স্কেলার গুণন: বণ্টনযোগ্যতা

There are two cases for the distributive property. For the first, let $p$ and $q$ be scalars and let $A$ be a matrix. Then $(p+q)A=pA+qA$.

For the second case, let $p$ be a scalar and let $A$ and $B$ be matrices of the same size. Then $p(A+B)=pA+pB$.

Example: Case 1

Let $p=2$, $q=3$ and $A=\begin{bmatrix}1&1\\1&1\end{bmatrix}$ $$\\\begin{align} (p+q)A&=(2+3)*\begin{bmatrix}1&1\\1&1\end{bmatrix}\\&=5*\begin{bmatrix}1&1\\1&1\end{bmatrix}\\&=\begin{bmatrix}5&5\\5&5\end{bmatrix}\end{align}\\$$ $$\\\begin{align} pA+qA&=2*\begin{bmatrix}1&1\\1&1\end{bmatrix}+3*\begin{bmatrix}1&1\\1&1\end{bmatrix}\\&=\begin{bmatrix}2&2\\2&2\end{bmatrix}+\begin{bmatrix}3&3\\3&3\end{bmatrix}\\&=\begin{bmatrix}5&5\\5&5\end{bmatrix}\end{align}$$

Example: Case 2

Let $p=2$, $A=\begin{bmatrix}1&1\\1&1\end{bmatrix}$, and $B=\begin{bmatrix}2&2\\2&2\end{bmatrix}$ $$\begin{align} p(A+B)&=2\Bigg(\begin{bmatrix}1&1\\1&1\end{bmatrix}+\begin{bmatrix}2&2\\2&2\end{bmatrix}\Bigg)\\&=2*\begin{bmatrix}3&3\\3&3\end{bmatrix}\\&=\begin{bmatrix}6&6\\6&6\end{bmatrix}\end{align}\\$$ $$\\\begin{align}pA+pB&=2*\begin{bmatrix}1&1\\1&1\end{bmatrix}+2*\begin{bmatrix}2&2\\2&2\end{bmatrix}\\&=\begin{bmatrix}2&2\\2&2\end{bmatrix}+\begin{bmatrix}4&4\\4&4\end{bmatrix}\\&=\begin{bmatrix}6&6\\6&6\end{bmatrix}\end{align}$$

(Video) Distributive Property of Matrix

by Arinjay Academy

This video by Arinjay Academy proves how we can use the distributive property when dealing with matrices.

Summary

Prove that $A(B+B) = AB + BC$, where $A$ and $B$ are matrices, and $C$ is a scalar.