Question

Let A be a square matrix all of whose entries are integers. Then, which one of the following is true?
A. If $\det A=\pm 1$, then ${{A}^{-1}}$ exists and all its entries are integers.
B. If $\det A=\pm 1$, then ${{A}^{-1}}$ need not exist.
C. If $\det A=\pm 1$, then ${{A}^{-1}}$ exists but all its entries are not necessarily integers.
D. if $\det A\ne \pm 1$, then ${{A}^{-1}}$ exists and all its entries are non-integers.

Answer 1

In this problem we need to find the correct option which is related to the given data. In the problem we have given that $A$ be a square matrix all of whose entries are integers. So, we will assume a square matrix of desired order and try to find the inverse of the matrix by calculating the values of $adj\left( A \right)$, $\left| A \right|$. From the values of ${{A}^{-1}}$, $\left| A \right|$ we will choose one correct option from the given options.

Complete step-by-step solution:
Given that the matrix $A$ is a square matrix all of whose entries are integers.
Let us assume the matrix $A$ as $A=\left[ \begin{matrix} -1 & 0 \\\ 0 & -1 \\\ \end{matrix} \right]$
Now the determinant of the matrix $A$ will be
$\Rightarrow \left| A \right|=\left| \begin{matrix} -1 & 0 \\\ 0 & -1 \\\ \end{matrix} \right|$
We know that the value of $\left| \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right|=ad-cb$, hence the determinant of the matrix $A$ will be
$\begin{aligned} & \Rightarrow \left| A \right|=-1\left( -1 \right)-0\left( 0 \right) \\\ & \Rightarrow \left| A \right|=1 \\\ \end{aligned}$
Here the value of $\left| A \right|$ is $1$. Now the value of $adj\left( A \right)$ will be
$\begin{aligned} & \Rightarrow adj\left( A \right)=\left[ \begin{matrix} -1 & -\left( 0 \right) \\\ -\left( 0 \right) & -1 \\\ \end{matrix} \right] \\\ & \Rightarrow adj\left( A \right)=\left[ \begin{matrix} -1 & 0 \\\ 0 & -1 \\\ \end{matrix} \right] \\\ \end{aligned}$
From the above values, the value of ${{A}^{-1}}$ will be
$\begin{aligned} & \Rightarrow {{A}^{-1}}=\dfrac{1}{\left| A \right|}adj\left( A \right) \\\ & \Rightarrow {{A}^{-1}}=\dfrac{1}{1}\left[ \begin{matrix} -1 & 0 \\\ 0 & -1 \\\ \end{matrix} \right] \\\ & \Rightarrow {{A}^{-1}}=\left[ \begin{matrix} -1 & 0 \\\ 0 & -1 \\\ \end{matrix} \right] \\\ \end{aligned}$
From the above value we can say that ‘If $\det A=\pm 1$, then ${{A}^{-1}}$ exists and all its entries are integers’.
Hence option – A is the correct one.

Note: For this problem we can directly write the answer without assuming the matrix because we have the formula for the inverse matrix as ${{A}^{-1}}=\dfrac{1}{\left| A \right|}adj\left( A \right)$. From this formula we can say that the inverse of an integer matrix should have integers entries.