Matrices.....?

Question

i have a matrix right and the equation of it is xI+BA+yA^2 = zero matrix. the A matrix is:
row one: 2, -5
row two: 3, 1
and I the identity matrix how do i find x B and y and they cant all be 0?

i would love if someone could sovle this with out negative powers.

ur only finding b how do i find x and y?

Answer 1

Negative powers are unavoidable, because A^(-1) is the inverse of the matrix A.

xI + BA + yA^2 = 0

To solve for B, first add -xI to both sides.

BA + yA^2 = -xI

Now, add (-yA^2) to both sides.

BA = -xI - yA^2

To isolate B, multiply both sides by A^(-1).

B = (-xI - yA^2)A^(-1)

Distribute.

B = -xIA^(-1) - yA^2 A^(-1)

Note that A(A^(-1)) = I, and A^2 = AA, so

B = -xIA^(-1) - yAA A^(-1)

B = -xIA^(-1) - yAI

The identity matrix I cancels the same way a 1 does for real numbers and multiplication.

B = -xA^(-1) - yA

This means you *have* to solve for A inverse, or A^(-1).

Since A =
[2 -5]
[3 1]

We calculate the inverse by putting it side by side with the identity, and then making the matrix into the identity.

[2 -5 | 1 0]
[3 1 | 0 1]

R2 -> (1/2)R2.

[1 -5/2 | 1/2 0]
[3 1 | 0 1]

R2 -> R2 - 3R1

[1 -5/2 | 1 0]
[0 17/2 | -3 1]

R2 -> (2/17) R2

[1 -5/2 | 1 0]
[0 1 | -6/17 2/17]

R1 -> R1 + (5/2)R2

[1 0 | 2/17 5/17]
[0 1 | -6/17 2/17]

That means A inverse is
[2/17 5/17]
[-6/17 2/17]

Since

B = -xA^(-1) - yA

We have

B = (-1)[2/17 5/17] [x1] - [2 -5] [y1]
. . . . . . [-6/17 2/17] [x2] . . .[3 1] [y2]

B = [-2/17 -5/17] [x1] + [-2 5] [y1]
. . . .[6/17 -2/17 ] [x2] . . .[-3 -1] [y2]