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What is the value of \(\left| {\begin{array}{*{20}{c}} {{b^2} + {c^2}}&{ab}&{ac}\\ {ba}&{{c^2} + {a^2}}&{bc}\\ {ca}&{cb}&{{a^2} + {b^2}} \end{array}} \right|\)

What is the value of \(\left| {\begin{array}{*{20}{c}} {{b^2} + {c^2}}&{ab}&{ac}\\ {ba}&{{c^2} + {a^2}}&{bc}\\ {ca}&{cb}&{{a^2} + {b^2}} \end{array}} \right|\) Correct Answer 4a<sup>2</sup>b<sup>2</sup>c<sup>2</sup>

Calculation:

Expanding the determinant along Row 1.

= (b2 + c2) – (ab) + (ac)

= (b2 + c2) (a4 + a2b2 + a2c2 + c2b2 – b2c2) – ab (a3b + b3a – c2ab) + ac (ab2c – a3c – c3a)

= (b2 + c2) (a4 + a2b2 + a2c2) – a4b2 - a2b4 + a2b2c2 + a2b2c2 - a4c2 – a2c4

= 4a2b2c2

∴ Value of determinant is  4a2b2c2  .

Related Questions

If a + b + c + d = 4, then find the value of $$\frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}}$$     + $$\frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}}$$     + $$\frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}}$$     + $$\frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}}$$     is?
If a + b + c + d = 4, then the value of $$\frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}}$$     + $$\frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}}$$     + $$\frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}}$$     + $$\frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}}$$     is?
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