Chapter 20: Problem 7
\(S^{-1}(T(z))=\frac{a z+b}{c z+d}\) where $$\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]=\operatorname{adj}\left(\left[\begin{array}{ll} 1 & -2 \\ 1 & -1 \end{array}\right]\right)\left[\begin{array}{ll} 2 & -3 \\ 1 & -3 \end{array}\right]=\left[\begin{array}{rr} 0 & -3 \\ -1 & 0 \end{array}\right].$$ Therefore, $$S^{-1}(T(z))=\frac{-3}{-z}=\frac{3}{z} \quad \text { and } \quad S^{-1}(w)=\frac{-w+2}{-w+1}=\frac{w-2}{w-1}.$$
Short Answer
Step by step solution
Understand the Problem
Calculate the Adjugate of the Given Matrix
Multiply the Adjugate Matrix by the Second Matrix
Determine Values of a, b, c, d
Substitute Values into the Fraction Expression
Analyze the Second Part of the Exercise
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