Question

If , p = [ R1 (10 - 2 ) R2 (-5 1) ]

Then find the value of P^(-1 )

Answer 1

$\bold{\huge{\underline{Question}}}$


if, $P = \left[\begin{array}{ccc}10& - 2\\ - 5&1\end{array}\right]$ then find the value of ${p}^{1}$



$\bold{\huge{\underline{Solution-}}}$



$\bold{\huge{\underline {\underline{\star\:METHOD-1}}}}$



$\bf In \: this \: type \: of \: question \: , remove \: \\ \bf \: the \: value \: of \: the \: baryakic \: \\ \bf \: if \: the \: value \: of \: the \: tabular \: is \: \\ \bf \: zero \: then \: P {}^{ - 1} will \: not \\ \bf \: come \: out.$




$\bf{|P| = \left[\begin{array}{ccc} 10& - 2\\ - 5&1\end{array}\right]} \\ \\ \\ \bf \implies \: 10 \times 1 - ( - 5)( - 2) \\ \\ \\ \bf \implies0 \\ \\ \\ \boxed{ \bf \underline{Here \:|P| \: will \: come \: 0 \: so \: we \: does \: not \: find \: the \: value \: of \: {p}^{ - 1} } }$






$\bold{\huge{\underline {\underline{\star\:METHOD-2}}}}$






$P \: = \left[\begin{array}{ccc}10& - 2\\ - 5&1\end{array}\right]$




$\bf \: By \: applying \: the \: initial \: column \: conversion \:$





$\bf \huge \underline{\star \: Note - } \\ \\ \bf \: When \: determining \: the \: first \: row \: \\ \bf \: or \: column \: by \: {A}^{ - 1} , \: if \: no \: \\ \bf line \: of \: viewer \: or \: any \: element \: of \: a \: \\ \bf \: column \: come s \: to \: zero \: then \: \\ \bf {A}^{ - 1} \: does \: not \: come \: out$





$\bf \: We \: Know \: that \: Formula \: \: \: \: \: \: \: \: \boxed{\boxed {\bf A = I \times \: A }}$





$\left[\begin{array}{ccc}10& - 2\\ - 5&1\end{array}\right] = \left[\begin{array}{ccc}1& 0\\ 0&1\end{array}\right] p\\ \\ \\ \\ \implies \: \bf \{c _{1} \: -> c _{1} + 5 \times c _{2} \} \\ \\ \\ \implies \: \left[\begin{array}{ccc}0& - 2\\ 0&1\end{array}\right] = \left[\begin{array}{ccc}1& 0\\ 5&1\end{array}\right]$





If the components of the first place are zero in the matrix of the left side then A^(-1) is does not exist