Let B=(1023)B = \begin{pmatrix} 1 & 0 \\ 2 & 3 \end{pmatrix}B=(1203)
What is B-1?
Given (1023)\begin{pmatrix} 1& 0 \\ 2 & 3 \end{pmatrix}(1203)
A=(abcd)A=\begin{pmatrix} a & b \\ c & d \end{pmatrix}A=(acbd)
A−1=1ad−bc(d−b−ca)A^{-1}=\frac{1}{ad-bc}\begin{pmatrix} d &- b \\ - c & a \end{pmatrix}A−1=ad−bc1(d−c−ba)
B−1=11×3−2×0(30−21)B^{-1}=\frac{1}{1×3-2×0}\begin{pmatrix} 3 & 0 \\ -2 & 1 \end{pmatrix}B−1=1×3−2×01(3−201)
B−1=13(30−21)B^{-1}=\frac{1}{3}\begin{pmatrix} 3 & 0\\ -2 & 1 \end{pmatrix}B−1=31(3−201)
B−1=(10−2313)B^{-1}=\begin{pmatrix} 1 & 0 \\ \frac{-2}{3} & \frac{1}{3} \end{pmatrix}B−1=(13−2031)
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