# $

{\text{Find }}{A^2}{\text{ if }}A = \left[ {\begin{array}{*{20}{c}}

1&2&5 \\

3&4&1 \\

1&{ - 1}&2

\end{array}} \right]

$

Answer

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\[

\Rightarrow {\text{Let }}A = \left[ {\begin{array}{*{20}{c}}

1&2&5 \\

3&4&1 \\

1&{ - 1}&2

\end{array}} \right]{\text{ (1)}} \\

{\text{So, now we have to find the value of }}{A^2}.{\text{ So,}} \\

\Rightarrow {A^2} = A*A{\text{ (2)}} \\

{\text{Putting value of }}A{\text{ in the RHS}}{\text{ of equation 2 we get,}} \\

\Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}

1&2&5 \\

3&4&1 \\

1&{ - 1}&2

\end{array}} \right]*\left[ {\begin{array}{*{20}{c}}

1&2&5 \\

3&4&1 \\

1&{ - 1}&2

\end{array}} \right] \\

{\text{As, we know that for a matrix }}X{\text{ if we have to find }}{X^2}{\text{ then we find it as given below}} \\

\Rightarrow {\text{Let }}X = \left[ {\begin{array}{*{20}{c}}

a&b&c \\

d&e&f \\

g&h&i

\end{array}} \right]{\text{ }} \\

{\text{So, }}{X^2} = X*X,{\text{ is calculated as,}} \\

\Rightarrow {X^2} = \left[ {\begin{array}{*{20}{c}}

{(a*a) + (b*d) + (c*g)}&{(a*b) + (b*e) + (c*h)}&{(a*c) + (b*f) + (c*i)} \\

{(d*a) + (e*d) + (f*g)}&{(d*b) + (e*e) + (f*h)}&{(d*c) + (e*f) + (f*i)} \\

{(g*a) + (h*d) + (i*g)}&{(g*b) + (h*e) + (i*h)}&{(g*c) + (h*f) + (i*i)}

\end{array}} \right] \\

{\text{So, like }}{X^2}{\text{ we can also find the value of }}{A^2}. \\

{\text{So, for calculating the value of }}{A^2}. \\

\Rightarrow {{\text{A}}^2} = \left[ {\begin{array}{*{20}{c}}

{(1*1) + (2*3) + (5*1)}&{(1*2) + (2*4) + (5*( - 1))}&{(1*5) + (2*1) + (5*2)} \\

{(3*1) + (4*3) + (1*1)}&{(3*2) + (4*4) + (1*( - 1))}&{(3*5) + (4*1) + (1*2)} \\

{(1*1) + ( - 1*3) + (2*1)}&{(1*2) + ( - 1*4) + (2*( - 1))}&{(1*5) + (( - 1)*1) + (2*2)}

\end{array}} \right]{\text{ So,}} \\

\Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}

{1 + 6 + 5}&{2 + 8 - 5}&{5 + 2 + 10} \\

{3 + 12 + 1}&{6 + 16 - 1}&{15 + 4 + 2} \\

{1 - 3 + 2}&{2 - 4 - 2}&{5 - 1 + 4}

\end{array}} \right]{\text{ So on solving this it becomes}} \\

{\text{Hence , }}{A^2} = \left[ {\begin{array}{*{20}{c}}

{12}&5&{17} \\

{16}&{21}&{21} \\

0&{ - 4}&8

\end{array}} \right] \\

{\text{NOTE: - Whenever you came up with this type of problem then make calculations proper}} \\

{\text{As, in this type of problem there are many chances of getting calculation mistakes}}{\text{.}} \\

\\

\]

\Rightarrow {\text{Let }}A = \left[ {\begin{array}{*{20}{c}}

1&2&5 \\

3&4&1 \\

1&{ - 1}&2

\end{array}} \right]{\text{ (1)}} \\

{\text{So, now we have to find the value of }}{A^2}.{\text{ So,}} \\

\Rightarrow {A^2} = A*A{\text{ (2)}} \\

{\text{Putting value of }}A{\text{ in the RHS}}{\text{ of equation 2 we get,}} \\

\Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}

1&2&5 \\

3&4&1 \\

1&{ - 1}&2

\end{array}} \right]*\left[ {\begin{array}{*{20}{c}}

1&2&5 \\

3&4&1 \\

1&{ - 1}&2

\end{array}} \right] \\

{\text{As, we know that for a matrix }}X{\text{ if we have to find }}{X^2}{\text{ then we find it as given below}} \\

\Rightarrow {\text{Let }}X = \left[ {\begin{array}{*{20}{c}}

a&b&c \\

d&e&f \\

g&h&i

\end{array}} \right]{\text{ }} \\

{\text{So, }}{X^2} = X*X,{\text{ is calculated as,}} \\

\Rightarrow {X^2} = \left[ {\begin{array}{*{20}{c}}

{(a*a) + (b*d) + (c*g)}&{(a*b) + (b*e) + (c*h)}&{(a*c) + (b*f) + (c*i)} \\

{(d*a) + (e*d) + (f*g)}&{(d*b) + (e*e) + (f*h)}&{(d*c) + (e*f) + (f*i)} \\

{(g*a) + (h*d) + (i*g)}&{(g*b) + (h*e) + (i*h)}&{(g*c) + (h*f) + (i*i)}

\end{array}} \right] \\

{\text{So, like }}{X^2}{\text{ we can also find the value of }}{A^2}. \\

{\text{So, for calculating the value of }}{A^2}. \\

\Rightarrow {{\text{A}}^2} = \left[ {\begin{array}{*{20}{c}}

{(1*1) + (2*3) + (5*1)}&{(1*2) + (2*4) + (5*( - 1))}&{(1*5) + (2*1) + (5*2)} \\

{(3*1) + (4*3) + (1*1)}&{(3*2) + (4*4) + (1*( - 1))}&{(3*5) + (4*1) + (1*2)} \\

{(1*1) + ( - 1*3) + (2*1)}&{(1*2) + ( - 1*4) + (2*( - 1))}&{(1*5) + (( - 1)*1) + (2*2)}

\end{array}} \right]{\text{ So,}} \\

\Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}

{1 + 6 + 5}&{2 + 8 - 5}&{5 + 2 + 10} \\

{3 + 12 + 1}&{6 + 16 - 1}&{15 + 4 + 2} \\

{1 - 3 + 2}&{2 - 4 - 2}&{5 - 1 + 4}

\end{array}} \right]{\text{ So on solving this it becomes}} \\

{\text{Hence , }}{A^2} = \left[ {\begin{array}{*{20}{c}}

{12}&5&{17} \\

{16}&{21}&{21} \\

0&{ - 4}&8

\end{array}} \right] \\

{\text{NOTE: - Whenever you came up with this type of problem then make calculations proper}} \\

{\text{As, in this type of problem there are many chances of getting calculation mistakes}}{\text{.}} \\

\\

\]

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