Question

I could use some more help

Answer 1

Answer:

$\left[\begin{array}{ccc}-30&22&8\\-4&-4&16\\30&-28&18\end{array}\right]$

Step-by-step explanation:

Given

$A = \left[\begin{array}{ccc}-1&9&2\\10&-10&2\\-5&6&-5\end{array}\right]$

$B = \left[\begin{array}{ccc}7&-1&-1\\6&-4&-3\\-10&10&-7\end{array}\right]$

Required

2A - 4B

To solve 2A - 4B, we first multiply matrix A by 2 and matrix B by 4

So, if

$A = \left[\begin{array}{ccc}-1&9&2\\10&-10&2\\-5&6&-5\end{array}\right]$

$2A = 2 *\left[\begin{array}{ccc}-1&9&2\\10&-10&2\\-5&6&-5\end{array}\right]$

$2A = \left[\begin{array}{ccc}-2&18&4\\20&-20&4\\-10&12&-10\end{array}\right]$

If

$B = \left[\begin{array}{ccc}7&-1&-1\\6&-4&-3\\-10&10&-7\end{array}\right]$

then

$4B = 4*\left[\begin{array}{ccc}7&-1&-1\\6&-4&-3\\-10&10&-7\end{array}\right]$

$4B = \left[\begin{array}{ccc}28&-4&-4\\24&-16&-12\\-40&40&-28\end{array}\right]$

So; 2A - 4B becomes

$\left[\begin{array}{ccc}-2&18&4\\20&-20&4\\-10&12&-10\end{array}\right] - \left[\begin{array}{ccc}28&-4&-4\\24&-16&-12\\-40&40&-28\end{array}\right]$

$\left[\begin{array}{ccc}-2-28&18-(-4)&4-(-4)\\20-24&-20-(-16)&4-(-12)\\-10-(-40)&12-40&-10-(-28)\end{array}\right]$

$\left[\begin{array}{ccc}-30&18+4&4+4\\20-24&-20+16&4+12\\-10+40&12-40&-10+28\end{array}\right]$

$\left[\begin{array}{ccc}-30&22&8\\-4&-4&16\\30&-28&18\end{array}\right]$

Hence, 2A - 4B is equivalent to

$\left[\begin{array}{ccc}-30&22&8\\-4&-4&16\\30&-28&18\end{array}\right]$