i posted two questions

[logistic_guy]

i posted two questions last month but i can't find them

this new one please solve it

X' = (6 1 1, 8 7 -1, 2 9 -1)X + (t, 10t, 6t)

 

[lookagain]

this new one please solve it

X' = (6 1 1, 8 7 -1, 2 9 -1)X + (t, 10t, 6t)

No, we are not here to solve the problems for you.

Please post the entire problem as given to you and write what you know
with it so that a helper may try to assist you with where you are stuck.

 

[topsquark]

i posted two questions last month but i can't find them

this new one please solve it

X' = (6 1 1, 8 7 -1, 2 9 -1)X + (t, 10t, 6t)

For clarity: is this
$\textbf{X}^{\prime} = \begin{pmatrix} 6 & 1 & 1 \\ 8 & 7 & -1 \\ 2 & 9 & -1 \end{pmatrix} \textbf{X} + \begin{pmatrix} t \\ 10 t \\ 6 t \end{pmatrix}$

Is the problem asking you to find $\textbf{X}$?

-Dan

 

[logistic_guy]

yes. can you show me the tap for this matrix please?

 

[topsquark]

yes. can you show me the tap for this matrix please?

I'm sorry, but I don't know what you mean by "tap."

Start by finding the eignenvalues and eigenvectors of the 3x3 matrix.

-Dan

 

[logistic_guy]

you write the matrix nicely. how my matrix be like your matrix? you don't know tap? if i reply posting to you there is tap for font size tab for font color tab for emojis. where is the tap for matrix. also i need the tap for lamda to write the eignvalue

i know to solve this
X' = (6 1, 8 7)X

idon't know to solve this
X' = (6 1, 8 7)X + (t, 10t)

i don't know to solve this
X' = (6 1 1, 8 7 -1, 2 9 -1)X + (t, 10t, 6t)

give me the tap for matrix and tap for lamda please

 

[topsquark]

you write the matrix nicely. how my matrix be like your matrix? you don't know tap? if i reply posting to you there is tap for font size tab for font color tab for emojis. where is the tap for matrix. also i need the tap for lamda to write the eignvalue

i know to solve this
X' = (6 1, 8 7)X

idon't know to solve this
X' = (6 1, 8 7)X + (t, 10t)

i don't know to solve this
X' = (6 1 1, 8 7 -1, 2 9 -1)X + (t, 10t, 6t)

give me the tap for matrix and tap for lamda please

We use LaTeX code here to write the Mathematics. Go back to my post and hit "reply." In the box you will see the code I used to make the matrix equation.

You find the eigenvalues for the 3x3 matrix the same way you do as for the 2x2: Given $M v = \lambda v$ then you find $det(M - I_{3x3} \lambda) = 0$ and solve for $\lambda$ (where $I_{3x3}$ is the 3x3 identity matrix.) Give it a try.

-Dan

 

[logistic_guy]

i can't solve this matrix today because i want to learn to write the matrix nicely. i understand better when i express my math correctly. i have to spend a couple of day to learning this Latex code. hold on topsquark. i'll get back to you soon. thank you so far.

 

[topsquark]

i can't solve this matrix today because i want to learn to write the matrix nicely. i understand better when i express my math correctly. i have to spend a couple of day to learning this Latex code. hold on topsquark. i'll get back to you soon. thank you so far.

Put your statements between the text [ imath ] and [ /imath ] (no spaces.)

\begin{pmatrix} ... \end{pmatrix} gives a matrix with "rounded" sides.

Inside of this, put the rows and columns: 1 & 2 & 3 is the first row, with three columns.

Put a \\ between the rows.

So the code
[ imath ] \begin{pmatrix} 6 & 1 & 1 \\ 8 & 7 & -1 \\ 2 & 9 & -1 \end{pmatrix} [ /imath ]

will do the trick.

$\begin{pmatrix} 6 & 1 & 1 \\ 8 & 7 & -1 \\ 2 & 9 & -1 \end{pmatrix}$

-Dan

 

[logistic_guy]

i'm back. i know little Latex now. i don't understand your writing post 7. i memorize to solve 2x2 matrix only. i will show you to find eignvalue of 2x2 matrix

$\displaystyle \begin{bmatrix}6-\lambda & 1 \\8 & 7-\lambda \end{bmatrix}$

$\displaystyle (6-\lambda)(7-\lambda) = 0$

$\displaystyle 67 - 6\lambda - \lambda 7 - \lambda^2 = 0$

$\displaystyle 42 - 7\lambda - \lambda^2 = 0$

i use quadrtic formula to solve this to get $\displaystyle \lambda_1 = -10, \lambda_2 = 3$

