Find an example of 2 by2 matrices with a12 = .5 for which:

a. A^2 = I
b. A^-1 = A^T
c. A^2 = A

By the way, a12 means the first row and second column.

I will do the first one:

let A =
a .5
b c

the A^2 =
(a^2+b/2) (a/2+c/2)
(ab + bc) (b/2+c^2)

then
a^2+b/2=1 ---> 2a^2 + b = 2 (#1)
a/2+c/2=0 ---> a+c=0 or a = -c (#2)
ab+bc=0 -----> b(a+c) = 0 (#3)
b/2+c^2=1 ---> b+2c^2=2 (#4)

from #3
b(a+c)=0 , but a+c=0
so b = 0
in #1
2a^2 + 0 = 2
a^2 = 1
a = 1
then c=-1

so A =
1 .5
0 -1

for the others, you should have learned a quick way to find the inverse of a 2by2
T stands for transpose,
to take the transpose, your rows of the first matrix become the columns of the the transpose.

To find examples of 2x2 matrices that satisfy the given conditions, we will break down each part separately:

a. A^2 = I

Let's start by considering a general 2x2 matrix A = [[a, b], [c, d]]. To satisfy A^2 = I, we need to square matrix A and set it equal to the identity matrix:

A^2 = [[a, b], [c, d]] * [[a, b], [c, d]] = [[1, 0], [0, 1]]

Expanding the matrix multiplication, we get:

[[a^2 + bc, ab + bd], [ac + cd, bc + d^2]] = [[1, 0], [0, 1]]

To satisfy this equation, we can set a^2 + bc = 1, ab + bd = 0, ac + cd = 0, and bc + d^2 = 1.

By substituting the value a12 = 0.5 into these equations, we find that a = -0.5, b = 1, c = -1, and d = 0.5. Thus, the matrix A that satisfies A^2 = I is:

A = [[-0.5, 1], [-1, 0.5]]

b. A^-1 = A^T

To find a matrix A that satisfies A^-1 = A^T, we need the inverse of A to be equal to its transpose. The inverse of A is given by:

A^-1 = 1/det(A) * adj(A)

where det(A) is the determinant of A, and adj(A) is the adjugate of A.

Calculating the determinant of A:

det(A) = (-0.5 * 0.5) - (1 * -1) = 0.5 + 1 = 1.5

Calculating the adjugate of A:

adj(A) = [[d, -b], [-c, a]] = [[0.5, -1], [1, -0.5]]

Finally, we can calculate the inverse of A:

A^-1 = (1/1.5) * [[0.5, -1], [1, -0.5]] = [[1/3, -2/3], [2/3, -1/3]]

Comparing A with A^T, we see that A = A^T, so the matrix A that satisfies A^-1 = A^T is:

A = [[1/3, -2/3], [2/3, -1/3]]

c. A^2 = A

To find a matrix A that satisfies A^2 = A, we need to find a matrix that, when squared, is equal to itself.

Let's consider a general 2x2 matrix A = [[a, b], [c, d]].

A^2 = [[a, b], [c, d]] * [[a, b], [c, d]] = [[a^2 + bc, ab + bd], [ac + cd, bc + d^2]]

Setting this equal to A, we get:

[[a^2 + bc, ab + bd], [ac + cd, bc + d^2]] = [[a, b], [c, d]]

By comparing the corresponding entries, we can set up the following equations:

a^2 + bc = a,
ab + bd = b,
ac + cd = c,
bc + d^2 = d

Plugging in the value a12 = 0.5 into these equations, we find that a = 1, b = 1, c = 0, and d = 0.5. Thus, the matrix A that satisfies A^2 = A is:

A = [[1, 1], [0, 0.5]]

So, to summarize:

a. A^2 = I: A = [[-0.5, 1], [-1, 0.5]]
b. A^-1 = A^T: A = [[1/3, -2/3], [2/3, -1/3]]
c. A^2 = A: A = [[1, 1], [0, 0.5]]