92. Let A = 960-6 04 5 3(a) Compute the characteristic polynomial PA(A).(b) Use your answer from part (a) to find all the cigenvalues of A.eigenspace.(c) For each eigenvalue from part (b), find a basis of the corresponding eigenspace.(d) Use your answer from part (c) to determine whether A is diagonalizable. If so, producean invertible matrix P and a diagonal matrix D such that A = PDP ¹. If not, explainwhy not in a complete sentence.T

Given A=9-609-60453 (a) The characteristic polynomial of A is A-λI=9-λ-609-6-λ0453-λ =3-λ9-λ-6-λ+54 =3-λλ2-3λ =3-λλλ-3 (b) The characteristic equation of A is λλ-32= 0 From this the eigen values of A are λ = 0, 3 , 3 (c) case(1): Let X1=xyz be the eigen vector of A corresponding eigen value of λ = 0 then A-λIX1=0 ⇒ 9-609-60453xyz=000 Apply R1 ↔R3 and R2 →R2-R1 4539-60000xyz=000 Apply R2→4R2-9R1 4530-69-27000xyz=000 Apply -13R2 4530139000xyz=000 From above Echlon form we get 4x+5y+3z=0 and 13y+9z=0 Let z = k the y=-913k 4x-4513k+3k=0 ⇒ x=-326 The basis of eigen space for eigen value λ=0 is X1=-326k-913kk case(2): Let X2=xyz be the eigen vector of A corresponding eigen value of λ = 3 then A-λIX2=0 ⇒ 6-609-90450xyz=000 Apply 16R1 , 19R3 1-101-10450xyz=000 Apply R2→R2-R1 and R3→R3-4R1 1-10000090xyz=000 Apply 19R3 and R2 ↔ R3 1-10010000xyz=000 From above Echlon form we get x-y=0 and y=0 Let z = p the x=p The basis of eigen space for eigen value λ=3 is X2=p0p (d) The algebraicity of eigen value 0 is 1 the algebraicity of eigen value 3 is 2 Sum of the algebraicity of A = 1 +2 = 3 The geometricity of eigen value of 0 is 1 ( dimension of eigen space is 1) The geometricity of eigen value of 3 is 1 ( dimension of eigen space is 1) Sum of the geometricity of A = 1 + 1 = 2 Hence Sum of the algebraicity of A ≠ Sum of the geometricity of A so A is not diagonalizable.