2.1 Q-3

Exercise 2.1 — Question 3

Problem

If $A=\left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right),B=\left(\begin{array}{cc}5& 6\\ 7& 8\end{array}\right)$ Find: (i) $A+B$, (ii) $A-B$, (iii) $AB$

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Key Concepts

Matrix Addition and Subtraction

Two matrices $A$ and $B$ can be added or subtracted only if they have the same order ($m\times n$).

If $A=[a_{ij}]$ and $B=[b_{ij}]$ are both $m\times n$ matrices: $A+B=[a_{ij}+b_{ij}],A-B=[a_{ij}-b_{ij}]$

Worked Solution (i) and (ii): $A+B=\left(\begin{array}{cc}1+5& 2+6\\ 3+7& 4+8\end{array}\right)=\left(\begin{array}{cc}6& 8\\ 10& 12\end{array}\right)$

$A-B=\left(\begin{array}{cc}1-5& 2-6\\ 3-7& 4-8\end{array}\right)=\left(\begin{array}{cc}-4& -4\\ -4& -4\end{array}\right)$

Matrix Multiplication

Matrix $A$ of order $m\times n$ can be multiplied by matrix $B$ of order $n\times p$ only when the inner dimensions match (columns of $A$ = rows of $B$). The product $AB$ has order $m\times p$.

The $(i,j)$-th element of $AB$ is: $(AB{)}_{ij}={\sum }_{k=1}^n{a}_{ik}\cdot b_{kj}$

Worked Solution (iii): $AB=\left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right)\left(\begin{array}{cc}5& 6\\ 7& 8\end{array}\right)$

$=\left(\begin{array}{cc}1\cdot 5+2\cdot 7& 1\cdot 6+2\cdot 8\\ 3\cdot 5+4\cdot 7& 3\cdot 6+4\cdot 8\end{array}\right)=\left(\begin{array}{cc}19& 22\\ 43& 50\end{array}\right)$

Operations with Complex Entries

The same rules apply when matrix entries are complex numbers. Addition and subtraction are performed element-wise — real and imaginary parts are handled separately.

Example: $\left(\begin{array}{cc}1+2i& 3\end{array}\right)+\left(\begin{array}{cc}2-i& 1+i\end{array}\right)=\left(\begin{array}{cc}(1+2)+(2-1)i& 3+1+i\end{array}\right)=\left(\begin{array}{cc}3+i& 4+i\end{array}\right)$

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Practice