$\displaystyle 16k_1 + 1k_2 = 0$
$\displaystyle 8k_1 + 17k_2 = 0$

if i solve system, i get matrix K1

$\displaystyle \bold{K_1} = \begin{bmatrix}k_1 \\k_2 \end{bmatrix}$

i repeat $\displaystyle \lambda_2$ to get $\displaystyle \bold{K_2}$

$\displaystyle \bold{X} = c_1\bold{X_1} + c_2\bold{X_2} = c_1\bold{K_1}e^{-10t} + c_2\bold{K_2}e^{3t}$

i learn Latex to show you this

 

[khansaheb]

i'm back. i know little Latex now. i don't understand your writing post 7. i memorize to solve 2x2 matrix only. i will show you to find eignvalue of 2x2 matrix

$\displaystyle \begin{bmatrix}6-\lambda & 1 \\8 & 7-\lambda \end{bmatrix}$

$\displaystyle (6-\lambda)(7-\lambda) = 0$

$\displaystyle 67 - 6\lambda - \lambda 7 - \lambda^2 = 0$

$\displaystyle 42 - 7\lambda - \lambda^2 = 0$

i use quadrtic formula to solve this to get $\displaystyle \lambda_1 = -10, \lambda_2 = 3$

$\displaystyle 16k_1 + 1k_2 = 0$
$\displaystyle 8k_1 + 17k_2 = 0$

if i solve system, i get matrix K1

$\displaystyle \bold{K_1} = \begin{bmatrix}k_1 \\k_2 \end{bmatrix}$

i repeat $\displaystyle \lambda_2$ to get $\displaystyle \bold{K_2}$

$\displaystyle \bold{X} = c_1\bold{X_1} + c_2\bold{X_2} = c_1\bold{K_1}e^{-10t} + c_2\bold{K_2}e^{3t}$

i learn Latex to show you this

you write:

$\displaystyle \begin{bmatrix}6-\lambda & 1 \\8 & 7-\lambda \end{bmatrix}$

$\displaystyle (6-\lambda)(7-\lambda) = 0$

That should be:

$\displaystyle (6-\lambda)(7-\lambda) - (1) *(8) = 0$

 

[mario99]

i posted two questions last month but i can't find them

this new one please solve it

X' = (6 1 1, 8 7 -1, 2 9 -1)X + (t, 10t, 6t)

You want to solve the system in post #3. The first step is to solve the homogenous version of that system. Rewrite the system as:

$\displaystyle \textbf{X}^{\prime} = \begin{pmatrix} 6 & 1 & 1 \\ 8 & 7 & -1 \\ 2 & 9 & -1 \end{pmatrix} \textbf{X}$

Now you can find the eigenvalues in the same way you did in the $2 \times 2$ system, but it would be a $3 \times 3$ system.

$\displaystyle \begin{pmatrix} 6 - \lambda & 1 & 1 \\ 8 & 7 - \lambda & -1 \\ 2 & 9 & -1 -\lambda \end{pmatrix} = 0$

How to solve the matrix above?

hint: Find the determinant of a $3 \times 3$ matrix.

When you find $\bold{K_1}, \bold{K_2}, \ \text{and} \ \bold{K_3}$, you basically found the complementary solution to the original system.

 

[topsquark]

i'm back. i know little Latex now. i don't understand your writing post 7. i memorize to solve 2x2 matrix only. i will show you to find eignvalue of 2x2 matrix

$\displaystyle \begin{bmatrix}6-\lambda & 1 \\8 & 7-\lambda \end{bmatrix}$

$\displaystyle (6-\lambda)(7-\lambda) = 0$

$\displaystyle 67 - 6\lambda - \lambda 7 - \lambda^2 = 0$

$\displaystyle 42 - 7\lambda - \lambda^2 = 0$

i use quadrtic formula to solve this to get $\displaystyle \lambda_1 = -10, \lambda_2 = 3$

$\displaystyle 16k_1 + 1k_2 = 0$
$\displaystyle 8k_1 + 17k_2 = 0$

if i solve system, i get matrix K1

$\displaystyle \bold{K_1} = \begin{bmatrix}k_1 \\k_2 \end{bmatrix}$

i repeat $\displaystyle \lambda_2$ to get $\displaystyle \bold{K_2}$

$\displaystyle \bold{X} = c_1\bold{X_1} + c_2\bold{X_2} = c_1\bold{K_1}e^{-10t} + c_2\bold{K_2}e^{3t}$

i learn Latex to show you this

Looks good (with the correction posted by khansaheb.)

Now you need to learn how to find the determinant of a 3x3 matrix. See here.

-Dan

 

[logistic_guy]

i'm mean

$\displaystyle (6-\lambda)(7-\lambda)- (1)(8) = 0$

i wached videos for determinant of 3x3 matrix but i didn't understand to do it myself. this is why i can't solve this problem

post 12 can you please show steps of solving 3x3 matrix?

 

[khansaheb]

Google the following

Calculate determinant of a matrix​

What did you find?

