Question

If A = [ First Row - 1 3 3 , Second Row - 1 4 3 , Third Row- 1 3 4 ], verify that A.adj = |A|.I and A^(-1) know ?

Answer 1

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


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




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



$\bf \: Firstly \: we \: extract \: co \: -factor \: of \: elements \: of \: |A| \:$




$\bf \: A _{11} = \left|\begin{array}{ccc}4&3\\3&4\end{array}\right| \: \: \: \: \: \: \: \: \: \: \: \: \: \: A _{12} \: = - \left|\begin{array}{ccc}1&3\\1&4\end{array}\right| \\ \\ \implies \: 16 - 9 = 7 \: \: \: \: \: \: \: \: \: \: \: \implies - (4 - 3) = - 1 \\ \\ \\ \bf \: A _{13} = \left|\begin{array}{ccc}1&4\\1&3\end{array}\right| \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: A _{21} = - \left|\begin{array}{ccc}3&3\\3&4\end{array}\right| \\ \\ \implies3 - 4 = - 1 \: \: \: \: \: \: \: \: \: \: \: \: \implies- (12 - 19) = - 3 \\ \\ \\ \bf \: A _{22} =\left|\begin{array}{ccc}1&3\\1&4\end{array}\right| \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: A _{23} = - \left|\begin{array}{ccc}1&3\\1&3\end{array}\right| \\ \\ \implies4 - 3 = 1 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \implies - (3 - 3) = 0 \\ \\ \\ \bf \:A _{31} = \left|\begin{array}{ccc}3&3\\4&3\end{array}\right| \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: A _{32} = - \left|\begin{array}{ccc}1&3\\1&3\end{array}\right| \\ \\ \implies9 - 12 = - 3 \: \: \: \: \: \: \: \: \: \: \: \implies - (3 - 3) = 0 \\ \\ \\ \bf \: A _{33} = \left|\begin{array}{ccc}1&3\\1&4\end{array}\right| \: \implies4 - 3 = 1$



$\bf \: Matrix \: formed \: from \: the \: coefficient \: of \: elements \: of \: |A|$




$\bf \: B = \left[\begin{array}{ccc}7& - 1& - 1\\ - 3&1&0\\ - 3&0&1\end{array}\right]$





${\: Change \: matrix \: of \: } adjA = B$



$= \left[\begin{array}{ccc}7& - 3& - 3\\ - 1&1&0\\ - 1&0&1\end{array}\right]$



$A.adj A= \left[\begin{array}{ccc}1&3&3\\1&4&3\\1&3&4\end{array}\right]\left[\begin{array}{ccc}7& - 3& - 3\\1&1&0\\1&0&1\end{array}\right] \\ \\ \implies\left[\begin{array}{ccc}7 - 3 - 3 \: \: &3 + 3 + 0 \: \: & - 3 + 0 + 3\\7 - 4 - 3 \: &3 + 4 + 0& - 3 + 0 + 3\\7 - 3 - 4& - 3 + 3 + 0& - 3 + 0 + 4\end{array}\right] \\ \\ \implies\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right] \\ \\ \\ \implies \: A = \left[\begin{array}{ccc}1&3&3\\1&4&3\\1&3&4\end{array}\right] \\ \\ \\ |A| = 1(16 - 9) - 3(4 - 3) + 3(3 - 4) \\ \: \: \: \: \: \: = 7 - 3 - 3 = 1 \\ \\ \\ |A| .I = 1.\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right] = \left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right] \\ \\ \: \: \: \: \: \: \: \: \: \: \: \boxed{ {A}^{ - 1} = \frac{1}{ |A| } .adjA }\: \\ \\ \\ = \frac{1}{1} \left[\begin{array}{ccc}7& - 3& - 3\\ - 1&1&0\\ - 1&0&1\end{array}\right] \\ \\ = \left[\begin{array}{ccc}7& - 3& - 3\\ - 1&1&0\\ - 1&0&1\end{array}\right]$