Question

A square matrix $A$ is invertible, if and only if

• A. Ais singular matrix i.e., $|A|\ne 0$
• B. Ais non-singular matrix i.e., $|A|\ne 0$
• C. A is non-singular matrix i.e., $|A|=0$
• D. A is singular matrix i.e., $|A|=0$

Answer 1

Answer: (B)

A square matrix $A$ is invertible, if and only if $A$ is non-singular matrix, i.e., $|A|\ne 0$.
Justification Let $A$ be invertible matrix of order $n$ and $/$ be the identity matrix of order $n$.
Then, there exists a square matrix $B$ of order $n$ such that $AB=BA=I$.
Now, $AB=I$. So, $|AB|=\mid I$ or $|A||B|=1$ (since, $|I|=1,|AB|=|A||B|$ )
This gives $|A|\ne 0$.
Hence, $A$ is non-singular.
Conversely, let $A$ be non-singular. Then, $|A|\ne 0$.
Now, $A(\mathrm{adj}A)=(\mathrm{adj}A)A=|A|I$
or $A\left(\frac{1}{|A|}\mathrm{adj}A\right)=\left(\frac{1}{|A|}\mathrm{adj}A\right)A=1$
or $AB=BA=I$, where $B=\frac{1}{|A|}$ adj $A$
Thus, $A$ is invertible and $A^{-1}=\frac{1}{|A|}$ adj $A$.