Determinant Calculation Explained

We need to find the value of the following determinant:

Strategy for Solving the Determinant

To simplify the calculation, we can use elementary row and column operations. A common strategy involves factoring out terms and then applying operations to create zeros or simplify the structure.

Step-by-Step Solution

1. First, multiply the first row by a, the second row by b, and the third row by c. To keep the value of the determinant unchanged, we divide the entire determinant by abc.

   $$ D = \frac{1}{abc} \left| {\begin{array}{} {a({b^2} + {c^2})}&{a(ab)}&{a(ac)}\\ {b(ba)}&{b({c^2} + {a^2})}&{b(bc)}\\ {c(ca)}&{c(cb)}&{c({a^2} + {b^2})} \end{array}} \right| $$ $$ D = \frac{1}{abc} \left| {\begin{array}{} {a{b^2} + a{c^2}}&{a^2b}&{a^2c}\\ {a{b^2}}&{b{c^2} + b{a^2}}&{b^2c}\\ {c{a^2}}&{c{b^2}}&{c{a^2} + c{b^2}} \end{array}} \right| $$

2. Now, factor out a from the first column, b from the second column, and c from the third column.

   $$ D = \frac{abc}{abc} \left| {\begin{array}{} {{b^2} + {c^2}}&{a^2}&{a^2}\\ {{b^2}}&{{c^2} + {a^2}}&{{b^2}}\\ {{c^2}}&{{c^2}}&{{a^2} + {b^2}} \end{array}} \right| $$

   Let this simplified determinant be denoted by D'.

   $$ D' = \left| {\begin{array}{} {{b^2} + {c^2}}&{a^2}&{a^2}\\ {{b^2}}&{{c^2} + {a^2}}&{{b^2}}\\ {{c^2}}&{{c^2}}&{{a^2} + {b^2}} \end{array}} \right| $$

3. Apply column operations to D': Replace Column 2 with (Column 2 - Column 1) and Column 3 with (Column 3 - Column 1).

   C_2 \rightarrow C_2 - C_1

   C_3 \rightarrow C_3 - C_1

   $$ D' = \left| {\begin{array}{} {{b^2} + {c^2}}&{a^2 - ({b^2} + {c^2})}&{a^2 - ({b^2} + {c^2})}\\ {{b^2}}&{({c^2} + {a^2}) - {b^2}}&{{b^2} - {b^2}}\\ {{c^2}}&{{c^2} - {c^2}}&{({a^2} + {b^2}) - {c^2}} \end{array}} \right| $$ $$ D' = \left| {\begin{array}{} {{b^2} + {c^2}}&{a^2 - b^2 - c^2}&{a^2 - b^2 - c^2}\\ {{b^2}}&{a^2 + c^2 - b^2}&{0}\\ {{c^2}}&{0}&{a^2 + b^2 - c^2} \end{array}} \right| $$

4. Expand the determinant D'. Let's expand along the first row:

   $$ D' = ({b^2} + {c^2}) \left| {\begin{array}{} {a^2 + c^2 - b^2}&{0}\\ {0}&{a^2 + b^2 - c^2} \end{array}} \right| - (a^2 - b^2 - c^2) \left| {\begin{array}{} {{b^2}}&{0}\\ {{c^2}}&{a^2 + b^2 - c^2} \end{array}} \right| + (a^2 - b^2 - c^2) \left| {\begin{array}{} {{b^2}}&{a^2 + c^2 - b^2}\\ {{c^2}}&{0} \end{array}} \right| $$

5. Calculate the minors:

   Minor 1: $({a^2} + {c^2} - {b^2})({a^2} + {b^2} - {c^2}) - 0 = ({a^2} + ({c^2} - {b^2})) ({a^2} - ({c^2} - {b^2})) = {a^4} - ({c^2} - {b^2})^2 = {a^4} - ({c^4} + {b^4} - 2{b^2}{c^2})$

   Minor 2: ${b^2}({a^2} + {b^2} - {c^2}) - 0 = {a^2}{b^2} + {b^4} - {b^2}{c^2}$

   Minor 3: $0 - {c^2}({a^2} + {c^2} - {b^2}) = -{a^2}{c^2} - {c^4} + {b^2}{c^2}$

6. Substitute the minors back into the expansion:

   $$ D' = ({b^2} + {c^2})({a^4} - {c^4} - {b^4} + 2{b^2}{c^2}) - (a^2 - b^2 - c^2)({a^2}{b^2} + {b^4} - {b^2}{c^2}) + (a^2 - b^2 - c^2)(-{a^2}{c^2} - {c^4} + {b^2}{c^2}) $$

   Let $X = a^2, Y = b^2, Z = c^2$. Then $D' = (Y+Z)(X^2 - (Y-Z)^2) - (X-Y-Z)(Y(X+Y-Z)) + (X-Y-Z)(-Z(X+Z-Y))$.

   Simplifying this expression leads to:

   $$ D' = X ( (Y+Z)^2 - (Y-Z)^2 ) $$ $$ D' = X ( Y^2 + 2YZ + Z^2 - (Y^2 - 2YZ + Z^2) ) $$ $$ D' = X ( 4YZ ) $$ $$ D' = 4XYZ $$

7. Substitute back $X = a^2$, $Y = b^2$, $Z = c^2$:

   $$ D' = 4a^2b^2c^2 $$

8. Since $D = D'$, the value of the determinant is $4a^2b^2c^2$.