Question

If A =,$\left[\begin{array}{cc}1/4& 3/4\\ a& b\end{array}\right]$ find a and b so that $A^2$ = I (where I = Identity matrix)

Answer 1

$A=\left[\begin{array}{cc}1/4& 3/4\\ a& b\end{array}\right]$

$A^2=\left[\begin{array}{cc}1/4& 3/4\\ a& b\end{array}\right]$$\left[\begin{array}{cc}1/4& 3/4\\ a& b\end{array}\right]$$=\left[\begin{array}{cc}\left(\frac{1}{16}\right)+\left(\frac{3a}{4}\right)& \left(\frac{3}{16}\right)+\left(\frac{3b}{4}\right)\\ \left(\frac{a}{4}\right)+\left(ab\right)& \left(\frac{3a}{4}\right)+(b.b)\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

Comparing the matrices, we get

(3/16) + (3b/4) = 0

3b/4 = -3/16

$\to$b = -1/4 …………………………………….. (1)

1/16 + 3a/4 = 1

3a/4 = 1 – (1/16)

3a/4 = 15/16

$\to$a = 5/4 …………………………………….. (2)

Therefore a = 5/4, b = -1/4