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Q.
A square matrix $A$ is invertible, if and only if
Determinants
Solution:
A square matrix $A$ is invertible, if and only if $A$ is non-singular matrix, i.e., $|A| \neq 0$.
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 $A B=B A=I$.
Now, $ A B=I$. So, $|A B|=\mid I$ or $|A||B|=1$ (since, $|I|=1,|A B|=|A||B|$ )
This gives $|A| \neq 0$.
Hence, $A$ is non-singular.
Conversely, let $A$ be non-singular. Then, $|A| \neq 0$.
Now, $ A(\operatorname{adj} A)=(\operatorname{adj} A) A=|A| I$
or $ A\left(\frac{1}{|A|} \operatorname{adj} A\right)=\left(\frac{1}{|A|} \operatorname{adj} A\right) A=1$
or $ A B=B A=I$, where $B=\frac{1}{|A|}$ adj $A$
Thus, $A$ is invertible and $A^{-1}=\frac{1}{|A|}$ adj $A$.