STEP III 1994 question 8 solution

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STEP III 1994 question 8 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP III 1994 question 8 solution

STEP III Question 8

$\begin{pmatrix}z_1&z_2\\-z_2^*&z_1^*\end{pmatrix} \begin{pmatrix}z_3&z_4\\-z_4^*&z_3^*\end{pmatrix}= \begin{pmatrix}z_1z_3-z_2z_4^*&z_1z_4+z_2z_3^*\\-z_2^*z_3-z_1^*z_4^*&-z_2^*z_4+z_1^*z_3^*\end{pmatrix}$ The determinant of this matrix is $(z_1z_3-z_2z_4^*)(z_1^*z_3^*-z_2^*z_4)-(z_1z_4+z_2z_3^*)(-z_2^*z_3-z_1^*z_4^*)= \newline z_1z_1^*z_3z_3^*-z_1z_3z_2^*z_4-z_2z_4^*z_1^*z_3^*+z_2z_2^*z_4z_4^*+z_1z_4z_2^*z_3+z_1z_1^*z_4z_4^*+z_2z_2^*z_3z_3^*+z_1^*z_2z_3^*z_4^* = \newline z_1z_1^*(z_3z_3^*+z_4z_4^*)+z_2z_2^*(z_3z_3^*+z_4z_4^*)= \newline (z_1z_1^*+z_2z_2^*)(z_3z_3^*+z_4z_4^*)=(|z_1|^2+|z_2|^2)(|z_3|^2+|z_4|^2)$

(In hindsight it would have been less algebra to use |AB|=|A||B|)

Now, chose e.g.

$z_1=p+qi\newline z_2=r+si \newline z_3=a+bi \newline\text{and} \newline z_4=c+di$

and the determinant of the matrix product will be the desired LHS, i.e. (a^2+b^2+c^2+d^2)(p^2+q^2+r^2+s^2) .

(Provided $z_1\not=-z_2 \text{and} z_3\not=-z_4$ , because then there is a zero-determinant, which we don't want)

Now, to get the RHS of the equality we are to prove, take a look at the matrix $\begin{pmatrix}z_1z_3-z_2z_4^*&z_1z_4+z_2z_3^*\\-z_2^*z_3-z_1^*z_4^*&-z_2^*z_4+z_1^*z_3^*\end{pmatrix}$ again. You can see that if we let w_1=z_1z_3-z_2z_4^* and w_2=z_1z_4+z_2z_3^* we have: ' $\begin{pmatrix}w_1& w_2\\-w_2^*&w_1^*\end{pmatrix}$

This matrix also has a real determinant (i.e. |w_1|^2+|w_2|^2 ). Now, if we set w_1=L_1+iL_2 and w_2=L_3+iL_4 this means the matrix has the determinant L_1^2+L_2^2+L_3^2+L_4^2 , which is the RHS in the equality.