Here are the proofs for the given properties of scalar multiplication and matrix addition. To prove these results, we will use the definitions of matrix addition and scalar multiplication. Let $A$ and $B$ be $m \times n$ matrices, and let $g$ and $k$ be scalars. Let $A = [a_{ij}]$ and $B = [b_{ij}]$, where $a_{ij}$ and $b_{ij}$ are the elements in the $i$-th row and $j$-th column of matrices $A$ and $B$, respectively. a) Prove $k(A+B) = kA + kB$ Step 1: Consider the left-hand side (LHS), $k(A+B)$. First, calculate $A+B$. $$A+B = [a_{ij}] + [b_{ij}] = [a_{ij} + b_{ij}]$$ Now, multiply by the scalar $k$. $$k(A+B) = k[a_{ij} + b_{ij}] = [k(a_{ij} + b_{ij})]$$ Using the distributive property of scalars over real numbers, $k(a_{ij} + b_{ij}) = ka_{ij} + kb_{ij}$. So, $$k(A+B) = [ka_{ij} + kb_{ij}]$$ Step 2: Consider the right-hand side (RHS), $kA + kB$. First, calculate $kA$. $$kA = k[a_{ij}] = [ka_{ij}]$$ Next, calculate $kB$. $$kB = k[b_{ij}] = [kb_{ij}]$$ Now, add $kA$ and $kB$. $$kA + kB = [ka_{ij}] + [kb_{ij}] = [ka_{ij} + kb_{ij}]$$ Step 3: Compare LHS and RHS. Since $[ka_{ij} + kb_{ij}] = [ka_{ij} + kb_{ij}]$, the LHS equals the RHS. Therefore, $k(A+B) = kA + kB$ is proven. $\boxed{\text{Proven}}$ b) Prove $(g+k)A = gA + kA$ Step 1: Consider the left-hand side (LHS), $(g+k)A$. $$ (g+k)A = (g+k)[a_{ij}] = [(g+k)a_{ij}] $$ Using the distributive property of scalars over real numbers, $(g+k)a_{ij} = ga_{ij} + ka_{ij}$. So, $$(g+k)A = [ga_{ij} + ka_{ij}]$$ Step 2: Consider the right-hand side (RHS), $gA + kA$. First, calculate $gA$. $$gA = g[a_{ij}] = [ga_{ij}]$$ Next, calculate $kA$. $$kA = k[a_{ij}] = [ka_{ij}]$$ Now, add $gA$ and $kA$. $$gA + kA = [ga_{ij}] + [ka_{ij}] = [ga_{ij} + ka_{ij}]$$ Step 3: Compare LHS and RHS. Since $[ga_{ij} + ka_{ij}] = [ga_{ij} + ka_{ij}]$, the LHS equals the RHS. Therefore, $(g+k)A = gA + kA$ is proven. $\boxed{\text{Proven}}$