Question

Without expanding evaluate the determinant: \[\left| {\begin{array}{*{20}{c}}
  {41}&1&5 \\
  {79}&7&9 \\
  {29}&5&3
\end{array}} \right|\].

Answer 1

Hint: We will apply column operations on the determinant because linear addition or subtraction of any row or column does not make any change in the value of determinant. Applying a relevant column operation, we will get two columns to be the same and hence, the answer will be 0.

Complete step-by-step answer:
First let us look at the given determinant carefully.
The given determinant is \[\left| {\begin{array}{*{20}{c}}
  {41}&1&5 \\
  {79}&7&9 \\
  {29}&5&3
\end{array}} \right|\].
We can clearly see that row 1 has extremely larger values as in comparison to other columns. So, we will just try to tone down its values.
Applying the column transformation of: ${C_1} \to {C_1} - 8{C_3}$.
We will get the determinant to be equal to
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  {41}&1&5 \\
  {79}&7&9 \\
  {29}&5&3
\end{array}} \right| \equiv \left| {\begin{array}{*{20}{c}}
  {41 - 8 \times 5}&1&5 \\
  {79 - 8 \times 9}&7&9 \\
  {29 - 8 \times 3}&5&3
\end{array}} \right|\]
Simplifying the values, we will get:-
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  {41}&1&5 \\
  {79}&7&9 \\
  {29}&5&3
\end{array}} \right| \equiv \left| {\begin{array}{*{20}{c}}
  {41 - 40}&1&5 \\
  {79 - 70}&7&9 \\
  {29 - 24}&5&3
\end{array}} \right|\]
Simplifying the values further, we will get:-
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  {41}&1&5 \\
  {79}&7&9 \\
  {29}&5&3
\end{array}} \right| \equiv \left| {\begin{array}{*{20}{c}}
  1&1&5 \\
  7&7&9 \\
  5&5&3
\end{array}} \right|\]
Now, since we know that if a determinant has any of the two or more rows/columns same, then the value of the determinant is always zero. This is among one of the important properties of determinant.
Hence, \[\left| {\begin{array}{*{20}{c}}
  {41}&1&5 \\
  {79}&7&9 \\
  {29}&5&3
\end{array}} \right| \equiv 0\].

Note: The students must remember that if they need to apply more than one operation before reaching the required answer, they must use only one type of transformation throughout. Using both row transformation and column transformation will lead you to the different values of determinant. So, like in this question, if we started out with column transformation and we will have to carry out every transformation like this only using columns.
In these types of questions, students can even check their answers for their satisfaction by expanding the determinant. Expansion should give you the same answer, if your answer is correct.
Let us now look at the property, why the determinant value is 0, when two of the rows or columns are exactly the same. Consider the example by taking the determinant: \[|A| = \left| {\begin{array}{*{20}{c}}
  a&a&d \\
  b&b&e \\
  c&c&f
\end{array}} \right|\].
Now, we will just expand this determinant using the third column.
We will get: \[|A| = d(bc - bc) - e(ac - ac) + f(ab - ab)\]
Now, simplifying the values on RHS, we will get:- \[|A| = d \times 0 - e \times 0 + f \times 0\].
Hence, \[|A| = 0\].