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- Question 16 & 17, Exercise 2.2
- 4 & 2 \\ \end{matrix} \right]$$ $$|A|=6+4$$ $$\Rightarrow |A|=10\ldots (1)$$ $$A^{-1}=\dfrac{1}{|A|}AdjA$$ ... & 11 \\ -1 & 1 \\ \end{matrix} \right]$$ $$\Rightarrow |AB|=4+11$$ $$\Rightarrow \,\,|AB|=15$$ $$AdjAB=\left[ \begin{matrix} 1 & -11 \\ 1 & 4 \\ \end{matrix... & -11 \\ 1 & 4 \\ \end{matrix} \right]$$ $$\Rightarrow ( AB )^{-1}=\left[ \begin{matrix} \dfrac{1}{15
- Question 1, Exercise 2.3
- 1 & 3 & \quad 2 \end{bmatrix}\text{ by } R_1\leftrightarrow R_2\\ \underset{\sim}{R}& \begin{bmatrix} 1 & -1 ... & 1 \\ 4 & 1 & 7 \end{bmatrix} \text{by}R_1\leftrightarrow R_2\\ \underset{\sim}{R}&\begin{bmatrix} 1 & 1 &
- Question 12, Exercise 2.2
- \lambda (1-(\lambda ^2-1))$$ $A$ is singular. $$\Rightarrow |A|=0$$ $$\lambda (1-(\lambda ^2-1))=0$$ $$\lamb
- Question 2, Exercise 2.3
- \ \end{matrix} \right. \right]\text{ by }R_1 \leftrightarrow R_1\\ \underset{\sim}{R}&\left[ \begin{matrix} 1
- Question 3, Exercise 2.3
- 1 \\ 3 & 1 & -4 \end{bmatrix} \text{ by }R_1\leftrightarrow R_3\\ \underset{\sim}{R}&\begin{bmatrix} 1 & -1 &
- Question 4, Exercise 2.3
- \ 9 & 10 & 11 & 12 \end{bmatrix} \text{by}R_1\leftrightarrow R_2\\ \underset{\sim}{R}&\begin{bmatrix} 1 & 1 &