Question

Prove that: \left| {\begin{array}{*{20}{c}} {{a^2} + 1}&{ab}&{ac} \\\ {ab}&{{b^2} + 1}&{bc} \\\ {ac}&{bc}&{{c^2} + 1} \end{array}} \right| = {a^2} + {b^2} + {c^2} + 1

Answer 1

Here we simply have to find determinant of a 3 by 3 matrix.

{{a_{11}}}&{{a_{22}}}&{{a_{33}}} \\\ {{b_{11}}}&{{b_{22}}}&{{b_{33}}} \\\ {{c_{11}}}&{{c_{22}}}&{{c_{33}}} \end{array}} \right| = {a_{11}}\left| {\begin{array}{*{20}{c}} {{b_{22}}}&{{b_{33}}} \\\ {{c_{22}}}&{{c_{33}}} \end{array}} \right| - {a_{22}}\left| {\begin{array}{*{20}{c}} {{b_{11}}}&{{b_{33}}} \\\ {{c_{11}}}&{{c_{33}}} \end{array}} \right| + {a_{33}}\left| {\begin{array}{*{20}{c}} {{b_{11}}}&{{b_{22}}} \\\ {{c_{11}}}&{{c_{22}}} \end{array}} \right|$$ _**Complete step-by-step answer:**_ Given that, $$\left| {\begin{array}{*{20}{c}} {{a^2} + 1}&{ab}&{ac} \\\ {ab}&{{b^2} + 1}&{bc} \\\ {ac}&{bc}&{{c^2} + 1} \end{array}} \right|$$ Now in order to determine the determinant value we will start with $${a_{11}}$$ ($${a^2} + 1$$) element, and we will drop first row and first column ,then $${a_{22}}$$ ($$ab$$) element and drop second row ,second column and at last take $${a_{33}}$$ ($$ac$$) element and drop third row and third column, as shown below. $$ \Rightarrow ({a^2} + 1)\left| {\begin{array}{*{20}{c}} {{b^2} + 1}&{bc} \\\ {bc}&{{c^2} + 1} \end{array}} \right| - ab\left| {\begin{array}{*{20}{c}} {ab}&{bc} \\\ {ac}&{{c^2} + 1} \end{array}} \right| + ac\left| {\begin{array}{*{20}{c}} {ab}&{{b^2} + 1} \\\ {ac}&{bc} \end{array}} \right|$$ Now solve the inner determinant as $$\left| {\begin{array}{*{20}{c}} a&c; \\\ b&d; \end{array}} \right| = ad - bc$$ Applying this to above determinant $$ \Rightarrow ({a^2} + 1)\left( {\left( {{b^2} + 1} \right)\left( {{c^2} + 1} \right) - \left( {bc} \right)\left( {bc} \right)} \right) - ab\left( {ab\left( {{c^2} + 1} \right) - acbc} \right) + ac\left( {abbc - ac\left( {{b^2} + 1} \right)} \right)$$ Multiplying the term sin the brackets, $$ \Rightarrow ({a^2} + 1)\left( {{b^2}{c^2} + {b^2} + {c^2} + 1 - {b^2}{c^2}} \right) - ab\left( {ab{c^2} + ab - ab{c^2}} \right) + ac\left( {a{b^2}c - ac{b^2} - ac} \right)$$ Now if we observe the terms in red colour obtained after solving the brackets can be cancelled $$ \Rightarrow ({a^2} + 1)\left( {{b^2} + {c^2} + 1} \right) - ab\left( {ab} \right) + ac\left( { - ac} \right)$$ Now multiply the terms outside the brackets with the terms inside respectively. $$ \Rightarrow {a^2}{b^2} + {a^2}{c^2} + {a^2} + {b^2} + {c^2} + 1 - {a^2}{b^2} - {a^2}{c^2}$$ Again after this operation if we observe the terms in red colour can be cancelled. $$ \Rightarrow {a^2} + {b^2} + {c^2} + 1$$ This is our final answer and it is equal to RHS. Thus we proved the statement. **Note:** Note that when we solve a determinant the order of elements is very important. If we miss a single element it will lead to a wrong answer. Always perform each and every step of this type of problem