Question

If a, b, c are integers such that b is always twice the sum of ‘a’ and ‘c’ lies between 1 and 9 and $6a^2, 5b^2, 4c^2 $ are three digit numbers then value of \[\left| \begin{matrix}
   6 & 6a^2 & a \\
   5 & 5b^2 & b \\
   4 & 4c^2 & c \\
\end{matrix} \right|\] equals?

Answer 1

Hint: These types of problems require some advanced knowledge of determinants and matrices as well as some fair idea of representation of numbers. In the given problem the first and foremost thing is to represent the second column of the determinant in terms of a proper number and after that we can break down the determinant into sums and evaluate each term separately. We should remember one law of determinant that, if two rows or columns of a matrix is equal then the value of the determinant of that matrix becomes zero. Moreover we break the determinant into more parts for easy solving of the problem.

Complete step by step solution:
Now we start off the solution to the given problem by writing that,
 \[\begin{align}
  & 6a^2=600+10a+2 \\
 & 5b^2=500+10b+2 \\
 & 4c^2=400+10c+2 \\
\end{align}\]
Substituting these values to our determinant, we write it as,
\[\left| \begin{matrix}
   6 & 600+10a+2 & a \\
   5 & 500+10b+2 & b \\
   4 & 400+10c+2 & c \\
\end{matrix} \right|\]
Now, we can break up the above determinant on the basis of the second column into three different determinants and then evaluate the values of the determinants separately. We write it as,
\[\left| \begin{matrix}
   6 & 600+10a+2 & a \\
   5 & 500+10b+2 & b \\
   4 & 400+10c+2 & c \\
\end{matrix} \right|=\left| \begin{matrix}
   6 & 600 & a \\
   5 & 500 & b \\
   4 & 400 & c \\
\end{matrix} \right|+\left| \begin{matrix}
   6 & 10a & a \\
   5 & 10b & b \\
   4 & 10c & c \\
\end{matrix} \right|+\left| \begin{matrix}
   6 & 2 & a \\
   5 & 2 & b \\
   4 & 2 & c \\
\end{matrix} \right|\]
Now, we first evaluate the first determinant on the right hand side as,
\[\left| \begin{matrix}
   6 & 600 & a \\
   5 & 500 & b \\
   4 & 400 & c \\
\end{matrix} \right|\]
We take 100 common from the second column of the determinant,
\[100\left| \begin{matrix}
   6 & 6 & a \\
   5 & 5 & b \\
   4 & 4 & c \\
\end{matrix} \right|\]
Now in this we can see that the first and the second column of the determinant is the same, thus this determinant value evaluates to zero.
Now we evaluate the second determinant value as,
\[\left| \begin{matrix}
   6 & 10a & a \\
   5 & 10b & b \\
   4 & 10c & c \\
\end{matrix} \right|\]
We take 10 common from the second column of the determinant and it becomes,
\[10\left| \begin{matrix}
   6 & a & a \\
   5 & b & b \\
   4 & c & c \\
\end{matrix} \right|\]
Here we can see that the second and the third column of the determinant is same, thus this determinant will evaluate to a value zero.
Now we evaluate the third determinant as,
\[\begin{align}
  & \left| \begin{matrix}
   6 & 2 & a \\
   5 & 2 & b \\
   4 & 2 & c \\
\end{matrix} \right|=6\left( 2c-2b \right)-2\left( 5c-4b \right)+a\left( 10-8 \right) \\
 & \Rightarrow \left| \begin{matrix}
   6 & 2 & a \\
   5 & 2 & b \\
   4 & 2 & c \\
\end{matrix} \right|=12c-12b-10c+8b+2a \\
 & \Rightarrow \left| \begin{matrix}
   6 & 2 & a \\
   5 & 2 & b \\
   4 & 2 & c \\
\end{matrix} \right|=2a-4b+2c \\
\end{align}\]
Thus the overall value of the determinant thus reduces to,
\[\left| \begin{matrix}
   6 & 600+10a+2 & a \\
   5 & 500+10b+2 & b \\
   4 & 400+10c+2 & c \\
\end{matrix} \right|=0+0+2a-4b+2c\]

Thus the answer to our problem is \[2a-4b+2c\].

Note: Problems like these needs in depth knowledge of determinant and matrices as well as representation of numbers in some other forms. We first undergo substitution of the number of the three digits, then break the determinant into smaller parts for easy solving of the problem. We need evaluate each of the parts of the determinant and then add it to get the final result of the original given determinant.