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Let A=[-1 1][0 -2] is able to express as B^3 + C^3 where B, C are two matrices then the value of |trace of B trace of C|=?
【Where A is a 2×2 matrix, [-1 1] is its first row and [0 -2] is its second row.】?
【Where A is a 2×2 matrix, [-1 1] is its first row and [0 -2] is its second row.】?
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Let A=[-1 1][0 -2] is able to express as B^3 + C^3 where B, C are two ...
Solution:
Step 1: Find Eigenvalues and Eigenvectors of A
Let λ be an eigenvalue of A and X be the corresponding eigenvector. Then
AX = λX
For λ = -1, we have
AX = -X
⇒ (A + I)X = 0
where I is the 2×2 identity matrix.
Solving (A + I)X = 0, we get
X = k[1 1]T, where k is a non-zero constant.
For λ = -2, we have
AX = -2X
⇒ (A + 2I)X = 0
Solving (A + 2I)X = 0, we get
X = k[1 0]T, where k is a non-zero constant.
Therefore, the eigenvalues and eigenvectors of A are
Eigenvalue λ1 = -1, eigenvector X1 = k1[1 1]T, k1 ≠ 0
Eigenvalue λ2 = -2, eigenvector X2 = k2[1 0]T, k2 ≠ 0
Step 2: Diagonalize A
Let P be the matrix whose columns are the eigenvectors of A. Then
P = [X1 X2] = [k1 k2][1 1][1 0]T
= [k1 k2][1 0][1 1]T
= [k1 k2][1 0][1 1]T
= [k1 k2][1 1][0 1]T
= [k1 k2 k1][1 1 0][0 1 1]
= [k1 k2 k1][1 0 1][0 1 1]T
= [k1 0 k1][-1 0 1][0 k2 k2]T
= [k1 0][0 k2][-1 0][0 1][0 1][1 0]T
Therefore, A = PDP-1, where D is the diagonal matrix
D = [-1 0][0 -2]
and P-1 is the inverse of P.
Step 3: Find B and C
Since A = PDP-1, we have
A^3 = (PDP-1)(PDP-1)(PDP-1)
= PD(P-1P)DP-1
= PD3P-1
Therefore, A can be expressed as A^3 = B^3C^3, where
B = PDP-1/3 and C = PDP-1/3
Step 4: Find |trace of B trace of C|
Since B and C are both diagonal matrices, we have
trace(B) = -1/3 + (-2/3) = -1
trace(C) = -1/3 + (-2/3) = -1
Therefore, |trace(B) trace(C)| = |-1 × -1| = 1
Hence, the value of |trace of B trace of C
Step 1: Find Eigenvalues and Eigenvectors of A
Let λ be an eigenvalue of A and X be the corresponding eigenvector. Then
AX = λX
For λ = -1, we have
AX = -X
⇒ (A + I)X = 0
where I is the 2×2 identity matrix.
Solving (A + I)X = 0, we get
X = k[1 1]T, where k is a non-zero constant.
For λ = -2, we have
AX = -2X
⇒ (A + 2I)X = 0
Solving (A + 2I)X = 0, we get
X = k[1 0]T, where k is a non-zero constant.
Therefore, the eigenvalues and eigenvectors of A are
Eigenvalue λ1 = -1, eigenvector X1 = k1[1 1]T, k1 ≠ 0
Eigenvalue λ2 = -2, eigenvector X2 = k2[1 0]T, k2 ≠ 0
Step 2: Diagonalize A
Let P be the matrix whose columns are the eigenvectors of A. Then
P = [X1 X2] = [k1 k2][1 1][1 0]T
= [k1 k2][1 0][1 1]T
= [k1 k2][1 0][1 1]T
= [k1 k2][1 1][0 1]T
= [k1 k2 k1][1 1 0][0 1 1]
= [k1 k2 k1][1 0 1][0 1 1]T
= [k1 0 k1][-1 0 1][0 k2 k2]T
= [k1 0][0 k2][-1 0][0 1][0 1][1 0]T
Therefore, A = PDP-1, where D is the diagonal matrix
D = [-1 0][0 -2]
and P-1 is the inverse of P.
Step 3: Find B and C
Since A = PDP-1, we have
A^3 = (PDP-1)(PDP-1)(PDP-1)
= PD(P-1P)DP-1
= PD3P-1
Therefore, A can be expressed as A^3 = B^3C^3, where
B = PDP-1/3 and C = PDP-1/3
Step 4: Find |trace of B trace of C|
Since B and C are both diagonal matrices, we have
trace(B) = -1/3 + (-2/3) = -1
trace(C) = -1/3 + (-2/3) = -1
Therefore, |trace(B) trace(C)| = |-1 × -1| = 1
Hence, the value of |trace of B trace of C
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Let A=[-1 1][0 -2] is able to express as B^3 + C^3 where B, C are two matrices then the value of |trace of B trace of C|=?【Where A is a 2×2 matrix, [-1 1] is its first row and [0 -2] is its second row.】?
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Let A=[-1 1][0 -2] is able to express as B^3 + C^3 where B, C are two matrices then the value of |trace of B trace of C|=?【Where A is a 2×2 matrix, [-1 1] is its first row and [0 -2] is its second row.】? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Let A=[-1 1][0 -2] is able to express as B^3 + C^3 where B, C are two matrices then the value of |trace of B trace of C|=?【Where A is a 2×2 matrix, [-1 1] is its first row and [0 -2] is its second row.】? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let A=[-1 1][0 -2] is able to express as B^3 + C^3 where B, C are two matrices then the value of |trace of B trace of C|=?【Where A is a 2×2 matrix, [-1 1] is its first row and [0 -2] is its second row.】?.
Let A=[-1 1][0 -2] is able to express as B^3 + C^3 where B, C are two matrices then the value of |trace of B trace of C|=?【Where A is a 2×2 matrix, [-1 1] is its first row and [0 -2] is its second row.】? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Let A=[-1 1][0 -2] is able to express as B^3 + C^3 where B, C are two matrices then the value of |trace of B trace of C|=?【Where A is a 2×2 matrix, [-1 1] is its first row and [0 -2] is its second row.】? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let A=[-1 1][0 -2] is able to express as B^3 + C^3 where B, C are two matrices then the value of |trace of B trace of C|=?【Where A is a 2×2 matrix, [-1 1] is its first row and [0 -2] is its second row.】?.
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