Knowing that 1 is eigenvalue of \begin{pmatrix} 1 & 0 & 0 \\ 2 & -2 & 1 \\ 4 & -6 & 3 \end{pmatrix}, check the matrix $P$ such that the product $P^{-1}AP$ is a diagonal matrix. Choose one option: $\begin{pmatrix} -4 & -4 & 3 \\ -1 & -6 & 6 \\ 6 & -5 & 2 \end{pmatrix}$ $\begin{pmatrix} -1 & 5 & 2 \\ -6 & 1 & -2 \\ 2 & -1 & 6 \end{pmatrix}$ $\begin{pmatrix} -5 & -1 & -4 \\ 4 & -6 & -3 \\ -4 & 5 & 6 \end{pmatrix}$ $\begin{pmatrix} 2 & 4 & 5 \\ 2 & 2 & -3 \\ 4 & -3 & 0 \end{pmatrix}$ $\begin{pmatrix} -1 & 3 & 0 \\ 0 & 2 & 1 \\ 2 & 0 & 2 \end{pmatrix}$
Added by Brittany M.
Close
Step 1
Start with the given information: the product of P and P transpose is equal to the identity matrix. This can be written as PP^T = I. Show more…
Show all steps
Your feedback will help us improve your experience
Babita Kumari and 98 other Geometry educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Show that $A=P D P^{-1},$ where $P$ is a matrix whose columns are the eigenvectors of $A,$ and $D$ is a diagonal matrix with the corresponding eigenvalues. $A=\left[ \begin{array}{lll}{0} & {0} & {1} \\ {0} & {1} & {0} \\ {0} & {0} & {2}\end{array}\right]$
Vectors and Matrix Models
Iterated Matrix Models
Show that $A=P D P^{-1},$ where $P$ is a matrix whose columns are the eigenvectors of $A,$ and $D$ is a diagonal matrix with the corresponding eigenvalues. $A=\left[ \begin{array}{ll}{1} & {2} \\ {2} & {1}\end{array}\right]$
Show that $A=P D P^{-1},$ where $P$ is a matrix whose columns are the eigenvectors of $A,$ and $D$ is a diagonal matrix with the corresponding eigenvalues. $A=\left[ \begin{array}{lll}{1} & {0} & {1} \\ {2} & {1} & {0} \\ {3} & {0} & {1}\end{array}\right]$
Recommended Textbooks
Geometry A Common Core Curriculum
Geometry
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD