Find the eigenvalues of the matrix:

[-43 0 80]
[40 -3 80]
[24 0 45]

-3

1 + 4 sqrt( 241 )

1 - 4 sqrt( 241 )

Step-by-step explanation:

We need minus lambda on the entries down the diagonal. I'm going to use m instead of the letter for lambda.

[-43-m 0 80]

[40 -3-m 80]

[24 0 45-m]

Now let's find the determinant

(-43-m)[(-3-m)(45-m)-0(80)]

-0[40(45-m)-80(24)]

+80[40(0)-(-3-m)(24)]

Let's simplify:

(-43-m)[(-3-m)(45-m)]

-0

+80[-(-3-m)(24)]

Continuing:

(-43-m)[(-3-m)(45-m)]

+80[-(-3-m)(24)]

I'm going to factor (-3-m) from both terms:

(-3-m)[(-43-m)(45-m)-80(24)]

Multiply the pair of binomials in the brackets and the other pair of numbers;

(-3-m)[-1935-2m+m^2-1920]

Simplify and reorder expression in brackets:

(-3-m)[m^2-2m-3855]

Set equal to 0 to find the eigenvalues

-3-m=0 gives us m=-3 as one eigenvalue

The other is a quadratic and looks scary because of the big numbers.

I guess I will use quadratic formula and a calculator.

(2 +/- sqrt( (-2)^2 - 4(1)(-3855) )/(2×1)

(2 +/- sqrt( 15424 )/(2)

(2 +/- sqrt( 64 )sqrt( 241 )/(2)

(2 +/- 8 sqrt( 241 )/(2)

1 +/- 4 sqrt( 241 )

hello,

b) 62.47% of the students are between 14 and 18 years old.

step-by-step explanation:

mean = 15 and standard deviation = 2

student between 14 to 18

formula used z-score

for age 14 , z score

using z-table , p(z> -0.5)=0.3085

for age 18 , z score

using z-table , p(z< 1.5)=0.9332

p(-0.5< z< 1.5)=0.9332-0.3085 = 0.6247

thus, 62.47% of the students are between 14 and 18 years old.

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