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- Question 1, Exercise 2.1
- \\-8 & -18 & -28 \end{bmatrix} \end{align} ====Go to ==== <text align="left"><btn type="success">[[
- Question 2, Exercise 2.1
- -20 & 1\\6 & 1 & 11 \end{bmatrix}\end{align} ====Go To==== <text align="left"><btn type="primary">[[e
- Question 3, Exercise 2.1
- rom (1) and (2), we have $$A(B-C)=AB-BC.$$ ====Go To==== <text align="left"><btn type="primary">[[m
- Question 4, Exercise 2.1
- mplies & \dfrac{1}{3}A^2-2A-9I=0 \end{align} ====Go To==== <text align="left"><btn type="primary">[[m
- Question 5 & 6, Exercise 2.1
- 3 & 3 \\ \end{matrix} \right].\end{align} ====Go To==== <text align="left"><btn type="primary">[[m
- Question 7, Exercise 2.1
- rom (1) and (2), we have $(A+B)^t=A^t+B^t.$ ====Go To==== <text align="left"><btn type="primary">[[m
- Question 8, Exercise 2.1
- \\ \end{matrix} \right]$$ $$AA^t\ne A^tA$$ ====Go To==== <text align="left"><btn type="primary">[[m
- Question 9, Exercise 2.1
- eft( AB \right)}^{t}}={{B}^{t}}{{A}^{t}}$$ ====Go To==== <text align="left"><btn type="primary">[[m
- Question 10, Exercise 2.1
- end{matrix} \right]$$ $$( A+B )^t=A^t+B^t$$ ====Go To==== <text align="left"><btn type="primary">[[m
- Question 11, Exercise 2.1
- nd{matrix} \right]$$ $$( A+B )^t=-( A+B )$$ ====Go To==== <text align="left"><btn type="primary">[[m
- Question 12, Exercise 2.1
- {matrix} \right]$$ $$( A-A^t )^t=-( A-A^t)$$ ====Go To==== <text align="left"><btn type="primary">[[m
- Question 13, Exercise 2.1
- ht)}^{t}}=-\left( A-{{A}^{t}} \right)$$ ====Go To==== <text align="left"><btn type="primary">[[m
- Question 1, Exercise 2.2
- -4 \right) \\ \implies|A|&=-14\end{align} ====Go to ==== <text align="left"><btn type="success">[[
- Question 2, Exercise 2.2
- nd add it in 2nd row of L.H.S to get R.H.S. ====Go To==== <text align="left"><btn type="primary">[[m
- Question 3, Exercise 2.2
- g (1) and (2), we have $$|A|=|{{A}^{t}}|.$$ ====Go To==== <text align="left"><btn type="primary">[[m