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

Given :

$A = \left[\begin{array}{cc}4&-2\\6&-3\end{array}\right]$

$B = \left[\begin{array}{cc}0&2\\1&-1\end{array}\right]$

$C = \left[\begin{array}{cc}-2&3\\1&-1\end{array}\right]$

Need to find : $A^2 - A + BC$

$A^2 =A*A = \left[\begin{array}{cc}4&-2\\6&-3\end{array}\right]*\left[\begin{array}{cc}4&-2\\6&-3\end{array}\right]$

⇒ $A^2 = \left[\begin{array}{cc}(4*4)-(2*6)&(4*-2)+(-2*-3)\\(6*4)-(3*6)&(6*-2)+(-3*-3)\end{array}\right]$

⇒ $A^2 = \left[\begin{array}{cc}4&-2\\6&-3\end{array}\right] = A$

$BC = \left[\begin{array}{cc}0&2\\1&-1\end{array}\right]*\left[\begin{array}{cc}-2&3\\1&-1\end{array}\right]$

⇒ $BC = \left[\begin{array}{cc}0+2&0-2\\-2-1&3+1\end{array}\right]$

$BC = \left[\begin{array}{cc}2&-2\\-3&4\end{array}\right]$

⇒ $A^2 - A + BC = A - A + BC$

∴  $A^2 -A +BC = BC = \left[\begin{array}{cc}2&-2\\-3&4\end{array}\right]$

HOPE THIS  HELPS YOU .!!

Answer 2

GiveN:-

$\sf A=\left[\begin{array}{ccc}\sf4&\sf-2\\ \\ \sf6&\sf-3\end{array}\right]$

$\sf B=\left[\begin{array}{ccc}\sf0&\sf2\\ \\ \sf1&\sf-1\end{array}\right]$

$\sf C=\left[\begin{array}{ccc}\sf-2&\sf3\\ \\ \sf1&\sf-1\end{array}\right]$

To FinD:-

A² - A + BC.

SolutioN:-

Analysis :

First we have to separately find the A², BC. After getting the matrices of A² and BC we will evaluate them as per the signs.

Solution :

First A² :

$\sf A=\left[\begin{array}{ccc}\sf4&\sf-2\\ \\\sf6&\sf-3\end{array}\right]$

So,

$\sf A^2=\left[\begin{array}{ccc}\sf4&-2 \\ \\ 6&-3\end{array}\right]\left[\begin{array}{ccc}\sf4&\sf-2\\ \\ \sf6&\sf-3\end{array}\right]$

$\\ \sf \: A^2=\left[\begin{array}{ccc}\sf4.4 +\sf(-2).6&&4.(-2)+(-2).(-3)\\ \\ \sf6.4 +\sf(-3).6&&6.(-2)+(-3).(-3)\end{array}\right]$

$\\ \sf \: A^2=\left[\begin{array}{ccc}\sf16+\sf(-12)&&(-8)+6\\ \\ \sf24+\sf(-18)&&(-12)+9\end{array}\right]$

$\\ \sf \: A^2=\left[\begin{array}{ccc}\sf16\sf-12&&-8+6\\ \\ \sf24\sf-18&&-12+9\end{array}\right]$

$\\ \therefore\boxed{\pink{\sf{ \: A^2=\left[\begin{array}{ccc}\sf4&-2\\ \\ \sf6&-3\end{array}\right]}}}$

Then A :

$\therefore\boxed{\pink{\sf{ A=\left[\begin{array}{ccc}\sf4&\sf-2\\ \\\sf6&\sf-3\end{array}\right]}}}$

Then BC :

$\\ \sf BC=\left[\begin{array}{ccc}\sf0&2 \\ \\ 1&-1\end{array}\right]\left[\begin{array}{ccc}\sf-2&\sf3\\ \\ \sf1&\sf-1\end{array}\right]$

$\\ \sf \: BC=\left[\begin{array}{ccc}\sf0.(-2)+\sf2.1&&1.(-2)+(-1).1\\ \\ \sf0.3+\sf2.(-1)&&1.3+(-1).(-1)\end{array}\right]$

$\\ \sf \: BC=\left[\begin{array}{ccc}\sf0+\sf2&&0+(-2)\\ \\ \sf(-2)+\sf(-1)&&3+1\end{array}\right]$

$\\ \sf \: BC=\left[\begin{array}{ccc}\sf0+\sf2&&0-2\\ \\ \sf-2\sf-1&&3+1\end{array}\right]$

$\\ \therefore\boxed{\pink{\sf{ \: BC=\left[\begin{array}{ccc}\sf2&-2\\ \\ \sf-3&4\end{array}\right]}}}$

Now A² - A + BC :

$\\ \sf A^2-A+BC=\left[\begin{array}{ccc}\sf4&-2 \\ \\ 6&-3\end{array}\right]-\left[\begin{array}{ccc}\sf4&\sf-2\\ \\ \sf6&\sf-3\end{array}\right]+\left[\begin{array}{ccc}\sf2&-2 \\ \\ \sf-3&4\end{array}\right]$

$\\ \sf A^2-A+BC=\left[\begin{array}{ccc}\sf4-4+2&(-2)-(-2)+(-2)\\ \\ 6-6+(-3)&-3-(-3)+4\end{array}\right]$

$\\ \sf A^2-A+BC=\left[\begin{array}{ccc}\sf4-4+2&-2+2-2\\ \\ 6-6-3&-3+3+4\end{array}\right]$

$\\ \sf A^2-A+BC=\left[\begin{array}{ccc}\sf\cancel{4}\cancel{-4}+2&\cancel{-2}\cancel{+2}-2\\ \\ \cancel{6}\cancel{-6}-3&\cancel{-3}\cancel{+3}+4\end{array}\right]$

$\\ \sf A^2-A+BC=\left[\begin{array}{ccc}\sf\:\:\:2&-2\\ \\ -3&\:\:\:4\end{array}\right]$

$\\ {\pink{\therefore{\boxed{\bf{A^2-A+BC=\left[\begin{array}{ccc}\sf\:\:\:2&-2\\ \\ -3&\:\:\:4\end{array}\right]}}}}}$