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Determinants — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDeterminants3 Marks
Question
$\begin{array}{l}\text { Verify } A(\text { adj. A) }=(\operatorname{adj} . A) A=|A| \text { : } \\ {\left[\begin{array}{ccc}1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3\end{array}\right]}\end{array}$
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Answer

Let $A = \left[ {\begin{array}{*{20}{c}} 1&{ - 1}&2 \\ 3&0&{ - 2} \\ 1&0&3 \end{array}} \right]$  
$\Rightarrow \left| A \right| = \left[ {\begin{array}{*{20}{c}} 1&{ - 1}&2 \\ 3&0&{ - 2} \\ 1&0&3 \end{array}} \right]$ 
$\therefore {A_{11}} =+ \left| {\begin{array}{*{20}{c}} 0&{ - 2} \\ 0&3 \end{array}} \right| = + 0 + 0 = 0,$${A_{12}} = - \left| {\begin{array}{*{20}{c}} 3&{ - 2} \\ 1&3 \end{array}} \right| = - \left( {9 + 2} \right) = - 11$ 
${A_{13}} = + \left| {\begin{array}{*{20}{c}} 3&0 \\ 1&0 \end{array}} \right| = + \left( {0 - 0} \right) = 0,$${A_{21}} = - \left| {\begin{array}{*{20}{c}} { - 1}&2 \\ 0&3 \end{array}} \right| - \left( { - 3 - 0} \right) = 3$  
${A_{22}} = + \left| {\begin{array}{*{20}{c}} 1&2 \\ 1&3 \end{array}} \right| = 3 - 2 = 1,$${A_{23}} = - \left| {\begin{array}{*{20}{c}} 1&{ - 1} \\ 1&0 \end{array}} \right| = - \left( {0 + 1} \right) = - 1$ 
${A_{31}} = + \left| {\begin{array}{*{20}{c}} { - 1}&2 \\ 0&{ - 2} \end{array}} \right| = 2 - 0 = 2,$ ${A_{32}} = - \left| {\begin{array}{*{20}{c}} 1&2 \\ 3&{ - 2} \end{array}} \right| = - \left( { - 2 - 6} \right) = 8$ 
${A_{33}} = + \left| {\begin{array}{*{20}{c}} 1&{ - 1} \\ 3&0 \end{array}} \right| = 3 + 0 = 3$ 
$\therefore adj.A = \left| {\begin{array}{*{20}{c}} 0&{ - 11}&0 \\ 3&1&{ - 1} \\ 2&8&3 \end{array}} \right|$ 
$= \left| {\begin{array}{*{20}{c}} 0&3&2 \\ { - 11}&1&8 \\ 0&{ - 1}&3 \end{array}} \right|$ 
$\therefore A.\left( {adj.A} \right) = \left[ {\begin{array}{*{20}{c}} 1&{ - 1}&2 \\ 3&0&{ - 2} \\ 1&0&3 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 0&3&2 \\ { - 11}&1&8 \\ 0&{ - 1}&3 \end{array}} \right]$ 
$\left[ {\begin{array}{*{20}{c}} {0 + 11 + 0}&{3 - 1 - 2}&{2 - 8 + 6} \\ {0 - 0 - 0}&{9 + 0 + 2}&{6 + 0 - 6} \\ {0 + 0 + 0}&{3 + 0 - 3}&{2 + 0 + 9} \end{array}} \right]$ 
$ = \left[ {\begin{array}{*{20}{c}} {11}&0&0 \\ 0&{11}&0 \\ 0&0&{11} \end{array}} \right]$...(i)
Again  (adj. A). A $= \left[ {\begin{array}{*{20}{c}} 0&3&2 \\ { - 11}&1&8 \\ 0&{ - 1}&3 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&{ - 1}&2 \\ 3&0&{ - 2} \\ 1&0&3 \end{array}} \right]$ 
$= \left[ {\begin{array}{*{20}{c}} {0 + 9 + 2}&{0 + 0 + 0}&{0 - 6 + 6} \\ { - 11 + 3 + 8}&{11 + 0 + 0}&{ - 22 - 2 + 24} \\ {0 - 3 + 3}&{0 - 0 + 0}&{0 + 2 + 9} \end{array}} \right]$ 
$= \left[ {\begin{array}{*{20}{c}} {11}&0&0 \\ 0&{11}&0 \\ 0&0&{11} \end{array}} \right]$….(ii)
And $\left| A \right| = \left| {\begin{array}{*{20}{c}} 1&{ - 1}&2 \\ 3&0&{ - 2} \\ 1&0&3 \end{array}} \right|$ 
$= 1\left( {0 - 0} \right) - \left( { - 1} \right)\left( {9 + 2} \right) + 2\left( {0 - 0} \right) = 0 + 11 + 0 = 11$ 
Also $\left| A \right|I = \left| A \right|{I_3} = 11\left[ {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {11}&0&0 \\ 0&{11}&0 \\ 0&0&{11} \end{array}} \right]$...(iii)
$\therefore$ From eq. (i), (ii) and (iii)  A. (adj. A) = (adj. A). A = |A|

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