Question

Recall that similarity of matrices is an equivalence relation, that is, the relation is reflexive, symmetric and transitive. Verify that $A = egin{bmatrix} 0 & -1 \ 3 & -5 end{bmatrix}$ is similar to itself by finding a $T$ such that $A = T^{-1}AT$. $T = egin{bmatrix} _ & _ \ _ & _ end{bmatrix}$ We know that $A$ and $B = egin{bmatrix} -3 & -1 \ -3 & -2 end{bmatrix}$ are similar since $A = P^{-1}BP$ where $P = egin{bmatrix} 1 & 0 \ -3 & 1 end{bmatrix}$. Verify that $B sim A$ by finding an $S$ such that $B = S^{-1}AS$. $S = egin{bmatrix} _ & _ \ _ & _ end{bmatrix}$ We also know that $B$ and $C = egin{bmatrix} -17 & -23 \ 9 & 12 end{bmatrix}$ are similar since $B = Q^{-1}CQ$ where $Q = egin{bmatrix} -1 & 3 \ 1 & -2 end{bmatrix}$. Verify that $A sim C$ by finding an $R$ such that $A = R^{-1}CR$. $R = egin{bmatrix} _ & _ \ _ & _ end{bmatrix}$

          Recall that similarity of matrices is an equivalence relation, that is, the relation is reflexive, symmetric and transitive.
Verify that $A = egin{bmatrix} 0 & -1 \ 3 & -5 end{bmatrix}$ is similar to itself by finding a $T$ such that $A = T^{-1}AT$.
$T = egin{bmatrix} _ & _ \ _ & _ end{bmatrix}$
We know that $A$ and $B = egin{bmatrix} -3 & -1 \ -3 & -2 end{bmatrix}$ are similar since $A = P^{-1}BP$ where $P = egin{bmatrix} 1 & 0 \ -3 & 1 end{bmatrix}$. Verify that $B sim A$ by finding an $S$ such that $B = S^{-1}AS$.
$S = egin{bmatrix} _ & _ \ _ & _ end{bmatrix}$
We also know that $B$ and $C = egin{bmatrix} -17 & -23 \ 9 & 12 end{bmatrix}$ are similar since $B = Q^{-1}CQ$ where $Q = egin{bmatrix} -1 & 3 \ 1 & -2 end{bmatrix}$. Verify that $A sim C$ by finding an $R$ such that $A = R^{-1}CR$.
$R = egin{bmatrix} _ & _ \ _ & _ end{bmatrix}$

Recall that similarity of matrices is an equivalence relation, that is, the relation is reflexive, symmetric and transitive.
Verify that A = eginbmatrix 0 -1 3 -5 endbmatrix is similar to itself by finding a T such that A = T^-1AT.
T = eginbmatrix endbmatrix
We know that A and B = eginbmatrix -3 -1 -3 -2 endbmatrix are similar since A = P^-1BP where P = eginbmatrix 1 0 -3 1 endbmatrix. Verify that B sim A by finding an S such that B = S^-1AS.
S = eginbmatrix endbmatrix
We also know that B and C = eginbmatrix -17 -23 9 12 endbmatrix are similar since B = Q^-1CQ where Q = eginbmatrix -1 3 1 -2 endbmatrix. Verify that A sim C by finding an R such that A = R^-1CR.
R = eginbmatrix endbmatrix

Added by Christina K.

Question

Please give Ace some feedback