Answer
(a) $\langle A, B \rangle =4$
(b) $\| A\| =2$
(c) $\| B \| =\sqrt{5}$
(d) $d(A,B)=1.$
Work Step by Step
$A=\left[\begin{array}{rr}{1} & {0} \\ {0} & {-1}\end{array}\right], \quad B=\left[\begin{array}{rr}{1} & {1} \\ {0} & {-1}\end{array}\right]$
(a) $\langle A, B \rangle =2a_{11} b_{11}+a_{12} b_{12}+a_{21} b_{21}+2a_{22} b_{22}=2+0+0+2=4$
(b) $\| A\| =\sqrt{\langle A, A\rangle}=\sqrt{a_{11}^{2}+a_{12}^{2}+a_{21}^{2}+2a_{22}^{2}}=\sqrt{2+0+0+2}=2$
(c) $\| B \| =\sqrt{\langle B, B\rangle}=\sqrt{2b_{11}^{2}+b_{12}^{2}+b_{21}^{2}+2b_{22}^{2}}=\sqrt{2+1+0+2}=\sqrt{5}$
(d) $d(A,B)=\| A-B \|=\| \left[\begin{array}{rr}{0} & {-1} \\ {0} & {0}\end{array}\right] \|=\sqrt{0+1+0+0}=1.$
