Prove that $∣ ∣ ∣ \begin{matrix} a + i b & c + i d - c + i d & a - i b \end{matrix} ∣ ∣ ∣ \times ∣ ∣ ∣ \begin{matrix} a - i β & γ - i δ - γ - i δ & a + i β \end{matrix} ∣ ∣ ∣$ ,
where $i = \sqrt{- 1}$ , can be written in the form
$∣ ∣ ∣ \begin{matrix} A - i B & C - i D - C - i D & A + i B \end{matrix} ∣ ∣ ∣$ ;
hence deduce the following theorem, due to Euler:
The product of two sums each of four squares can be expressed as the sum of four squares.

Open in App

Solution

$∣ ∣ ∣ \begin{matrix} a + i b & c + i d - c + i d & a - i b \end{matrix} ∣ ∣ ∣ \times ∣ ∣ ∣ \begin{matrix} α - i β & γ - i δ - γ - i δ & α + i β \end{matrix} ∣ ∣ ∣$

$= ∣ ∣ ∣ \begin{matrix} (a α + b β - c γ + d δ) - i (a β - b α + c δ + d γ) & (a γ + b δ - c α - d β) - i (a δ - b γ + c β + d α) - (a γ + b δ - c α - d β) - i (a δ - b γ + c β + d α) & (a α + b β - c γ + d δ) + i (a β - b α + c δ + d γ) \end{matrix} ∣ ∣ ∣$

$= ∣ ∣ ∣ \begin{matrix} A - i B & C - i D - C - i D & A + i B \end{matrix} ∣ ∣ ∣$ , where

$A = a α + b β - c γ + d δ B = a β - b α + c δ + d γ C = a γ + b δ - c α - d β D = a δ - b γ + c β + d α$

By expanding the determinants, we can also see that,
$(a^{2} + b^{2} + c^{2} + d^{2}) \times (α^{2} + β^{2} + γ^{2} + δ^{2}) = (A^{2} + B^{2} + C^{2} + D^{2})$
Or $(a^{2} + b^{2} + c^{2} + d^{2}) (α^{2} + β^{2} + γ^{2} + δ^{2}) = (a α + b β + c γ + d δ)^{2} + (a β - b α + c δ - d γ)^{2} + (a γ - b δ - c α + d β)^{2} + (a δ + b γ - c β - d α)^{2}$

Suggest Corrections

Similar questions

∣ ∣ ∣ \begin{matrix} a + i b & c + i d - c + i d & a - i b \end{matrix} ∣ ∣ ∣

|\begin{matrix} a + i b & c + i d \\ - c + i d & a - i b \end{matrix}| .

A = [\begin{matrix} a + i b & c + i d - c + i d & a - i b \end{matrix}]

a^{2} + b^{2} + c^{2} + d^{2} = 1

A = [\begin{matrix} a + i b & c + i d - c + i d & a - i b \end{matrix}]

a^{2} + b^{2} + c^{2} + d^{2} = 1

A^{- 1}

a, b, c

d

A = ∣ ∣ ∣ \begin{matrix} a + i b & c + i d - c + i d & a - i b \end{matrix} ∣ ∣ ∣

A^{- 1} =

Join BYJU'S Learning Program