Matrix, Diagonalize calculation

[Nemanjavuk69]

Hello

Can someone explain to me how the final calculation in Example two is being calculated?

I have put a red circle around the part which I don't quite understand is being calculated. I am familiar with the "row$\cdot$column" (dot product), however, it does not quite look like the dot product?

An example is $$2\cdot5^k-3^k$$
When I do the dot product I get $5^k\cdot2+0\cdot(-1)$ where is the $3^k$ coming from?

 

[Dr.Peterson]

Hello

Can someone explain to me how the final calculation in Example two is being calculated?

I have put a red circle around the part which I don't quite understand is being calculated. I am familiar with the "row$\cdot$column" (dot product), however, it does not quite look like the dot product?

An example is $$2\cdot5^k-3^k$$
When I do the dot product I get $5^k\cdot2+0\cdot(-1)$ where is the $3^k$ coming from?

View attachment 34679

There is no dot product here, just two ordinary matrix multiplications. Multiply $PD^k$, then multiply the result by $P^{-1}$.

Show your work, so we can see what you are doing wrong.

 

[Nemanjavuk69]

There is no dot product here, just two ordinary matrix multiplications. Multiply $PD^k$, then multiply the result by $P^{-1}$.

Show your work, so we can see what you are doing wrong.

Got ya, I did not know that. I will do my work and post it here. Thanks for letting me know!

 

[BigBeachBanana]

There is no dot product here, just two ordinary matrix multiplications. Multiply $PD^k$, then multiply the result by $P^{-1}$.

Show your work, so we can see what you are doing wrong.

Matrix multiplication should go right to left i.e. $D^kP^{-1}$, then $P$?

 

[Nemanjavuk69]

There is no dot product here, just two ordinary matrix multiplications. Multiply $PD^k$, then multiply the result by $P^{-1}$.

Show your work, so we can see what you are doing wrong.

I just did the calculatios, I got the exact same results beside I multiplied my $5^k$ and $3^k$ so I got the following matrix
$$\begin{bmatrix} 10^k-3^k & 5^k-3^k\\ -10^k+6^k & -5^k+6^k \end{bmatrix}$$
Is there a reason to why the textbook is not multiplying the numbers together?

 

[Nemanjavuk69]

Matrix multiplication should go right to left i.e. $D^kP^{-1}$, then $P$?

I was under the same impression, however, following @Dr.Peterson I do get the right answer

 

[Nemanjavuk69]

Matrix multiplication should go right to left i.e. $D^kP^{-1}$, then $P$?

I just verified by using the "Matrix Multiplication Theorem", you can go left to right or right to left AS LONG as you do not switch the order of the matrices. So doing$(P \cdot D) \cdot P^{-1}$ is the same as doing $P \cdot (D \cdot P^{-1})$

 

[BigBeachBanana]

I just verified by using the "Matrix Multiplication Theorem", you can go left to right or right to left AS LONG as you do not switch the order of the matrices. So doing$(P \cdot D) \cdot P^{-1}$ is the same as doing $P \cdot (D \cdot P^{-1})$

Yes, you're correct. I've forgotten about that property. I tend to view matrix multiplication as function compositions. More natural to go right to left.

 

[Dr.Peterson]

I just verified by using the "Matrix Multiplication Theorem", you can go left to right or right to left AS LONG as you do not switch the order of the matrices. So doing$(P \cdot D) \cdot P^{-1}$ is the same as doing $P \cdot (D \cdot P^{-1})$

In other words, matrix multiplication is associative, though not commutative.

I just did the calculatios, I got the exact same results beside I multiplied my $5^k$ and $3^k$ so I got the following matrix
$$\begin{bmatrix} 10^k-3^k & 5^k-3^k\\ -10^k+6^k & -5^k+6^k \end{bmatrix}$$
Is there a reason to why the textbook is not multiplying the numbers together?

You are referring to $2\cdot5^k$, and so on. Keep the order of operations in mind: you changed it to $(2\cdot5)^k$, which is different. You should know this long before you learn about matrices!

As it stands, it can't be simplified further than what they showed in the answer.

 

[Nemanjavuk69]

In other words, matrix multiplication is associative, though not commutative.

You are referring to $2\cdot5^k$, and so on. Keep the order of operations in mind: you changed it to $(2\cdot5)^k$, which is different. You should know this long before you learn about matrices!

As it stands, it can't be simplified further than what they showed in the answer.

Yes, I see. Thank you @Dr.Peterson