CBSE

State and prove Bernoullis theorem class 11 physics CBSE

1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

In which part of the body the blood is purified oxygenation class 11 biology CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells
