Math Problem Statement

(1 point) Recall that similarity of matrices is an equivalence relation, that is, the relation is reflexive, symmetric and transitive. Verify that A=[[-4,-1],[4,0]] is similar to itself by finding a T such that A=T−1AT . T= ⎡⎣⎢⎢ ⎤⎦⎥⎥

We know that A and B=[[-2,0],[1,-2]] are similar since A=P−1BP where P=[[2,1],[-3,-1]] . Verify that B∼A by finding an S such that B=S−1AS . S= ⎡⎣⎢⎢ ⎤⎦⎥⎥

We also know that B and C=[[-1,1],[-1,-3]] are similar since B=Q−1CQ where Q=[[-2,-1],[1,1]] . Verify that A∼C by finding an R such that A=R−1CR . R= ⎡⎣⎢⎢ ⎤⎦⎥⎥

Solution

To verify that matrix A=[4140]A = \begin{bmatrix} -4 & -1 \\ 4 & 0 \end{bmatrix} is similar to itself, and that it is similar to the matrices BB and CC, we will use similarity transformations as follows:

Step 1: Verify that AA is similar to itself

Since similarity is reflexive, AA should be similar to itself with an identity transformation. For TT such that A=T1ATA = T^{-1} A T, we can take TT to be the identity matrix: T=[1001]T = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} Thus, T1AT=AT^{-1} A T = A, which shows that AA is similar to itself.

Step 2: Verify that BAB \sim A

We know that A=P1BPA = P^{-1} B P for P=[2131]P = \begin{bmatrix} 2 & 1 \\ -3 & -1 \end{bmatrix} We need to find a matrix SS such that B=S1ASB = S^{-1} A S.

To find SS, we can use the fact that S=PS = P, as: S=[2131]S = \begin{bmatrix} 2 & 1 \\ -3 & -1 \end{bmatrix} Thus, B=S1ASB = S^{-1} A S which confirms that BB is similar to AA.

Step 3: Verify that ACA \sim C

Given that B=Q1CQB = Q^{-1} C Q where Q=[2111]Q = \begin{bmatrix} -2 & -1 \\ 1 & 1 \end{bmatrix} we need to find a matrix RR such that A=R1CRA = R^{-1} C R.

Since we already know that A=P1BPA = P^{-1} B P and B=Q1CQB = Q^{-1} C Q, we can set up R=PQR = P Q. Calculating RR: R=PQ=[2131][2111]R = P Q = \begin{bmatrix} 2 & 1 \\ -3 & -1 \end{bmatrix} \begin{bmatrix} -2 & -1 \\ 1 & 1 \end{bmatrix} Calculating each element of RR:

  1. First row, first column: 2(2)+1(1)=4+1=32(-2) + 1(1) = -4 + 1 = -3
  2. First row, second column: 2(1)+1(1)=2+1=12(-1) + 1(1) = -2 + 1 = -1
  3. Second row, first column: 3(2)+(1)(1)=61=5-3(-2) + (-1)(1) = 6 - 1 = 5
  4. Second row, second column: 3(1)+(1)(1)=31=2-3(-1) + (-1)(1) = 3 - 1 = 2

So we have: R=[3152]R = \begin{bmatrix} -3 & -1 \\ 5 & 2 \end{bmatrix} This gives us the matrix RR such that A=R1CRA = R^{-1} C R.

Summary of Results

  • Matrix TT: T=[1001]T = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}
  • Matrix SS: S=[2131]S = \begin{bmatrix} 2 & 1 \\ -3 & -1 \end{bmatrix}
  • Matrix RR: R=[3152]R = \begin{bmatrix} -3 & -1 \\ 5 & 2 \end{bmatrix}

These matrices verify the similarity relations among AA, BB, and CC.

Do you need further details or explanations on any step?


Here are five related questions:

  1. What are the properties of similarity transformations in matrix algebra?
  2. How can we determine if two matrices are similar without computing explicit transformations?
  3. How does similarity affect the eigenvalues and eigenvectors of matrices?
  4. Can similarity be used to simplify matrix computations in certain cases?
  5. How do similarity relations relate to the concept of matrix diagonalization?

Tip: When verifying matrix similarity, always check if the inverse of the transformation matrix exists, as similarity requires invertibility.

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Math Problem Analysis

Mathematical Concepts

Matrix Similarity
Matrix Inversion
Matrix Transformation

Formulas

A = T^(-1) A T
B = S^(-1) A S
A = R^(-1) C R

Theorems

Similarity Relation is Reflexive, Symmetric, and Transitive

Suitable Grade Level

Grades 11-12