જો $A$ એ $3\times3$ શ્રેણિક છે કે જેથી
$A\left[ {\begin{array}{*{20}{c}}
1&2&3 \\
0&2&3 \\
0&1&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
0&0&1 \\
1&0&0 \\
0&1&0
\end{array}} \right]$
તો $A^{-1}$ મેળવો.
JEE MAIN 2014, Difficult
Download our app for free and get started
Given $A\left[ {\begin{array}{*{20}{c}}
1&2&3\\
0&2&3\\
0&1&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
0&0&1\\
1&0&0\\
0&1&0
\end{array}} \right]$
1&2&3\\
0&2&3\\
0&1&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
0&0&1\\
1&0&0\\
0&1&0
\end{array}} \right]$
Applying ${C_1} \leftrightarrow {C_3}$
$A\left[ {\begin{array}{*{20}{c}}
3&2&1\\
3&2&0\\
1&1&0
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1&0&0\\
0&0&1\\
0&1&0
\end{array}} \right]$
Again Applying ${C_2} \leftrightarrow {C_3}$
$A\left[ {\begin{array}{*{20}{c}}
3&1&2\\
3&0&2\\
1&0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1&0&0\\
0&1&0\\
0&0&1
\end{array}} \right]$
pre-multiplying both sides by ${A^{ - 1}}$
${A^{ - 1}}A\left[ {\begin{array}{*{20}{c}}
3&1&2\\
3&0&2\\
1&0&1
\end{array}} \right] = {A^{ - 1}}\left[ {\begin{array}{*{20}{c}}
1&0&0\\
0&1&0\\
0&0&1
\end{array}} \right]$
$I\left[ {\begin{array}{*{20}{c}}
3&1&2\\
3&0&2\\
1&0&1
\end{array}} \right] = {A^{ - 1}}I = {A^{ - 1}}$
($\because$ ${A^{ - 1}}A = I$ and $I=$ Identity matrix)
Hence, ${A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}
3&1&2\\
3&0&2\\
1&0&1
\end{array}} \right]$
Download our appand get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
Similar Questions
- 1જો $\left| {\,\begin{array}{*{20}{c}}a&b&{a\alpha + b}\\b&c&{b\alpha + c}\\{a\alpha + b}&{b\alpha + c}&0\end{array}\,} \right| = 0$ તો $a,b,c$ એ . . . .શ્રેણીમાં છે .View Solution
- 2$\lambda$ અને $\mu$ ની કિમંત મેળવો કે જેથી સમીકરણ સંહતિ $x+y+z=6,3 x+5 y+5 z=26, x+2 y+\lambda z=\mu$ નો ઉકેલગણ ખાલીગણ થાય.View Solution
- 3$\left| {\,\begin{array}{*{20}{c}}{265}&{240}&{219}\\{240}&{225}&{198}\\{219}&{198}&{181}\end{array}\,} \right| =$View Solution
- 4જો $A = \left[ {\begin{array}{*{20}{c}}{\cos \alpha }&{ - \sin \alpha }\\{\sin \alpha }&{\cos \alpha }\end{array}} \right]$ અને $B = \left[ {\begin{array}{*{20}{c}}{\cos \beta }&{ - \sin \beta }\\{\sin \beta }&{\cos \beta }\end{array}} \right]$, તો સાચો સંબંધ મેળવો.View Solution
- 5જો $A=\left[\begin{array}{cc}8 & 0 \\ 4 & -2 \\ 3 & 6\end{array}\right]$ અને $B=\left[\begin{array}{cc}2 & -2 \\ 4 & 2 \\ -5 & 1\end{array}\right]$ તો $2 \mathrm{A}+3 \mathrm{X}=5 \mathrm{B}$ થાય એવો શ્રેણિક $X$ શોધો.View Solution
- 6$\left| {\,\begin{array}{*{20}{c}}{{b^2} - ab}&{b - c}&{bc - ac}\\{ab - {a^2}}&{a - b}&{{b^2} - ab}\\{bc - ac}&{c - a}&{ab - {a^2}}\end{array}\,} \right| = $View Solution
- 7બે સમાન કક્ષાના ચોરસ શ્રેણિક $A$ અને $B$ માટે આપેલ પૈકી ક્યૂ વિધાન સત્ય છે .View Solution
- 8જો $A=$ $\left[ {\begin{array}{*{20}{c}}{5a}&{ - b}\\3&2\end{array}} \right]$ અને $A\;adj\;A = A\;{A^T},$તો $5a+b= $. . . . .View Solution
- 9જો $A, B, C$ એ ત્રિકોણના ખૂણા હોય તો નિશ્ચાયક $\left| {\begin{array}{*{20}{c}}View Solution
{\sin \,2A}&{\sin \,C}&{\sin \,B} \\
{\sin \,C}&{\sin \,2B}&{\sin A} \\
{\sin \,B}&{\sin \,A}&{\sin \,2C}
\end{array}} \right|$ ની કિમંત મેળવો. - 10$\Delta \text{ABC}$ માં , જો $\left| {\,\begin{array}{*{20}{c}}1&a&b\\1&c&a\\1&b&c\end{array}\,} \right| = 0$, તો ${\sin ^2}A + {\sin ^2}B + {\sin ^2}C = $View Solution