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Exercise 1.3.3

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\begin{matrix} y_{1} = B_{1} & y_{1} = B_{1} \\ y_{1} + y_{2} = B_{2} & gives y_{2} = - B_{1} + B_{2} \\ y_{1} + y_{2} + y_{3} = B_{3} & y_{3} = - B_{2} + B_{3} \end{matrix} = [\begin{matrix} 1 & 0 & 0 \\ - 1 & 1 & 0 \\ 0 & - 1 & 1 \end{matrix}] [\begin{matrix} B_{1} \\ B_{2} \\ B_{3} \end{matrix}]

The inverse of $S = [\begin{matrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{matrix}]$ is $A = [\begin{matrix} 1 & 0 & 0 \\ - 1 & 1 & 0 \\ 0 & - 1 & 1 \end{matrix}] :$ independent columns in $A$ and $S!$

mathchakra

2021-11-20 14:13

Exercise 1.3.3

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