Answer
See the explanation below.
Work Step by Step
Let $A=\left[\begin{array}{rr}{0} & {3} \\ {2} & {1}\end{array}\right], \quad B=\left[\begin{array}{rr}{-3} & {1} \\ {4} & {3}\end{array}\right]$, $\langle A, B \rangle =a_{11} b_{11}+a_{12} b_{12}+a_{21} b_{21}+a_{22} b_{22}$
Then, we have
$\langle A, B \rangle =0+3+8+3=14$ $\| A\| =\sqrt{\langle A, A\rangle}=\sqrt{a_{11}^{2}+a_{12}^{2}+a_{21}^{2}+a_{22}^{2}}=\sqrt{0+9+4+1}=\sqrt{14}$ $\| B \| =\sqrt{\langle B, B\rangle}=\sqrt{b_{11}^{2}+b_{12}^{2}+b_{21}^{2}+b_{22}^{2}}=\sqrt{9+1+16+9}=\sqrt{35}$ $\| A+B \|=\| \left[\begin{array}{rr}{-3} & {4} \\ {6} & {4}\end{array}\right] \|=\sqrt{9+16+36+16}=\sqrt{77}.$
Now, we get
(a) Cauchy-Schwarz Inequality:
$|\langle A, B \rangle|=14\leq\| A \|\| B \|=\sqrt{14}\sqrt{35}=(3.74) (5.91)=22.12$
(b) The triangle inequality:
$\| A+B \| =\sqrt{77}=8.77\leq\| A \|+\| B \|=\sqrt{14}+\sqrt{35}=9.65$.
