Question

$please \: answer \: only \: the \: circle \: ones \: please$
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Answer 1

$: \rightarrow \underline{ \underline{ \mathfrak{Required \: \: \: Solution :-}}}$

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$\sf \: If \: 2\left [ \begin{array}{c c } \sf 3& \sf x \\\\ \sf \: 0& \sf \: 1 \end{array} \right] + 3 \left[ \begin{array}{c c } \sf 1& \sf 3 \\\\ \sf \: y& \sf \: 2 \end{array} \right] =\left[ \begin{array}{c c } \sf z& \sf - 7 \\\\ \sf \: 15& \sf \: 8\end{array} \right] ;$

find the values of x , y and z.

${\huge{⇾}}\left[ \begin{array}{c c } \sf 6& \sf 2x \\\\ \sf \: 0& \sf \: 2 \end{array} \right] + \left[ \begin{array}{c c } \sf 3& \sf 9 \\\\ \sf \: 3y& \sf \: 6 \end{array} \right] = \left[ \begin{array}{c c } \sf z& \sf - 7\\\\ \sf \: 15& \sf \: 8 \end{array} \right] \\ \\ \\ {\huge{⇾}} \left[ \begin{array}{c c } \sf 9& \sf 2x + 9 \\\\ \sf \: 3y& \sf \: 8 \end{array} \right] = \left[ \begin{array}{c c } \sf z& \sf - 7 \\\\ \sf \: 15& \sf \: 8 \end{array} \right] \\ \\ \\ \boxed{ \green{ \therefore \sf z = 9 }}\\ \\ \sf {\huge{⇾}}2x + 9 = - 7 \\ \sf {\huge{⇾}}2x = - 7 - 9 \\ \sf {\huge{⇾}}2x = - 16 \\ \sf {\huge{⇾}}x = \frac{ - \cancel{16} {}^{8} }{ \cancel{2}} \\ \\ \boxed{ \green{ \therefore \sf \:x = - 8 }}\\ \\ \sf {\huge{⇾}}3y = 15 \\ \sf {\huge{⇾}}y = \frac{ \cancel{15} {}^{5} }{ \cancel{3}} \\ \\ \boxed{ \green{ \sf \therefore \: y = 5}}$

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$\sf \:8. \: Given \: A = \left[ \begin{array}{c c } \sf 1& \sf 1 \\\\ \sf \: - 2& \sf \: 0 \end{array} \right] \: and \: B = \left[ \begin{array}{c c } \sf 2& \sf - 1 \\\\ \sf \: 1& \sf \: 1 \end{array} \right]$

i) X + 2A = B

$\sf \: Let \: \: \: X = \left[ \begin{array}{c c } \sf a& \sf b \\\\ \sf \: c& \sf \: d \end{array} \right]$

${ \huge{ ⇾}} \: \left[ \begin{array}{c c } \sf a& \sf b\\\\ \sf \: c& \sf \: d \end{array} \right] + 2 \left[ \begin{array}{c c } \sf 1& \sf 1 \\\\ \sf \: - 2& \sf \: 0 \end{array} \right] = \left[ \begin{array}{c c } \sf 2& \sf - 1 \\\\ \sf \: 1& \sf \: 1 \end{array} \right] \\ \\ \\ { \huge{ ⇾}} \left[ \begin{array}{c c } \sf a& \sf b\\\\ \sf \: c& \sf \: d \end{array} \right] + \left[ \begin{array}{c c } \sf 2& \sf 2\\\\ \sf \: - 4& \sf \: 0 \end{array} \right] = \left[ \begin{array}{c c } \sf 2& \sf - 1 \\\\ \sf \: 1& \sf \: 1 \end{array} \right] \\ \\ \\ { \huge{ ⇾}} \: \left[ \begin{array}{c c } \sf a& \sf b\\\\ \sf \: c& \sf \: d \end{array} \right] = \left[ \begin{array}{c c } \sf 2& \sf - 1 \\\\ \sf \: 1& \sf \: 1 \end{array} \right] - \left[ \begin{array}{c c } \sf 2& \sf 2\\\\ \sf \: - 4& \sf \: 0 \end{array} \right] \\ \\ \\ { \huge{ ⇾}} \: \left[ \begin{array}{c c } \sf a& \sf b\\\\ \sf \: c& \sf \: d \end{array} \right] = \left[ \begin{array}{c c } \sf 0& \sf - 3\\\\ \sf \: 5& \sf \: 1 \end{array} \right] \\ \\ \\ \green{\sf \therefore \: X= \left[ \begin{array}{c c } \sf 0& \sf - 3\\\\ \sf \: 5& \sf \: 1 \end{array} \right]}$

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Like this, we'll have to solve 8 ii) and iii)..

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@MissSolitary ✌️

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