Find A if (A-1-3I)T= 2 [-1 2
5 4]
Given,
(A−1−3I)T=2[−1254]⇒(A−1)T−(3I)T=[−24108]⇒(AT)−1−3I=[−24108]⇒(AT)−1=[−24108]+3I⇒(AT)−1=[−24108]+[3003]⇒(AT)−1=[141011]⇒(AT)=[141011]−1⇒(AT)=111−40[11−4−101]⇒(AT)=−129[11−4−101]⇒A=−129[11−4−101]T⇒A=−129[11−10−41](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}(A−1−3I)T=2[−1524]⇒(A−1)T−(3I)T=[−21048]⇒(AT)−1−3I=[−21048]⇒(AT)−1=[−21048]+3I⇒(AT)−1=[−21048]+[3003]⇒(AT)−1=[110411]⇒(AT)=[110411]−1⇒(AT)=11−401[11−10−41]⇒(AT)=−291[11−10−41]⇒A=−291[11−10−41]T⇒A=−291[11−4−101]
#339914 on Dec 2023