Question

plz answer correctly ..
wrong answer will be reported​

Answer 1

Answer:

This is a simple question of Matrix multiplication.

Given:

$\implies \left[\begin{array}{cc}a&-b\\-b&-a\end{array}\right] \: \left[\begin{array}{cc}-a&b\\b&a\end{array}\right]\\\\\\\text{Multiplying the corresponding elements we get:}\\\\\\\implies \left[\begin{array}{cc}(a\times-a) + (-b \times b)&(a\times b) + (-b \times a)\\(-b \times -a) + (-a \times b)&(-b \times b) + (-a \times a)\end{array}\right] \\\\\\\\\implies \left[\begin{array}{cc}(-a^2-b^2)&(ab-ab)\\(ab-ab)&(-b^2-a^2)\end{array}\right]$

$\implies \left[\begin{array}{cc}(-a^2-b^2)&(0)\\(0)&(-b^2-a^2)\end{array}\right] = \left[\begin{array}{cc}(-1)&(0)\\(0)&(-1)\end{array}\right]$

Comparing the respective elements of LHS and RHS we get:

⇒ ( -a² - b² ) = - 1

⇒ - [ a² + b² ] = - 1

Cancelling out the negative sign, we get:

⇒ ( a² + b² ) = 1

Hence the value of ( a² + b² ) is equal to 1.