2.2 Q-7

Exercise 2.2 — Question 7

Problem

If $A=\left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right)$ and $B=\left(\begin{array}{cc}2& 0\\ 1& 3\end{array}\right)$, verify that $(AB{)}^T=B^T{A}^T$.

---

Key Concepts

Matrix Multiplication: For matrices $A$ (of order $m\times n$) and $B$ (of order $n\times p$), the product $AB$ is of order $m\times p$. The element in row $i$, column $j$ of $AB$ is: $(AB{)}_{ij}={\sum }_{k=1}^n{a}_{ik}{b}_{kj}$

Transpose of a Matrix: The transpose $A^T$ is obtained by interchanging rows and columns of $A$.

Reversal Law for Transposes: $(AB{)}^T=B^T{A}^T$

---

Worked Solution

Step 1: Compute $AB$

$AB=\left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right)\left(\begin{array}{cc}2& 0\\ 1& 3\end{array}\right)$

$=\left(\begin{array}{cc}(1)(2)+(2)(1)& (1)(0)+(2)(3)\\ (3)(2)+(4)(1)& (3)(0)+(4)(3)\end{array}\right)=\left(\begin{array}{cc}4& 6\\ 10& 12\end{array}\right)$

Step 2: Compute $(AB{)}^T$

$(AB{)}^T=\left(\begin{array}{cc}4& 10\\ 6& 12\end{array}\right)$

Step 3: Compute $B^T$ and $A^T$

$B^T=\left(\begin{array}{cc}2& 1\\ 0& 3\end{array}\right),A^T=\left(\begin{array}{cc}1& 3\\ 2& 4\end{array}\right)$

Step 4: Compute $B^T{A}^T$

$B^T{A}^T=\left(\begin{array}{cc}2& 1\\ 0& 3\end{array}\right)\left(\begin{array}{cc}1& 3\\ 2& 4\end{array}\right)$

$=\left(\begin{array}{cc}(2)(1)+(1)(2)& (2)(3)+(1)(4)\\ (0)(1)+(3)(2)& (0)(3)+(3)(4)\end{array}\right)=\left(\begin{array}{cc}4& 10\\ 6& 12\end{array}\right)$

Conclusion: $(AB{)}^T=B^T{A}^T=\left(\begin{array}{cc}4& 10\\ 6& 12\end{array}\right)$ ✓

---

Important Rules for Matrix Operations