I find:

Calculate determinant of a matrix​

The determinant of a matrix is a scalar value that can be calculated using various methods. Here are some ways to calculate the determinant of a matrix:​

For a 2x2 Matrix: The determinant of a 2x2 matrix is calculated using the formula:​

|a b| = ad - bc​

Where a, b, c, and d are the elements of the matrix.​

For a 3x3 Matrix: The determinant of a 3x3 matrix can be calculated using the formula:​

|a b c| = a(ei - fh) - b(di - fg) + c(dh - eg)​

Where a, b, c, d, e, f, g, and h are the elements of the matrix.​

Can you follow the instructions given above?

Where are you getting stuck?

 

[MaxWong]

i'm back. i know little Latex now. i don't understand your writing post 7. i memorize to solve 2x2 matrix only. i will show you to find eignvalue of 2x2 matrix

$\displaystyle \begin{bmatrix}6-\lambda & 1 \\8 & 7-\lambda \end{bmatrix}$

$\displaystyle (6-\lambda)(7-\lambda) = 0$

$\displaystyle 67 - 6\lambda - \lambda 7 - \lambda^2 = 0$

$\displaystyle 42 - 7\lambda - \lambda^2 = 0$

i use quadrtic formula to solve this to get $\displaystyle \lambda_1 = -10, \lambda_2 = 3$

$\displaystyle 16k_1 + 1k_2 = 0$
$\displaystyle 8k_1 + 17k_2 = 0$

if i solve system, i get matrix K1

$\displaystyle \bold{K_1} = \begin{bmatrix}k_1 \\k_2 \end{bmatrix}$

i repeat $\displaystyle \lambda_2$ to get $\displaystyle \bold{K_2}$

$\displaystyle \bold{X} = c_1\bold{X_1} + c_2\bold{X_2} = c_1\bold{K_1}e^{-10t} + c_2\bold{K_2}e^{3t}$

i learn Latex to show you this

You can use a similar approach for $3 \times 3$ matrices.

You subtract all numbers on the diagonal by $\lambda$ and find the values of $\lambda$ such that the determinant of this matrix is 0.

$$\det \begin{bmatrix}6-\lambda&1&1\\8&7-\lambda&-1\\2&9&-1-\lambda\end{bmatrix} = 0$$
To find the determinant of a $3 \times 3$ matrix, you use the following formula:

$$\det \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33}\end{bmatrix} = a_{11} \det \begin{bmatrix} a_{22} & a_{23}\\a_{32} & a_{33}\end{bmatrix} - a_{12} \det \begin{bmatrix}a_{21} & a_{23} \\ a_{31} & a_{33}\end{bmatrix} + a_{13} \det \begin{bmatrix} a_{21} & a_{22} \\ a_{31} & a_{32}\end{bmatrix}$$
This is called Laplace expansion. You look through a row, and for each element $a_{rc}$ in the $r$-th row, you cover up the $r$-th row and the $c$-th column, and evaluate the determinant. Then multiply that to $a_{rc}(-1)^{r+c}$, and then sum up the results through the row. This works for any row, and the formula I gave for the special case of $3 \times 3$ determinants is the Laplace expansion of first row the determinant.

After using the formula and doing some work, you will find the eigenvalues. Finding eigenvectors is just the same.

 

[Steven G]

i'm back. i know little Latex now. i don't understand your writing post 7. i memorize to solve 2x2 matrix only. i will show you to find eignvalue of 2x2 matrix

$\displaystyle \begin{bmatrix}6-\lambda & 1 \\8 & 7-\lambda \end{bmatrix}$

$\displaystyle (6-\lambda)(7-\lambda) = 0$

Come on, you're studying linear algebra and need to use the quadratic formula to solve $\displaystyle (6-\lambda)(7-\lambda)=0$?? The answers to your (incorrect) equation is naturally $\displaystyle \lambda=6\ and\ \lambda=7=0$ Not only did you use the quadratic formula but you got the incorrect solutions!

 

[logistic_guy]

post 16 i'll try to do this on paper. hold on MaxWong. i'll get back to you. thank you

post 17 this is differential equation. i learn to use quadratic when i have function $\displaystyle \lambda^2$

 

[Otis]

i wached videos for determinant of 3x3 matrix but i didn't understand

Did you watch the video that topsquark linked in post#13? It shows how to easily write out what MaxWong explained in post#16, using an actual matrix.

If you did use topsquark's link, then please describe the confusing parts in that video.

If you didn't watch it, then I suggest you do.

 

[khansaheb]

post 17 this is differential equation. i learn to use quadratic when i have function $\displaystyle \lambda^2$

But you do have $\displaystyle \lambda^2$ !!!

$\displaystyle (6-\lambda)(7-\lambda) - 8 = 0$

$\displaystyle \lambda^2 - 13*\lambda + 34 = 0$...............This is the quadratic equation that @Steven G was referring to.