I have labeled where the confusion of the solution is. I don't need an answer to the problem (also attached). I just need an explanation on how did we get this matrix that I've added a note in red? from where did the [ 2/3 1/3 ... ] came?? what was done in order to get it? Thank you :) -[-2-1Let A3(a) Find the eigenvalues and corresponding eigenvectors of A.(b) Find a nonsingular matrix P such that D = P-' AP is diagonal. [2 -11. Let A-2Observe first thatt – 21Aa(t) = |-= (t – 2)(t – 3) – 2 = t2 – 5t + 6 – 2 = t? – 5t + 4 = (t – 4)(t – 1).%3D|t – 3(a) Eigenvalues and corresponding eigenvectors of A.Since A4(t) = (t – 4)(t – 1), the eigenvalues areA= 1, 1 = 4.--[1with echelon-1The eigenspace for A = 1 is the null space of the matrix2 -2-11formA vectoris in E if and only if -x + y = 0, so y = x. Hence0 0{[:]B, - {{:}E1| xERand a basis for E1 is B12 1The eigenspace for A= 4 is the null space of the matrixwith echelon form1A vectoris in E4 if and only if 2x + y =0, so y = -2x. Hence0 0{[ }}1E4|x ER-2xand a basis for E4 is B4-2(b) Nonsingular matrir P such that D = P-'AP is diagonal.[11Using the vectors in the bases for E1 and E4, we have P1 -2How 2222/31/3hence P-1and1/3 -1/32/31/32-11D = P-'AP1/3 -1/3 ]1 -2-231 00 42/31/314= D.1/3 -1/31 -8(c) f(A), where f(t) = t – 4t3 + 3t – 7.We have D = P-'AP, so A = PDP-l and f(A) = f(PDP-1) = Pf(D)P-1.(See diagonal factorization, text page 298.)Next,41 00 4111 00 41 0+30 41 00 1f(D)-73+0 -2561-40 256+0 12-7-7 00 51- 4+3 - 7256 – 256 + 12 – 7and so1/3[-7 00 52/3f(A) = Pf(D)P-11 -21/3 -1/3-752/31/3-3-4= f(A).-7 -101/3 -1/3-81

Question

I have labeled where the confusion of the solution is. I don't need an answer to the problem (also attached). I just need an explanation on how did we get this matrix that I've added a note in red? from where did the [ 2/3  1/3 ... ] came?? what was done in order to get it? Thank you :) -[-2-1Let A3(a) Find the eigenvalues and corresponding eigenvectors of A.(b) Find a nonsingular matrix P such that D = P-' AP is diagonal. [2 -11. Let A-2Observe first thatt – 21Aa(t) = |-= (t – 2)(t – 3) – 2 = t2 – 5t + 6 – 2 = t? – 5t + 4 = (t – 4)(t – 1).%3D|t – 3(a) Eigenvalues and corresponding eigenvectors of A.Since A4(t) = (t – 4)(t – 1), the eigenvalues areA= 1, 1 = 4.--[1with echelon-1The eigenspace for A = 1 is the null space of the matrix2 -2-11formA vectoris in E if and only if -x + y = 0, so y = x. Hence0 0{[:]B, - {{:}E1| xERand a basis for E1 is B12 1The eigenspace for A= 4 is the null space of the matrixwith echelon form1A vectoris in E4 if and only if 2x + y =0, so y = -2x. Hence0 0{[ }}1E4|x ER-2xand a basis for E4 is B4-2(b) Nonsingular matrir P such that D = P-'AP is diagonal.[11Using the vectors in the bases for E1 and E4, we have P1 -2How 2222/31/3hence P-1and1/3 -1/32/31/32-11D = P-'AP1/3 -1/3 ]1 -2-231 00 42/31/314= D.1/3 -1/31 -8(c) f(A), where f(t) = t – 4t3 + 3t – 7.We have D = P-'AP, so A = PDP-l and f(A) = f(PDP-1) = Pf(D)P-1.(See diagonal factorization, text page 298.)Next,41 00 4111 00 41 0+30 41 00 1f(D)-73+0 -2561-40 256+0 12-7-7 00 51- 4+3 - 7256 – 256 + 12 – 7and so1/3[-7 00 52/3f(A) = Pf(D)P-11 -21/3 -1/3-752/31/3-3-4= f(A).-7 -101/3 -1/3-81

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(b) Nonsingular matrir P such that D = P-1AP is diagonal. 1 Using the vectors in the bases for E and E4, we have P = 1 -2 How ?22 1/3 2/3 1/3 -1/3 hence P-1 and