Question
Verify $A(\operatorname{adj} . A)=(\operatorname{adj} . A) A=|A| \mid$ : $\left[\begin{array}{cc} 2 & 3 \\ -4 & -6 \end{array}\right] $
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Answer
Let $A = \left[ {\begin{array}{*{20}{c}} 2&3 \\ { - 4}&{ - 6} \end{array}} \right]$
$\Rightarrow adj.A = \left[ {\begin{array}{*{20}{c}} { - 6}&{ - 3} \\ 4&2 \end{array}} \right]$
$ \Rightarrow A.\left( {adj.A} \right) = \left[ {\begin{array}{*{20}{c}} 2&3 \\ { - 4}&{ - 6} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { - 6}&{ - 3} \\ 4&2 \end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}} { - 12 + 12}&{ - 6 + 6} \\ {24 - 24}&{12 - 12} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&0 \\ 0&0 \end{array}} \right]$ ...(i)
Again (adj. A). A $ = \left[ {\begin{array}{*{20}{c}} { - 6}&{ - 3} \\ 4&2 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 2&3 \\ { - 4}&{ - 6} \end{array}} \right]$
$= \left[ {\begin{array}{*{20}{c}} { - 12 + 12}&{ - 18 + 18} \\ {8 + 8}&{12 - 12} \end{array}} \right]=\left[ {\begin{array}{*{20}{c}} 0&0 \\ 0&0 \end{array}} \right]$…..(ii)
And $\left| A \right| = \left| {\begin{array}{*{20}{c}} 2&3 \\ { - 4}&{ - 6} \end{array}} \right| = 2\left( { - 6} \right) - 3\left( { - 4} \right) = - 12 + 12 = 0$
Again $\left| A \right|I = \left| A \right|{I_2} = \left( 0 \right)\left[ {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&0 \\ 0&0 \end{array}} \right]$....(iii)
From eq. (i), (ii) and (iii) A. (adj. A) = (adj. A). A = |A|I
$\Rightarrow adj.A = \left[ {\begin{array}{*{20}{c}} { - 6}&{ - 3} \\ 4&2 \end{array}} \right]$
$ \Rightarrow A.\left( {adj.A} \right) = \left[ {\begin{array}{*{20}{c}} 2&3 \\ { - 4}&{ - 6} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { - 6}&{ - 3} \\ 4&2 \end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}} { - 12 + 12}&{ - 6 + 6} \\ {24 - 24}&{12 - 12} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&0 \\ 0&0 \end{array}} \right]$ ...(i)
Again (adj. A). A $ = \left[ {\begin{array}{*{20}{c}} { - 6}&{ - 3} \\ 4&2 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 2&3 \\ { - 4}&{ - 6} \end{array}} \right]$
$= \left[ {\begin{array}{*{20}{c}} { - 12 + 12}&{ - 18 + 18} \\ {8 + 8}&{12 - 12} \end{array}} \right]=\left[ {\begin{array}{*{20}{c}} 0&0 \\ 0&0 \end{array}} \right]$…..(ii)
And $\left| A \right| = \left| {\begin{array}{*{20}{c}} 2&3 \\ { - 4}&{ - 6} \end{array}} \right| = 2\left( { - 6} \right) - 3\left( { - 4} \right) = - 12 + 12 = 0$
Again $\left| A \right|I = \left| A \right|{I_2} = \left( 0 \right)\left[ {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&0 \\ 0&0 \end{array}} \right]$....(iii)
From eq. (i), (ii) and (iii) A. (adj. A) = (adj. A). A = |A|INeed a full question paper?
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