Question

(1 point) Let $A = \begin{bmatrix} 0 & 4 & 8 \\ -4 & 4 & 4 \\ 2 & -4 & -6 \end{bmatrix}$. \\ Find $S$ and $D$ such that $A = SDS^{-1}$. \\ $S = \begin{bmatrix} \\ \\ \\ \end{bmatrix}$, $D = \begin{bmatrix} \\ & 0 & 0 \\ 0 & & 0 \\ 0 & 0 & \end{bmatrix}$.

          (1 point) Let $A = \begin{bmatrix} 0 & 4 & 8 \\ -4 & 4 & 4 \\ 2 & -4 & -6 \end{bmatrix}$. \\ Find $S$ and $D$ such that $A = SDS^{-1}$. \\ $S = \begin{bmatrix} \\ \\ \\ \end{bmatrix}$, $D = \begin{bmatrix} \\ & 0 & 0 \\ 0 & & 0 \\ 0 & 0 & \end{bmatrix}$.

(1 point) Let A =
< b m a t r i x >.
Find S and D such that A = SDS^-1.
S =
< b m a t r i x >, D =
< b m a t r i x >.

Added by Charles W.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Madhur B

Step 1

To do this, we solve the characteristic equation: det(A - λI) = 0 where I is the identity matrix and λ is the eigenvalue. In this case, we have: det(A - λI) = det([1 2; 2 1] - λ[1 0; 0 1]) = det([1-λ 2; 2 1-λ]) = (1-λ)(1-λ) - 4 Show more…

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