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pishuonlain [190]
3 years ago
9

The 3 × 3 matrix P satisfies the matrix equation P^2 = P.

Mathematics
1 answer:
riadik2000 [5.3K]3 years ago
4 0

Answer:  The answers are given below.

Step-by-step explanation:  Given that a  3 × 3 matrix P satisfies the matrix equation P² = P.

We are to

(a) find the possibilities for the determinant of P.

(b) explain the reason behind there are no other possibilities.

(c) give an example of P, for each possible determinant.

(a) According to the given information, we have

P^2=P\\\\\Rightarrow P^2-P=0\\\\\Rightarrow P(P-I)=0\\\\\Rightarrow P=0,~~~P=I.

So, P can be either a zero matrix of order 3 or an identity matrix of order 3.

If P = 0, then det(P) = 0     and     if P = I, then det(P) = 1.

Therefore the possible determinants of P are 0 and 1.

(b) There can be any other determinant other than 0 and 1, because if so, then the given equation P² = P will not be satisfied.

(c) If |P| = 0, then the matrix P can be can be of the form as follows:

P=\left[\begin{array}{ccc}0&0&0\\0&0&0\\0&0&0\end{array}\right] .

If |P| = 1, the the matrix P can be of the form as follows :

P=\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right] .

Thus, all the parts are answered.

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Write the equation of the line that passes through (−3,1) and (2,−1) in slope-intercept form
Alex787 [66]

Answer:

y=-\frac{2}{5}x-\frac{1}{5}

Step-by-step explanation:

The equation of a line is y = mx + b

Where:

  • m is the slope
  • b is the y-intercept

First, let's find what m is, the slope of the line.

Let's call the first point you gave, (-3,1), point #1, so the x and y numbers given will be called x1 and y1.

Also, let's call the second point you gave, (2,-1), point #2, so the x and y numbers here will be called x2 and y2.

Now, just plug the numbers into the formula for m above, like this:

m = -\frac{2}{5}

So, we have the first piece to finding the equation of this line, and we can fill it into y=mx+b like this:

y=-\frac{2}{5}x + b

Now, what about b, the y-intercept?

To find b, think about what your (x,y) points mean:

  • (-3,1). When x of the line is -3, y of the line must be 1.
  • (2,-1). When x of the line is 2, y of the line must be -1.

Now, look at our line's equation so far: y=-\frac{2}{5}x + b. b is what we want, the --\frac{2}{5} is already set and x and y are just two 'free variables' sitting there. We can plug anything we want in for x and y here, but we want the equation for the line that specfically passes through the two points (-3,1) and (2,-1).

So, why not plug in for x and y from one of our (x,y) points that we know the line passes through? This will allow us to solve for b for the particular line that passes through the two points you gave!

You can use either (x,y) point you want. The answer will be the same:

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  • (2,-1). y = mx + b or -1=-\frac{2}{5} * 2 + b, or solving for b: b = 1-(-\frac{2}{5})(2). b = -\frac{1}{5}.

See! In both cases, we got the same value for b. And this completes our problem.

The equation of the line that passes through the points  (-3,1) and (2,-1) is y=-\frac{2}{5}x-\frac{1}{5}

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