Question

solve this.....................​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

$\: \: \: \: \: \: \: \: \: \underline{ \underline{{ \huge{ \mathfrak { \: S}}}{ \textbf{ \textsf{OLUTION - \: }}}}}$

${ \tt{A = diag[a_1 , a_2 , a_3]}}$

${ \tt{n \geqslant 1}}$

${ \tt{To \: prove :-}} \\ { \tt{A = diag[a_1 ^{n} , a_2 {}^{n} , a_3 {}^{n} ]}}$

Now,

${ \tt{A = diag[a_1 , a_2 , a_3]}} \\ \\ \implies{ \tt{A = \left[ \begin{array}{c c } \sf {a}_{1} & \sf 0 & \sf 0\\\\ \sf 0& \sf \: {a}_{2}& \sf 0 \\ \\ \sf 0& \sf 0 & \sf {a}_{3} \end{array} \right]_{ 3×3}}} \\ \\ {\tt{L.H.S ,}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \implies{ \tt{ {A}^{n} =\left[ \begin{array}{c c } \sf {a}_{1} ^{n} & \sf 0 & \sf 0\\\\ \sf 0& \sf \: {a}_2^{n}& \sf 0 \\ \\ \sf 0& \sf 0 & \sf {a}_3^{n} \end{array} \right] }} \\ \\ \\ \implies{ \tt{diag[a_1 {}^{n} , a_2 {}^{n} , a_3 {}^{n} ]}}$

$\implies{ \tt{R.H.S = diag[a_1^n , a_2^n , a_3^n]}} \\ \\ \\ \therefore{ \tt{ L.H.S= R.H.S \: \: \: \: \: \: \: (proved)}}$

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