Question

4. Find the sum of the following using the colum (a) a² + 2ab + b², a? - 2ab - bº, a? + b2​

Answer 1

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Answer 2

Answer:

$\mathrm{\left[\begin{array}{ccc}a&0&0\\0&a&0\\0&0&a\end{array}\right] \times X = \left[\begin{array}{ccc}a^2&2ab&b^2\\a&2ab&a\\a&b&2\end{array}\right]}$

Step-by-step explanation:

$\left[\begin{array}{ccc}a_1_1&a_1_2&a_1_3\\a_2_1&a_2_2&a_2_3\\a_3_1&a_3_2&a_3_3\end{array}\right] = \mathrm{ +a_1_1 \times a_2_2\times a_3_3}$

                              $\mathrm{ +a_1_2 \times a_2_3\times a_3_1}$

                             $\mathrm{ +a_1_3 \times a_2_1\times a_3_2}$

                             $\mathrm{ -a_1_3 \times a_2_2\times a_3_1}$

                             $\mathrm{ -a_1_1 \times a_2_3\times a_3_2}$

                           $\mathrm{ -a_1_2 \times a_2_1\times a_3_3}$

$\mathrm{\left[\begin{array}{ccc}a&0&0\\0&a&0\\0&0&a\end{array}\right] aaa+0\times0\times0+0\times0\times0-0a\times0-0\times0\times a-a\times0\times0 = a^3}$

$\mathrm{\left[\begin{array}{ccc}a&0&0\\0&a&0\\0&0&a\end{array}\right] ^{-1}}$

$\frac{1}{a^3} \times \mathrm{\left[\begin{array}{ccc}a^2&0&0\\0&a^2&0\\0&0&a^2\end{array}\right] } = \mathrm{\left[\begin{array}{ccc}\frac{1}{a} &0&0\\0&\frac{1}{a} &0\\0&0&\frac{1}{a} \end{array}\right] }$

$= \mathrm{\left[\begin{array}{ccc}a&2b&\frac{b^2}{a} \\\\1&2b&\frac{b}{a} \\\\1&\frac{b}{a} &\frac{2}{a} \end{array}\right]}$