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Question Number 171064 by 119065 last updated on 07/Jun/22

$$When{A}^{-1}=\left[\begin{array}{cc}3& 1\\ 8& 4\end{array}\right]$$$$findtheA=?,\mid A^{-1}\mid \cdot A=?$$

Answered by som(math1967) last updated on 07/Jun/22

$$Adj{A}^{-1}={\left[\begin{array}{cc}4& -8\\ -1& 3\end{array}\right]}^T=\left[\begin{array}{cc}4& -1\\ -8& 3\end{array}\right]$$$$\mid A^{-1}\mid =12-8=4$$$$A=\frac{1}{4}\left[\begin{array}{cc}4& -1\\ -8& 3\end{array}\right]$$$$\mid A^{-1}\mid A=4\times \frac{1}{4}\left[\begin{array}{cc}4& -1\\ -8& 3\end{array}\right]=\left[\begin{array}{cc}4& -1\\ -8& 3\end{array}\right]$$

Commented by 119065 last updated on 07/Jun/22

$$nicesolution$$

Answered by manxsol last updated on 05/Feb/23

$$teoria$$$$A^{-1}=\frac{AdjA}{\mid A\mid }\Rightarrow A=\frac{{AdjA}^{-1}}{\mid A^{-1}\mid }=\frac{{\left({cofactA}^{-1}\right)}^T}{\mid A^{-1}\mid }$$$$A.\mid A^{-1}\mid ={\left({cofactA}^{-1}\right)}^T$$$$$$

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