Ravi Kant Kumar

Evaluate Row × Matrix × Column Evaluate Row × Matrix × Column Question: Evaluate: \[ [1\;\;2\;\;3] \begin{bmatrix} 1 & 0 & 2 \\ 2 & 0 & 1 \\ 0 & 1 & 2 \end{bmatrix} \begin{bmatrix} 2 \\ 4 \\ 6 \end{bmatrix} \] Solution: Step 1: Multiply first two matrices \[ [1\;\;2\;\;3] \begin{bmatrix} 1 & […]

Evaluate (A + B)C Evaluate \((A + B)C\) Question: Evaluate: \[ \left( \begin{bmatrix}1 & 3 \\ -1 & -4\end{bmatrix} + \begin{bmatrix}3 & -2 \\ -1 & 1\end{bmatrix} \right) \begin{bmatrix}1 & 3 & 5 \\ 2 & 4 & 6\end{bmatrix} \] Solution: Step 1: Add matrices \[ A+B = \begin{bmatrix} 1+3 & 3+(-2) \\ -1+(-1) &

Show AB ≠ BA Show That \(AB \ne BA\) Question: Given \[ A=\begin{bmatrix} 1 & 3 & -1 \\ 2 & -1 & -1 \\ 3 & 0 & -1 \end{bmatrix}, \quad B=\begin{bmatrix} -2 & 3 & -1 \\ -1 & 2 & -1 \\ -6 & 9 & -4 \end{bmatrix} \] show that \(AB

Row Column Matrix Product Row × Column Matrix Products Question: Compute: \[ [a\;\; b]\begin{bmatrix}c\\ d\end{bmatrix}, \quad [a\;\; b\;\; c\;\; d]\begin{bmatrix}a\\ b\\ c\\ d\end{bmatrix} \] Solution: Part 1: \( [a\;\; b]\begin{bmatrix}c\\ d\end{bmatrix} \) This is a dot product: \[ = ac + bd \] Part 2: \( [a\;\; b\;\; c\;\; d]\begin{bmatrix}a\\ b\\ c\\ d\end{bmatrix} \) \[

Row and Column Matrix Multiplication Compute AB and BA Question: Given \[ A = \begin{bmatrix}1 & -1 & 2 & 3\end{bmatrix}, \quad B = \begin{bmatrix}0 \\ 1 \\ 3 \\ 2\end{bmatrix} \] compute \(AB\) and \(BA\). Solution: Step 1: Check order \(A\) is \(1 \times 4\), \(B\) is \(4 \times 1\) \(AB\): Possible → \(1×1\)

Compute AB and BA Compute AB and BA Question: Given \[ A=\begin{bmatrix}3 & 2 \\ -1 & 0 \\ -1 & 1\end{bmatrix}, \quad B=\begin{bmatrix}4 & 5 & 6 \\ 0 & 1 & 2\end{bmatrix} \] compute \(AB\) and \(BA\), wherever possible. Solution: Step 1: Check order \(A\) is \(3 \times 2\), \(B\) is \(2 \times

Compute AB and BA Compute AB and BA Question: Given \[ A=\begin{bmatrix}1 & -2 \\ 2 & 3\end{bmatrix}, \quad B=\begin{bmatrix}1 & 2 & 3 \\ 2 & 3 & 1\end{bmatrix} \] Compute \(AB\) and \(BA\), wherever possible. Solution: Step 1: Check order \(A\) is \(2 \times 2\), \(B\) is \(2 \times 3\) \(AB\): Possible (2×2

Show AB ≠ BA (Example) Show That \(AB \ne BA\) Question: Given \[ A=\begin{bmatrix}1 & 3 & 0 \\ 1 & 1 & 0 \\ 4 & 1 & 0\end{bmatrix}, \quad B=\begin{bmatrix}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 5 & 1\end{bmatrix} \] show that \(AB \ne BA\). Solution:

Show AB ≠ BA (3×3) Show That \(AB \ne BA\) Question: Given \[ A=\begin{bmatrix}-1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4\end{bmatrix}, \quad B=\begin{bmatrix}1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0\end{bmatrix} \] show that \(AB \ne BA\). Solution:

Show AB ≠ BA Show That \(AB \ne BA\) Question: Given \[ A=\begin{bmatrix}5 & -1 \\ 6 & 7\end{bmatrix}, \quad B=\begin{bmatrix}2 & 1 \\ 3 & 4\end{bmatrix} \] show that \(AB \ne BA\). Solution: Step 1: Compute \(AB\) \[ AB = \begin{bmatrix} 5(2)+(-1)(3) & 5(1)+(-1)(4) \\ 6(2)+7(3) & 6(1)+7(4) \end{bmatrix} \] \[ = \begin{bmatrix} 10-3