$\left[\begin{array}{ccc}5&7\\5&-8\\3&-9\end{array}\right] \left[\begin{array}{ccc}c\\d\end{array}\right] =\left[\begin{array}{ccc}-16\\3\\-15\end{array}\right]$

This tells us that:

$A=\left[\begin{array}{ccc}5&7\\5&-8\\3&-9\end{array}\right]$

$b=\left[\begin{array}{ccc}-16\\3\\-15\end{array}\right]$

Step-by-step explanation:

So we are saying we have scalars, c and d, such that:

$c\left[\begin{array}{ccc}5\\5\\ 3\end{array}\right]+d\left[\begin{array}{ccc}7\\-8\\-9\end{array}\right]=\left[\begin{array}{ccc}-16\\3\\-15\end{array}\right]$ .

So we want to find a way to express this as:

Ax=b where x is the scalar vector, $\left[\begin{array}{ccc}c\\d\end{array}\right]$ .

So we can write this as:

$\left[\begin{array}{ccc}5&7\\5&-8\\3&-9\end{array}\right] \left[\begin{array}{ccc}c\\d\end{array}\right] =\left[\begin{array}{ccc}-16\\3\\-15\end{array}\right]$

3 0

2 years ago

Find equations of the line that is parallel to the z-axis and passes through the midpoint between the two points (0, −4, 9) and

Lorico [155]

Answer: x = -4, y = 0.5, z = 5 +t

Hi!

The line L whose direction is parallel to vector V a passes through point A

is parametrized

$P_L = Vt + A$

Where t, is a real number, and $P_L$ is a any point on line L.

In this case the direction is that of the z-axis , so V = (0, 0, 1)

A is the midpoint between points B = (0, -4, 9) and C=(-8, 5, 1)

The midpoint is A = (B + C)/2 = (-4, 0.5, 5)

Then the line is:

$P_L = (0,0,1) t + (-4, 0.5, 5)$

The equations for each coordinate are:

$x = -4\\y = 0.5\\z = 5 + t$

7 0

2 years ago

I need help with 1 and 2

oee [108]

1) -149, -1; -148, -2; -147, -3
2) No, according to the commutative property of addition, it does not matter the order they are put in.

Hope this helps!

6 0

3 years ago

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