જો A = left[ {begin{array}{*{20}{c}}3&{ - 3}&42&{ - 3}&40&{ - 1}&1end{array}} right], તો {A^{

Q: જો $A = \left[ {\begin{array}{*{20}{c}}3&{ - 3}&4\\2&{ - 3}&4\\0&{ - 1}&1\end{array}} \right]$, તો ${A^{ - 1}}$=

Here, $C_{11}=1, C_{12}=-2, C_{13}$ $=-2\ C_{21}=1, C_{22}=-2, C_{23}=3$ $C_{31}=1, C_{32}=-2, C_{33}=-3$ $\Rightarrow\ \text{det}\ A=|A| = \left[ {\begin{array}{*{20}{c}} 3&{ - 3}&4 \\ 2&{ - 3}&4 \\ 0&{ - 1}&1 \end{array}} \right]$ $\Rightarrow\ A^{-1} = {1\over{|A|}} .(\text{Adj } A) = {1\over1}\ \left[ {\begin{array}{*{20}{c}} {{C_{11}}}&{{C_{21}}}&{{C_{31}}} \\ {{C_{12}}}&{{C_{22}}}&{{C_{32}}} \\ {{C_{13}}}&{{C_{23}}}&{{C_{33}}} \end{array}} \right]$ Now, ${A^2} = \left[ {\,\begin{array}{*{20}{c}}3&{ - 4}&4\\0&{ - 1}&0\\{ - 2}&2&{ - 3}\end{array}\,} \right]$ and ${A^3} = {A^2}.A = \left[ {\begin{array}{*{20}{c}}3&{ - 4}&4\\0&{ - 1}&0\\{ - 2}&2&{ - 3}\end{array}} \right]\, \times \,\left[ {\begin{array}{*{20}{c}}3&{ - 3}&4\\2&{ - 3}&4\\0&{ - 1}&1\end{array}} \right]$ $ = \left[ {\,\begin{array}{*{20}{c}}1&{ - 1}&0\\{ - 2}&3&{ - 4}\\{ - 2}&3&{ - 3}\end{array}\,} \right] = {A^{ - 1}}$.

MCQ

ગુજરાતી માધ્યમ

જો $A = \left[ {\begin{array}{*{20}{c}}3&{ - 3}&4\\2&{ - 3}&4\\0&{ - 1}&1\end{array}} \right]$, તો ${A^{ - 1}}$=

A$A$
B${A^2}$
C${A^3}$
D${A^4}$

Easy

ધોરણ 12 સાયન્સ
ગણિત
3 and 4 . determinant and metrices
JEE

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