Question

guys please help me with this question and whoever will give me the answer I will mark them as brainliest but I don't want useless answers​

Answer 1

Step-by-step explanation:

I think this answer is helpful for you

Answer 2

GiveN :

$\\ \sf A=\left[\begin{array}{ccc}2&3 \\ \\5&7\end{array}\right]$

$\\ \sf B=\left[\begin{array}{ccc}0&4 \\ \\-1&7\end{array}\right]$

$\\ \sf C=\left[\begin{array}{ccc}1&0 \\ \\-1&4\end{array}\right]$

To FinD :

AC + B² - 10C.

SolutioN :

Analysis :

First we have to solve AC and then B² then 10C. After taking out all the variables separately we will simplify AC + B² - 10C.

ExplanatioN :

AC :

$\\ \sf A=\left[\begin{array}{ccc}2&3 \\ \\5&7\end{array}\right]$

$\\ \sf C=\left[\begin{array}{ccc}1&0 \\ \\-1&4\end{array}\right]$

$\\ :\normalsize\implies{\sf AC=\left[\begin{array}{ccc}2&3 \\ \\5&7\end{array}\right]\left[\begin{array}{ccc}1&0 \\ \\-1&4\end{array}\right]}$

$\\ :\normalsize\implies{\sf AC=\left[\begin{array}{ccc}2.1+3.-1&&2.0+3.4\\ \\5.1+7.-1&&5.0+7.4\end{array}\right]}$

$\\ :\normalsize\implies{\sf AC=\left[\begin{array}{ccc}2+(-3)&&0+12\\ \\5+(-7)&&0+28\end{array}\right]}$

$\\ :\normalsize\implies{\sf AC=\left[\begin{array}{ccc}2-3&&0+12\\ \\5-7&&0+28\end{array}\right]}$

$\\ \normalsize\therefore\boxed{\pink{\sf AC=\left[\begin{array}{ccc}-1&12\\ \\-2&28\end{array}\right]}}$

B² :

$\\ \sf B=\left[\begin{array}{ccc}0&4 \\ \\-1&7\end{array}\right]$

$\\ :\normalsize\implies{\sf B^2=\left[\begin{array}{ccc}0&4 \\ \\-1&7\end{array}\right]\left[\begin{array}{ccc}0&4 \\ \\-1&7\end{array}\right]}$

$\\ :\normalsize\implies{\sf B^2=\left[\begin{array}{ccc}0.0+4.-1&&0.4+4.7\\ \\-1.0+7.-1&&-1.4+7.7\end{array}\right]}$

$\\ :\normalsize\implies{\sf B^2=\left[\begin{array}{ccc}0+(-4)&&0+28\\ \\0+(-7)&&(-4)+49\end{array}\right]}$

$\\ :\normalsize\implies{\sf B^2=\left[\begin{array}{ccc}0-4&&0+28\\ \\0-7&&-4+49\end{array}\right]}$

$\\ \normalsize\therefore\boxed{\pink{\sf B^2=\left[\begin{array}{ccc}-4&28\\ \\-7&45\end{array}\right]}}$

10C :

$\\ \sf C=\left[\begin{array}{ccc}1&0 \\ \\-1&4\end{array}\right]$

$\\ :\normalsize\implies{\sf 10C=10\left[\begin{array}{ccc}1&0 \\ \\-1&4\end{array}\right]}$

$\\ :\normalsize\implies{\sf 10C=\left[\begin{array}{ccc}10.1&10.0\\ \\10.-1&10.4\end{array}\right]}$

$\\ \normalsize\therefore\boxed{\pink{\sf 10C=\left[\begin{array}{ccc}10&0\\ \\-10&40\end{array}\right]}}$

Now AC + B² - 10C :

$\\ {\sf AC=\left[\begin{array}{ccc}-1&12\\ \\-2&28\end{array}\right]}$

$\\ {\sf B^2=\left[\begin{array}{ccc}-4&28\\ \\-7&45\end{array}\right]}$

$\\ {\sf 10C=\left[\begin{array}{ccc}10&0\\ \\-10&40\end{array}\right]}$

$\\ :\normalsize\implies{\sf AC+B^2-10C=\left[\begin{array}{ccc}-1&12\\ \\-2&28\end{array}\right]+\left[\begin{array}{ccc}-4&28\\ \\ -7&45\end{array}\right]-\left[\begin{array}{ccc}10&0\\ \\-10&40\end{array}\right]}$

$\\ :\normalsize\implies{\sf AC+B^2-10C=\left[\begin{array}{ccc}-1+(-4)-10&&12+28-0\\ \\ -2+(-7)-(-10)&&28+45-40\end{array}\right]}$

$\\ :\normalsize\implies{\sf AC+B^2-10C=\left[\begin{array}{ccc}-1-4-10&&12+28-0\\ \\ -2-7+10&&28+45-40\end{array}\right]}$

$\\ :\normalsize\implies{\sf AC+B^2-10C=\left[\begin{array}{ccc}-5-10&&40-0\\ \\ -9+10&&73-40\end{array}\right]}$

$\\ :\normalsize\implies{\sf AC+B^2-10C=\left[\begin{array}{ccc}-15&40\\ \\ \:\:1&33\end{array}\right]}$

$\\ \\ \normalsize{\pink{\therefore{\boxed{\bf AC+B^2-10C=\left[\begin{array}{ccc}-15&40\\ \\ \:\:1&33\end{array}\right]}}}}$