Question #289811

Find A if (A-1-3I)T= 2 [-1 2

5 4]


1
Expert's answer
2022-01-25T10:48:29-0500

Solution:

Given,

(A13I)T=2[1254](A1)T(3I)T=[24108](AT)13I=[24108](AT)1=[24108]+3I(AT)1=[24108]+[3003](AT)1=[141011](AT)=[141011]1(AT)=11140[114101](AT)=129[114101]A=129[114101]TA=129[111041](A^{-1}-3I)^T=2\begin{bmatrix} -1 & 2\\5 &4\end{bmatrix} \\ \Rightarrow (A^{-1})^T-(3I)^T=\begin{bmatrix} -2 & 4\\10 &8\end{bmatrix} \\ \Rightarrow (A^T)^{-1}-3I=\begin{bmatrix} -2 & 4\\10 &8\end{bmatrix} \\ \Rightarrow (A^T)^{-1}=\begin{bmatrix} -2 & 4\\10 &8\end{bmatrix}+3I \\ \Rightarrow (A^T)^{-1}=\begin{bmatrix} -2 & 4\\10 &8\end{bmatrix}+\begin{bmatrix} 3 & 0\\0 &3\end{bmatrix} \\ \Rightarrow (A^T)^{-1}=\begin{bmatrix} 1 & 4\\10 &11\end{bmatrix} \\ \Rightarrow (A^T)=\begin{bmatrix} 1 & 4\\10 &11\end{bmatrix}^{-1} \\ \Rightarrow (A^T)=\dfrac{1}{11-40}\begin{bmatrix} 11 & -4\\-10 &1\end{bmatrix} \\ \Rightarrow (A^T)=-\dfrac{1}{29}\begin{bmatrix} 11 & -4\\-10 &1\end{bmatrix} \\ \Rightarrow A=-\dfrac{1}{29}\begin{bmatrix} 11 & -4\\-10 &1\end{bmatrix}^T \\ \Rightarrow A=-\dfrac{1}{29}\begin{bmatrix} 11 & -10\\-4 &1\end{bmatrix}


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