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- Question 1, Exercise 2.2
- 2 & 0 \\ 0 & -2 \\ \end{matrix} \right|$$ $$\implies{{A}_{11}}=-4$$ $${{A}_{21}}={{\left( -1 \right)}^... 3 & 0 \\ 1 & -2 \\ \end{matrix} \right|$$ $$\implies{{A}_{21}}=6$$ $${{A}_{23}}={{\left( -1 \right)}^{... 1 & 3 \\ 2 & 0 \\ \end{matrix} \right|$$ $$\implies{{A}_{23}}=6$$ $${{A}_{31}}={{\left( -1 \right)}^{... 3 & 1 \\ 2 & 0 \\ \end{matrix} \right|$$ $$\implies{{A}_{31}}=-2$$ $${{A}_{32}}={{\left( -1 \right)}^
- Question 3, Exercise 2.1
- \\-5+8 & 0+20 & -2+8 \\\end{matrix} \right]\\ \implies A( B+C )&=\left[ \begin{matrix}11 & 15 & 8 \\3 &... 5\\-2+5 & 15+5 & 11-5\\ \end{matrix} \right] \\ \implies AB+AC&=\left[ \begin{matrix} 11 & 15 & 8 \\ 3 & ... \\ 0-2 & 4-1 & 2-0 \\ \end{matrix} \right]\\ \implies B-C&=\left[ \begin{matrix}-1 & 2 & -8 \\ -2 & 3 ... 6 \\ 1-8 & -2+12 & 8+8\\ \end{matrix} \right]\\ \implies A(B-C)&=\left[ \begin{matrix} -7 & 11 & -2 \\ -7
- Question 5 & 6, Exercise 2.1
- t]$$ Since $A$ is given to be symmetirc, $A^t=A$, implies $$\left[ \begin{matrix} 0 & 3 & 3a \\ 2b &... ht]$$ This gives $$ 3a=-2 \text{ and } 2b=3,$$ $$\implies a=-\dfrac{2}{3} \text{ and } b=\dfrac{3}{2}.$$
- Question 4, Exercise 2.1
- & 0 \\ 0 & 0 & 0 \\ \end{matrix} \right] \\ \implies & \dfrac{1}{3}A^2-2A-9I=0 \end{align} ====Go To=
- Question 7, Exercise 2.1
- 2 \\2+1 & 5+4 & 6-1 \\ \end{matrix} \right]\\ \implies A^t+B^t&=\left[ \begin{matrix}3 & 4 & 3 \\-1 & 4
- Question 8, Exercise 2.1
- 2 & 0 \\3 & -1 & 4 \\ \end{matrix} \right]\\ \implies( A^t)^t&=A. \end{align} =====Question 8(ii)=====
- Question 2, Exercise 2.3
- & -\dfrac{6}{49} & \dfrac{8}{49} \end{bmatrix}\\ \implies \quad A^{-1}&=\dfrac{1}{49}\begin{bmatrix} 3 & 16