Question #169345

Determine whether the matrix (

1/√3 1/√6 −1/√2

1/√3 −2/√6 0

1/√3 1/√6 1/√2

) is Orthogonal or not.


1
Expert's answer
2021-03-09T02:00:51-0500

Solution.

A=(13161213260131612)A=\begin{pmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{3}} & -\frac{2}{\sqrt{6}} & 0 \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{2}} \\ \end{pmatrix}


An orthogonal matrix is ​​a nondegenerate square matrix AA (usually with real elements) such that

AAT=ATA=I,AA^{T}=A^{T}A=I, where ATA^{T} is transposed matrix to matrix AA, and II is the identity matrix.

detA=13161316=10.\det A=-\frac{1}{3}-\frac{1}{6}-\frac{1}{3}-\frac{1}{6}=-1\neq0.


AT=(13131316261612012)A^{T}=\begin{pmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{6}} & -\frac{2}{\sqrt{6}} & \frac{1}{\sqrt{6}} \\ -\frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\ \end{pmatrix}AAT=(13161213260131612)(13131316261612012)==(100010001)=I.AA^{T}=\begin{pmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{3}} & -\frac{2}{\sqrt{6}} & 0 \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{2}} \\ \end{pmatrix} \cdot \begin{pmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{6}} & -\frac{2}{\sqrt{6}} & \frac{1}{\sqrt{6}} \\ -\frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\ \end{pmatrix}=\newline =\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}=I.ATA=I.A^{T}A=I.

Answer. Yes, AA is orthogonal matrix.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS