Question

Convert [1 3][2 4] in to upper triangular matrix using row transformations.
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Answer 1

$\tiny \text{Find the inverse of the matrix \(A=\left[\begin{array}{ll}1 & 3 \\ 2 & 4\end{array}\right] \),using elementary row transformations.}$

Given,

$A=\left[\begin{array}{ll}3 & 1 \\ 4 & 2\end{array}\right]$

$\tt\[ |A|=\left|\begin{array}{ll} 3 & 1 \\ 4 & 2 \end{array}\right|=6-4=2 \neq 0 \Rightarrow A^{-1} \text { exist. } \]$

Co-factors of |A| are :

$\color{blue}{\begin{array}{l} \tt A_{11}=(-1)^{2}(2)=2 \\ \\ \tt A_{12}=(-1)^{3}(4)=-4 \\ \\ \tt A_{21}=(-1)^{3}(1)=-1 \\ \\ \tt A_{22}=(-1)^{4}(3)=3 \end{array}}$

$\red{ \therefore \quad \operatorname{AdjA} \tt=\left[\begin{array}{ll}A_{11} & A_{12} \\ A_{21} & A_{22}\end{array}\right]^{T}\left[\begin{array}{ll}A_{11} & A_{21} \\ A_{12} & A_{2}\end{array}\right]=\left[\begin{array}{cc}2 & -1 \\ -4 & 3\end{array}\right] }$

Now,

$\tt A^{-1}=\frac{1}{|A|} \operatorname{adj}(A)=\frac{1}{2}\left[\begin{array}{cc}2 & -1 \\ -4 & 3\end{array}\right]$

$\color{darkcyan} \therefore \quad \tt A^{-1}=\left[\begin{array}{cc}1 & - \dfrac{1}{2} \\ \\ -2 & \: \: \: \dfrac{3}{2} \end{array}\right]$