Question

Evaluate the determinant to the closest integer: $A=\left[ \begin{matrix} {{\log }_{3}}512 & {{\log }_{4}}3 \\\ {{\log }_{3}}8 & {{\log }_{4}}9 \\\ \end{matrix} \right]$.

Answer 1

We start solving the problem by recalling the determinant of the matrix $\left[ \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right]$ as $\left| \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right|=\left( a\times d \right)-\left( b\times c \right)$. We use this definition for the matrix and make use of the result ${{\log }_{{{a}^{m}}}}{{b}^{n}}=\dfrac{n}{m}{{\log }_{a}}b$ to proceed through the problem. We then make use of the result ${{\log }_{a}}b\times {{\log }_{b}}a=1$ and make necessary calculations to find the required value of the determinant.

Complete step by step answer:
According to the problem, we need to find the determinant of the given matrix $A=\left[ \begin{matrix} {{\log }_{3}}512 & {{\log }_{4}}3 \\\ {{\log }_{3}}8 & {{\log }_{4}}9 \\\ \end{matrix} \right]$ and find its closest integer.
We know that the determinant of the matrix $\left[ \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right]$ is defined as $\left| \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right|=\left( a\times d \right)-\left( b\times c \right)$. Let us use this definition for finding the determinant of matrix A.
So, we have $\left| A \right|=\left| \begin{matrix} {{\log }_{3}}512 & {{\log }_{4}}3 \\\ {{\log }_{3}}8 & {{\log }_{4}}9 \\\ \end{matrix} \right|$.
$\Rightarrow \left| A \right|=\left( {{\log }_{3}}512\times {{\log }_{4}}9 \right)-\left( {{\log }_{3}}8\times {{\log }_{4}}3 \right)$.
$\Rightarrow \left| A \right|=\left( {{\log }_{3}}{{2}^{9}}\times {{\log }_{{{2}^{2}}}}{{3}^{2}} \right)-\left( {{\log }_{3}}{{2}^{3}}\times {{\log }_{{{2}^{2}}}}3 \right)$ ---(1).
We know that ${{\log }_{{{a}^{m}}}}{{b}^{n}}=\dfrac{n}{m}{{\log }_{a}}b$. We use this result in equation (1).
$\Rightarrow \left| A \right|=\left( \left( \dfrac{9}{1}{{\log }_{3}}2 \right)\times \left( \dfrac{2}{2}{{\log }_{2}}3 \right) \right)-\left( \left( \dfrac{3}{1}{{\log }_{3}}2 \right)\times \left( \dfrac{1}{2}{{\log }_{2}}3 \right) \right)$.
$\Rightarrow \left| A \right|=\left( \left( 9{{\log }_{3}}2 \right)\times \left( {{\log }_{2}}3 \right) \right)-\left( \left( 3{{\log }_{3}}2 \right)\times \left( \dfrac{1}{2}{{\log }_{2}}3 \right) \right)$.
$\Rightarrow \left| A \right|=9\left( {{\log }_{3}}2\times {{\log }_{2}}3 \right)-\dfrac{3}{2}\left( {{\log }_{3}}2\times {{\log }_{2}}3 \right)$ ---(2).
We know that ${{\log }_{a}}b\times {{\log }_{b}}a=1$. We use this result in equation (2).
$\Rightarrow \left| A \right|=9\left( 1 \right)-\dfrac{3}{2}\left( 1 \right)$.
$\Rightarrow \left| A \right|=9-\dfrac{3}{2}$.
$\Rightarrow \left| A \right|=\dfrac{18-3}{2}$.
$\Rightarrow \left| A \right|=\dfrac{15}{2}$.
$\Rightarrow \left| A \right|=7.5$.
We know that the closest integer(s) to 7.5 is 7 or 8.

∴ The closest integer(s) to the determinant of matrix $A=\left[ \begin{matrix} {{\log }_{3}}512 & {{\log }_{4}}3 \\\ {{\log }_{3}}8 & {{\log }_{4}}9 \\\ \end{matrix} \right]$ is 7 or 8.

Note: We can prove ${{\log }_{a}}b\times {{\log }_{b}}a=1$ by using the fact ${{\log }_{a}}b=\dfrac{{{\log }_{e}}b}{{{\log }_{e}}a}$. We can see that the given problem contains a heavy amount of calculation, so we need to perform each step carefully. We should not confuse $\left| A \right|$ with modulus of A and take only positive value for the determinant which is the most common mistake done by students. We can see that the obtained determinant is the middle value of the integers 7 and 8 which is the reason why we had taken both integers as closest.