Question

solve this.....................​

Answer 1

Answer:

Q-3. fill in the blanks si that following statements become true : i) 3+ _______= 0 ii). -2 +(-7)=(-7)+_______ iii)8+(-8)=_______ iv)-51+______=-51. v): 2+[(-7)+8] =(3+(____) ) vi). [3+(-8)] +______=3 +[(-8)+7]

Step-by-step explanation:

Q-3. fill in the blanks si that following statements become true : i) 3+ _______= 0 ii). -2 +(-7)=(-7)+_______ iii)8+(-8)=_______ iv)-51+______=-51. v): 2+[(-7)+8] =(3+(____) ) vi). [3+(-8)] +______=3 +[(-8)+7] exl rbi ucch tv gum rb ucch

Answer 2

$\orange{ \underline{ \boxed{ \boxed{ \underline{ { \huge{ \mathtt{ \: \: \: \: \: \: M}}}{ \mathtt{ATRIX - \: \: \: \: \: \: \: \: \: }}}}}}}$

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

$\tt{I = \left[ \begin{array}{c c } \sf 1& \sf 0 \\\\ \sf 0& \sf \: 1 \end{array} \right] \: \: \: \: \: \: \: \: \: \: \: \: \: E = \left[ \begin{array}{c c } \sf 0& \sf 1 \\\\ \sf 0& \sf \: 0 \end{array} \right] } \\ \\ \\ \mathtt{(aI + bE)³ = a³I + 3a²bE} \\ \\ L.H.S \implies{ \mathtt{ {\huge{(}}a\left[ \begin{array}{c c } \sf 1& \sf 0 \\\\ \sf 0& \sf \: 1 \end{array} \right] + b\left[ \begin{array}{c c } \sf 0& \sf 1 \\\\ \sf 0& \sf \: 0 \end{array} \right] { \huge{)}} ^{3} }} \\ \\ \\ \implies{ \mathtt{{ \huge{(}}\left[ \begin{array}{c c } \sf a& \sf 0 \\\\ \sf 0& \sf \: a \end{array} \right] +\left[ \begin{array}{c c } \sf 0& \sf b \\\\ \sf 0& \sf \: 0 \end{array} \right] { \huge{)}} ^{3} }} \\ \\ \\ \implies{ \mathtt{{ \huge{(}}\left[ \begin{array}{c c } \sf a& \sf b \\\\ \sf 0& \sf \: a \end{array} \right]{ \huge{)}} ^{3} }} \\ \\ \\ \implies{ \mathtt{\left[ \begin{array}{c c } \sf a& \sf b \\\\ \sf 0& \sf \: a \end{array} \right].\left[ \begin{array}{c c } \sf a& \sf b \\\\ \sf 0& \sf \:a \end{array} \right].\left[ \begin{array}{c c } \sf a& \sf b \\\\ \sf 0& \sf \: a \end{array} \right]}} \\ \\ \\ \implies{ \mathtt{\left[ \begin{array}{c c } \sf {a}^{2} & \sf 2ab \\\\ \sf 0& \sf \: {a}^{2} \end{array} \right].\left[ \begin{array}{c c } \sf a& \sf b \\\\ \sf 0& \sf \: a \end{array} \right]}} \\ \\ \\ \red{ \implies{ \mathtt{\left[ \begin{array}{c c } \sf {a}^{3} & \sf 3 {a}^{2} b \\\\ \sf 0& \sf \: {a}^{3} \end{array} \right]}}}$

${ \mathtt{R.H.S = a³I + 3a²bE}} \\ \\ \implies{ \mathtt{ {a}^{3} \left[ \begin{array}{c c } \sf 1& \sf 0 \\\\ \sf 0& \sf \: 1 \end{array} \right] + 3 {a}^{2}b \left[ \begin{array}{c c } \sf 0& \sf 1 \\\\ \sf 0& \sf \: 0 \end{array} \right]}} \\ \\ \\ \implies{ \mathtt{\left[ \begin{array}{c c } \sf {a}^{3} & \sf 0 \\\\ \sf 0& \sf \: {a}^{3} \end{array} \right] +\left[ \begin{array}{c c } \sf 0& \sf 3 {a}^{2} b\\\\ \sf 0& \sf \: 0 \end{array} \right] }} \\ \\ \\ \red{ \implies{ \tt{\left[ \begin{array}{c c } \sf {a}^{3} & \sf 3 {a}^{2} b\\\\ \sf 0& \sf \: {a}^{3} \end{array} \right]}}}$

therefore,

L.H.S = R.H.S (proved)

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