Question

Given, A = $A= \left[\begin{array}{ccc}2&1\\0&-2\end{array}\right] B = \left[\begin{array}{ccc}4&1\\-3&-2\end{array}\right] and C = \left[\begin{array}{ccc}-3&-2\\-1&4\end{array}\right]$ , find A² + AC – 5B.

Answer 1

Answer:

Step-by-step explanation:

Hope it works !!!

Answer 2

Solution:

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

$A^{2}= \left[\begin{array}{ccc}2&1\\0&-2\end{array}\right] \times \left[\begin{array}{ccc}2&1\\0&-2\end{array}\right]$

$A^{2}= \left[\begin{array}{ccc}4+0&2-2\\0-0&2+2\end{array}\right]$
$A^{2}= \left[\begin{array}{ccc}4&0\\0&4\end{array}\right]$

by the same way find AC

$AC= \left[\begin{array}{ccc}2&1\\0&-2\end{array}\right]\times \left[\begin{array}{ccc}-3&-2\\-1&4\end{array}\right]$

$AC= \left[\begin{array}{ccc}-6-1&-4+4\\0+2&0-8\end{array}\right]$

$AC= \left[\begin{array}{ccc}-7&0\\2&-8\end{array}\right]$

Now to solve

A² + AC – 5B=

$=\left[\begin{array}{ccc}4&0\\0&4\end{array}\right]+\left[\begin{array}{ccc}-7&0\\2&-8\end{array}\right]-5\left[\begin{array}{ccc}4&1\\-3&-2\end{array}\right]$

$=\left[\begin{array}{ccc}4-7-20&0+0-5\\0+2+15&4-8+10\end{array}\right]$

$=\left[\begin{array}{ccc}-23&-5\\17&6\end{array}\right]$

Hope it helps you