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Sia ? 6 years, 4 months ago
According to the question we are required to find matrix A such that {tex} \left[ \begin{array} { c c } { 2 } & { - 1 } \\ { 1 } & { 0 } \\ { - 3 } & { 4 } \end{array} \right] A = \left[ \begin{array} { c c } { - 1 } & { - 8 } \\ { 1 } & { - 2 } \\ { 9 } & { 22 } \end{array} \right]{/tex} .
Let us take the order of matrix A as m{tex}\times{/tex}n {tex}{/tex},therefore, m = 2, n = 2.
Let {tex}A = \left[ \begin{array} { l l } { x } & { y } \\ { s } & { t } \end{array} \right]{/tex} ...(i)
{tex}\therefore \left[ \begin{array} { c c } { 2 } & { - 1 } \\ { 1 } & { 0 } \\ { - 3 } & { 4 } \end{array} \right] A = \left[ \begin{array} { c c } { - 1 } & { - 8 } \\ { 1 } & { - 2 } \\ { 9 } & { 22 } \end{array} \right]{/tex}
{tex}\Rightarrow \left[ \begin{array} { c c } { 2 } & { - 1 } \\ { 1 } & { 0 } \\ { - 3 } & { 4 } \end{array} \right] \left[ \begin{array} { l l } { x } & { y } \\ { s } & { t } \end{array} \right] = \left[ \begin{array} { c c } { - 1 } & { - 8 } \\ { 1 } & { - 2 } \\ { 9 } & { 22 } \end{array} \right]{/tex}
{tex}\Rightarrow \left[ \begin{array} { c c } { 2 x - s } & { 2 y - t } \\ { x } & { y } \\ { - 3 x + 4 s } & { - 3 y + 4 t } \end{array} \right] = \left[ \begin{array} { c c } { - 1 } & { - 8 } \\ { 1 } & { - 2 } \\ { 9 } & { 22 } \end{array} \right]{/tex}
Therefore,on equating corresponding elements both sides, we get,
2 - s = -1, x = 1, y = -2 and 2y - t = -8
At x = 1, 2x - s = -1 {tex}\Rightarrow{/tex} 2{tex}\times{/tex}1 - s = -1
{tex}\Rightarrow{/tex} -s = -1 - 2 {tex}\Rightarrow{/tex} s = 3 and at y = -2, 2y - t = -8,
{tex}\Rightarrow{/tex} 2{tex}\times{/tex}(-2) - t = -8 {tex}\Rightarrow{/tex} -4 - t = -8 {tex}\Rightarrow{/tex} t = 4
Therefore,on putting x = 1, y = -2, s = 3 and t = 4 in Eq. (i),
we get, {tex}A = \left[ \begin{array} { c c } { 1 } & { - 2 } \\ { 3 } & { 4 } \end{array} \right]{/tex}
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Sia ? 6 years, 4 months ago
Modulus Function f : R {tex}\rightarrow{/tex} R, given by f(x) = |x| + 2
One-one: f(1) = |1| + 2 = 3 and f(2) = |2| + 2 = 4,so distinct elements have same image.So, f is not one-one.
Onto: f takes only positive values, so range(f) = set of positive real numbers {tex}\neq{/tex} R, codomain. So, f is not onto.
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Sia ? 6 years, 4 months ago
{tex}s = \pi {r^2} + \pi rl{/tex} (given)
{tex}\Rightarrow{/tex}{tex}l = \frac{{s - \pi {r^2}}}{{\pi r}}{/tex}
Let v be the volume
{tex}v = \frac{1}{3}\pi {r^2}h{/tex}
{tex}\Rightarrow{/tex}{tex}{v^2} = \frac{1}{9}{\pi ^2}{r^4}{h^2}\,\,\left[ {{h^2} = {l^2} - {r^2}} \right]{/tex}
{tex}\Rightarrow{/tex}{tex}{v^2} = \frac{1}{9}{\pi ^2}{r^4}\,\left( {{l^2} - {r^2}} \right){/tex}
{tex}\Rightarrow{/tex}{tex}{v^2} = \frac{1}{9}{\pi ^2}{r^4}\left[ {\,{{\left( {\frac{{s - \pi {r^2}}}{{\pi r}}} \right)}^2} - {r^2}} \right]{/tex}
{tex}= \frac{1}{9}{\pi ^2}{r^4}\left[ {\frac{{\left( {s - \pi {r^2}} \right)}^2}{{{\pi ^2}{r^2}}} - \frac{{{r^2}}}{1}} \right]{/tex}
{tex}= \frac{1}{9}{r^2}\left[ {{{\left( {s - \pi {r^2}} \right)}^2} - {\pi ^2}{r^4}} \right]{/tex}
{tex}= \frac{1}{9}{r^2}\left[ {{s^2} + {\pi ^2}{r^4} - 2s\pi {r^2} - {\pi ^2}{r^4}} \right]{/tex}
{tex}= \frac{1}{9}{r^2}\left[ {{s^2} - 2s\pi {r^2}} \right]{/tex}
{tex}z = \frac{1}{9}\left[ {{s^2}{r^2} - 2s\pi {r^4}} \right]{/tex}
{tex}\left[ {\because {v^2} = z} \right]{/tex}
Now {tex}\frac{{dz}}{{dr}} = \frac{1}{9}\,\left[ {2r{s^2} - 8s\pi {r^3}} \right]{/tex}
{tex}0 = \frac{1}{9}\,\left[ {2r{s^2} - 8s\pi {r^3}} \right]{/tex}
{tex}8s\pi {r^2} = 2r{s^2}{/tex}
{tex}\implies{/tex} {tex}4\pi {r^2} = s{/tex}
Now {tex}\frac{{{d^2}z}}{{d{x^2}}} = \frac{1}{9}\,\left[ {2{s^2} - 24s\pi {r^2}} \right]{/tex}
{tex}{\left. {\frac{{{d^2}z}}{{d{x^2}}}} \right]_{{r^2} = \frac{s}{{4\pi }}}} = \frac{1}{9}\,\left[ {{{25}^2} - 24\pi .\frac{5}{{4\pi }}} \right]{/tex}
= + ve
Hence minimum
Now {tex}s = 4\pi {r^2}{/tex}
We have {tex}s = \pi rl + \pi {r^2}{/tex}
{tex}4\pi {r^2} = \pi rl + \pi {r^2}{/tex}
{tex}\Rightarrow{/tex} {tex}3\pi {r^2} = \pi rl{/tex}
{tex}\Rightarrow{/tex}3 r = l
{tex}\Rightarrow{/tex}{tex}\frac{r}{l} = \frac{1}{3}{/tex}
{tex}\Rightarrow{/tex}{tex}\sin \alpha = \frac{1}{3}{/tex}
{tex}\therefore{/tex}{tex}\alpha = {\sin ^{ - 1}}\left( {\frac{1}{3}} \right){/tex}
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Shalini Sharma 6 years, 4 months ago
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