# If $A$ is unimodular, then which of the following matrices are unimodular?

(a) $-A$

(b) ${{A}^{-1}}$

(c) $adj\left( A \right)$

(d) $\omega A$, where $\omega $ is cube root of unity

Last updated date: 29th Mar 2023

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Hint: A matrix $A$ is said to be unimodular if the determinant of this matrix $\left| A \right|$ is equal to +1 or -1. Check each option and use the properties of matrices and determinants to find the determinant of the matrix given in each option.

Before proceeding with the question, we must know the formulas and the definitions that will be required to solve this question.

For a matrix $A$, if the determinant of this matrix $\left| A \right|$ is equal to +1 or -1, then it is said to be unimodular. In the question, it is given that$A$ is unimodular. This means that,

$\left| A \right|=1$ or $\left| A \right|=-1...........\left( 1 \right)$

In matrices, for a matrix $A$ of any order n, we have some formulas and properties,

$\left| -kA \right|={{\left( -k \right)}^{n}}\left| A \right|..................\left( 2 \right)$, here k is a constant.

$\begin{align}

& {{A}^{-1}}=\dfrac{1}{\left| A \right|}adj\left( A \right) \\

& \Rightarrow \left| {{A}^{-1}} \right|=\left| \dfrac{1}{\left| A \right|}adj\left( A \right) \right| \\

& \Rightarrow \left| {{A}^{-1}} \right|=\left| \dfrac{1}{\left| A \right|} \right|\left| adj\left( A \right) \right| \\

& \Rightarrow \left| {{A}^{-1}} \right|=\dfrac{1}{\left| A \right|}\left| adj\left( A \right) \right|..................\left( 3 \right) \\

\end{align}$

If we take the adjoint of a matrix, the determinant of the adjoint of that matrix is equal to the determinant of that matrix.

$\Rightarrow \left| adj\left( A \right) \right|=\left| A \right|................\left( 4 \right)$

In this question, we are given a matrix $A$ which is a unimodular matrix. From definition $\left( 1 \right)$, we can say that,

$\left| A \right|=+1$ or $\left| A \right|=-1$

Now, we are required to check which of the options contains a unimodular matrix.

(a) To check whether $-A$ is unimodular or not, let us find it’s determinant. The determinant of $-A$ is $\left| -A \right|$. Substituting k=-1 in equation $\left( 2 \right)$, we get,

$\left| -A \right|={{\left( -1 \right)}^{n}}\left| A \right|$ , where n is the order of matrix $A$

Since n is the order of the matrix, it is an integer. Also, we have found that $\left| A \right|=+1$ or $\left| A \right|=-1$. So, we can say that,

$\left| -A \right|=+1$ or $\left| -A \right|=-1$

Hence, $-A$ is a unimodular matrix.

(b) Let us find the determinant of ${{A}^{-1}}$ i.e. $\left| {{A}^{-1}} \right|$. From formula $\left( 3 \right)$, we have,

$\left| {{A}^{-1}} \right|=\dfrac{1}{\left| A \right|}\left| adj\left( A \right) \right|$

We have found that $\left| A \right|=+1$ or $\left| A \right|=-1$. Also, from formula $\left( 4 \right)$, we have $\left| adj\left( A \right) \right|=\left| A \right|$. So, we can say that $\left| adj\left( A \right) \right|=+1$ or $\left| adj\left( A \right) \right|=-1$. Substituting these values of $\left| adj\left( A \right) \right|$ and $\left| A \right|$, we get,

$\left| {{A}^{-1}} \right|=+1$ or $\left| {{A}^{-1}} \right|=-1$

Hence, ${{A}^{-1}}$ is a unimodular matrix.

(c) Let us find the determinant of $adj\left( A \right)$ i.e. $\left| adj\left( A \right) \right|$. From formula $\left( 4 \right)$, we have $\left| adj\left( A \right) \right|=\left| A \right|$. So, we can say that $\left| adj\left( A \right) \right|=+1$ or $\left| adj\left( A \right) \right|=-1$.

Hence, $adj\left( A \right)$ is a unimodular matrix.

(d) Let us find the determinant of $\omega A$. The determinant of $\omega A$ is $\left| \omega A \right|$. Substituting $k=\omega $ in equation $\left( 2 \right)$, we get,

$\left| \omega A \right|={{\left( \omega \right)}^{n}}\left| A \right|$ , where n is the order of matrix $A$

We have found that $\left| A \right|=+1$ or $\left| A \right|=-1$.

In complex numbers, we have ${{\omega }^{n}}=1$ only when n is a multiple of 3. So, we can say that \[\left| \omega A \right|=1\] or \[\left| \omega A \right|=-1\] only when n is a multiple of 3.

Hence, \[\omega A\] is unimodular only if n is a multiple of 3.

