Question

By using elementary row transformations, find the inverse of
2 0 - 1
A = 5 1 0
0 1 3​

Answer 1

Answer:

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→$\left[\begin{array}{ccc}2&0&-1\\5&1&0\\0&1&3\end{array}\right]$=  A,

$\left[\begin{array}{ccc}2&0&-1\\5&1&0\\0&1&3\end{array}\right] =\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]$,

→$\large\boxed{R_1=\frac{R_2}{2}}$,

→$\left[\begin{array}{ccc}1&0&\frac{-1}{2} \\5&1&0\\0&1&3\end{array}\right] =\left[\begin{array}{ccc}\frac{1}{2} &0&0\\0&1&0\\0&0&1\end{array}\right]$,

→$\large\boxed{R_1=R_2-5R_1}$

→$\left[\begin{array}{ccc}1&0&\frac{-1}{2} \\0&1&\frac{5}{2} \\0&1&3\end{array}\right] =\left[\begin{array}{ccc}\frac{1}{2} &0&0\\\frac{-5}{2} &1&0\\0&0&1\end{array}\right]$,

→$\large\boxed{R_3-R_3-R_2}$,

→$\left[\begin{array}{ccc}1&0&\frac{-1}{2} \\0&1&\frac{5}{2} \\0&0&\frac{1}{2} \end{array}\right] =\left[\begin{array}{ccc}\frac{1}{1} &0&0\\\frac{-5}2} &0&0\\\frac{5}{2} &-1&1\end{array}\right]$,

→$\large\boxed{R_3=2R_3}$

→$\left[\begin{array}{ccc}1&0&\frac{-1}{2} \\0&1&\frac{5}{2} \\0&0&1\end{array}\right] =\left[\begin{array}{ccc}\frac{1}{2} &0&0\\\frac{-5}{2} &1&0\\5&-2&2\end{array}\right]$,

→$\large\boxed{{R_1=R_1+\frac{R_3}{2} };R_2=R_2-\frac{5}{2} R_3}$,

→$\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right] = \left[\begin{array}{ccc}3&-1&1\\-15&6&-5\\5&-2&2\end{array}\right]$,

→$So, A^-1=\left[\begin{array}{ccc}3&-1&-1\\-15&6&-5\\5&-2&-2\end{array}\right].$

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