Question

Find the value of the determinant \left| {\begin{array}{*{20}{c}} {15}&{11}&7 \\\ {11}&{17}&{14} \\\ {10}&{16}&{13} \end{array}} \right|.

Answer 1

First apply the property of the determinant to reduce it in a simple form. Subtract the third row from the second row putting the result in the second row. This will make all the elements of the second row 1. Now again make two of them 0 by subtracting columns. Finally expand the determinant to get its value.

Complete step-by-step answer:
According to the question, we have to calculate the value of the determinant \left| {\begin{array}{*{20}{c}} {15}&{11}&7 \\\ {11}&{17}&{14} \\\ {10}&{16}&{13} \end{array}} \right|. Let its value is $D$. So we have:
\Rightarrow D = \left| {\begin{array}{*{20}{c}} {15}&{11}&7 \\\ {11}&{17}&{14} \\\ {10}&{16}&{13} \end{array}} \right|
First we will apply the property of determinant to make it simpler and then we will expand it to get the final value.
So to start with, subtracting third row from the second row and putting the result in the second row, the value of determinant will remain unchanged. This is shown below:
{R_2} \to {R_2} - {R_3} \\\ \Rightarrow D = \left| {\begin{array}{*{20}{c}} {15}&{11}&7 \\\ {11 - 10}&{17 - 16}&{14 - 13} \\\ {10}&{16}&{13} \end{array}} \right| \\\ \Rightarrow D = \left| {\begin{array}{*{20}{c}} {15}&{11}&7 \\\ 1&1&1 \\\ {10}&{16}&{13} \end{array}} \right| \\\
Now, subtracting first column from the second column and putting it in second column and also subtracting first column from the third column and putting it in third column, we will get:
{C_2} \to {C_2} - {C_1}{\text{ and }}{C_3} \to {C_3} - {C_1} \\\ \Rightarrow D = \left| {\begin{array}{*{20}{c}} {15}&{11 - 15}&{7 - 15} \\\ 1&{1 - 1}&{1 - 1} \\\ {10}&{16 - 10}&{13 - 10} \end{array}} \right| \\\ \Rightarrow D = \left| {\begin{array}{*{20}{c}} {15}&{ - 4}&{ - 8} \\\ 1&0&0 \\\ {10}&6&3 \end{array}} \right| \\\
Now we can easily expand the determinant by using the second row. Doing so, we’ll get:
$\Rightarrow D = - 1\left[ {\left( { - 4} \right) \times 3 - 6 \times \left( { - 8} \right)} \right] + 0\left[ {15 \times 3 - 10 \times \left( { - 8} \right)} \right] - 0\left[ {15 \times 6 - 10 \times \left( { - 4} \right)} \right] \\\ \Rightarrow D = - 1 \times \left( { - 12 + 48} \right) + 0 - 0 \\\ \Rightarrow D = - 36 \\\$

Thus the value of the determinant is -36.

Note: This determinant can also be solved by expanding from the first step, without applying property. But this method is not recommended because this method will require a lot of unnecessary calculations and it will also take much time.
Another important point is that, if we somehow manage to make any row or column entirely zero by applying properties then the value of determinant becomes zero straightaway. For example,
\Rightarrow \left| {\begin{array}{*{20}{c}} x&0&a; \\\ y&0&b; \\\ z&0&c; \end{array}} \right| = 0
This property was not used in the above problem but it is widely used in many other determinant problems.