Answer
see the proof below.
Work Step by Step
For any $A=\left[\begin{array}{ll}{a_{11}} & {a_{12}} \\ {a_{21}} & {a_{22}}\end{array}\right], B=\left[\begin{array}{ll}{b_{11}} & {b_{12}} \\ {b_{21}} & {b_{22}}\end{array}\right], C=\left[\begin{array}{ll}{c_{11}} & {c_{12}} \\ {c_{21}} & {c_{22}}\end{array}\right] \in M_{2,2}$ and $k\in R$.
(1) $\langle A, A\rangle=2 a_{11}^{2}+a_{12}^{2}+a_{21}^{2}+2a_{22}^{2} \geq 0$ for all $a_{11}, a_{12}, a_{21}, a_{22} \in \mathbb{R},$ and
$\langle A, A\rangle= 2a_{11}^{2}+a_{12}^{2}+a_{21}^{2}+2a_{22}^{2}=0$ iff $a_{11}=a_{12}=a_{21}=a_{22}=0$ i.e. $A$
is the zero matrix.
(2)
\begin{aligned}\langle A, B\rangle &= 2a_{11} b_{11}+a_{12} b_{12}+a_{21} b_{21}+2a_{22} b_{22} \\ &=2b_{11} a_{11}+b_{12} a_{12}+b_{21} a_{21}+2b_{22} a_{22} \\ &=\langle B, A\rangle \end{aligned}
(3) \begin{aligned}\langle k A, B\rangle &=\left\langle\left[\begin{array}{ll}{k a_{11}} & {k a_{12}} \\ {k a_{21}} & {k a_{22}}\end{array}\right],\left[\begin{array}{ll}{b_{11}} & {b_{12}} \\ {b_{21}} & {b_{22}}\end{array}\right]\right\rangle \\ &= 2k a_{11} b_{11}+k u_{12} b_{12}+k a_{21} b_{21}+2k a_{22} b_{22} \\ &=k\left(2a_{11} b_{11}+a_{12} b_{12}+a_{21} b_{21}+2a_{22} b_{22}\right) \\ &=k\langle A, B\rangle \end{aligned}
(4) \begin{aligned}&\langle A+B, C\rangle =\left\langle\left[\begin{array}{cc}{a_{11}+b_{11}} & {a_{12}+b_{12}} \\ {a_{21}+b_{21}} & {a_{22}+b_{22}}\end{array}\right],\left[\begin{array}{cc}{c_{11}} & {c_{12}} \\ {c_{21}} & {c_{22}}\end{array}\right]\right\rangle \\ &=2\left(a_{11}+b_{11}\right) c_{11}+\left(a_{12}+b_{12}\right) c_{12}+\left(a_{21}+b_{21}\right) c_{21}+2\left(a_{22}+b_{22}\right) c_{22} \\ &=\left(2a_{11} c_{11}+a_{12} c_{12}+a_{21} c_{21}+2a_{22} c_{22}\right)+\left(2b_{11} c_{11}+b_{12} c_{12}+b_{21} c_{21}+2b_{22} c_{22}\right) \\ &=\langle A, C\rangle+\langle B, C\rangle \end{aligned}