Hence, the answer is (a), (b), (c).

Note: There is a possibility that one may include option (d) as the correct answer. In the question, it is asked that “which of the following is unimodular?”. Since the matrix in option (d) is not always unimodular, we will not include option (d) in our answer.

Before proceeding with the question, we must know the formulas and the definitions that will be required to solve this question.

For a matrix $A$, if the determinant of this matrix $\left| A \right|$ is equal to +1 or -1, then it is said to be unimodular. In the question, it is given that$A$ is unimodular. This means that,

$\left| A \right|=1$ or $\left| A \right|=-1...........\left( 1 \right)$

In matrices, for a matrix $A$ of any order n, we have some formulas and properties,

$\left| -kA \right|={{\left( -k \right)}^{n}}\left| A \right|..................\left( 2 \right)$, here k is a constant.

$\begin{align}

& {{A}^{-1}}=\dfrac{1}{\left| A \right|}adj\left( A \right) \\

& \Rightarrow \left| {{A}^{-1}} \right|=\left| \dfrac{1}{\left| A \right|}adj\left( A \right) \right| \\

& \Rightarrow \left| {{A}^{-1}} \right|=\left| \dfrac{1}{\left| A \right|} \right|\left| adj\left( A \right) \right| \\

& \Rightarrow \left| {{A}^{-1}} \right|=\dfrac{1}{\left| A \right|}\left| adj\left( A \right) \right|..................\left( 3 \right) \\

\end{align}$

If we take the adjoint of a matrix, the determinant of the adjoint of that matrix is equal to the determinant of that matrix.

$\Rightarrow \left| adj\left( A \right) \right|=\left| A \right|................\left( 4 \right)$

In this question, we are given a matrix $A$ which is a unimodular matrix. From definition $\left( 1 \right)$, we can say that,

$\left| A \right|=+1$ or $\left| A \right|=-1$

Now, we are required to check which of the options contains a unimodular matrix.

(a) To check whether $-A$ is unimodular or not, let us find it’s determinant. The determinant of $-A$ is $\left| -A \right|$. Substituting k=-1 in equation $\left( 2 \right)$, we get,

$\left| -A \right|={{\left( -1 \right)}^{n}}\left| A \right|$ , where n is the order of matrix $A$

Since n is the order of the matrix, it is an integer. Also, we have found that $\left| A \right|=+1$ or $\left| A \right|=-1$. So, we can say that,

$\left| -A \right|=+1$ or $\left| -A \right|=-1$

Hence, $-A$ is a unimodular matrix.

(b) Let us find the determinant of ${{A}^{-1}}$ i.e. $\left| {{A}^{-1}} \right|$. From formula $\left( 3 \right)$, we have,

$\left| {{A}^{-1}} \right|=\dfrac{1}{\left| A \right|}\left| adj\left( A \right) \right|$

We have found that $\left| A \right|=+1$ or $\left| A \right|=-1$. Also, from formula $\left( 4 \right)$, we have $\left| adj\left( A \right) \right|=\left| A \right|$. So, we can say that $\left| adj\left( A \right) \right|=+1$ or $\left| adj\left( A \right) \right|=-1$. Substituting these values of $\left| adj\left( A \right) \right|$ and $\left| A \right|$, we get,

$\left| {{A}^{-1}} \right|=+1$ or $\left| {{A}^{-1}} \right|=-1$

Hence, ${{A}^{-1}}$ is a unimodular matrix.

(c) Let us find the determinant of $adj\left( A \right)$ i.e. $\left| adj\left( A \right) \right|$. From formula $\left( 4 \right)$, we have $\left| adj\left( A \right) \right|=\left| A \right|$. So, we can say that $\left| adj\left( A \right) \right|=+1$ or $\left| adj\left( A \right) \right|=-1$.

Hence, $adj\left( A \right)$ is a unimodular matrix.

(d) Let us find the determinant of $\omega A$. The determinant of $\omega A$ is $\left| \omega A \right|$. Substituting $k=\omega $ in equation $\left( 2 \right)$, we get,

$\left| \omega A \right|={{\left( \omega \right)}^{n}}\left| A \right|$ , where n is the order of matrix $A$

We have found that $\left| A \right|=+1$ or $\left| A \right|=-1$.

In complex numbers, we have ${{\omega }^{n}}=1$ only when n is a multiple of 3. So, we can say that \[\left| \omega A \right|=1\] or \[\left| \omega A \right|=-1\] only when n is a multiple of 3.

Hence, \[\omega A\] is unimodular only if n is a multiple of 3.

Hence, the answer is (a), (b), (c).

Note: There is a possibility that one may include option (d) as the correct answer. In the question, it is asked that “which of the following is unimodular?”. Since the matrix in option (d) is not always unimodular, we will not include option (d) in our answer.

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