i posted two questions

[Steven G]

i learn to use quadratic when i have function $\displaystyle \lambda^2$

If someone actually told you to solve factored quadratic equations, then that person doesn't know what they're doing.
You should know then whenever you have (x-a)(x-b)=0, then the solutions are x=a and x=b.
I'm wondering how you would solve the following cubic equation (x-1)(x-3)(x-11)=0? Please post back your preferred method on solving such an equation.

 

[Steven G]

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.

Actually I wasn't. The OP tried solving $\displaystyle (6-\lambda)(7-\lambda)=0$ by using the quadratic equation.
Fair enough, it was the wrong equation that s/he tried to solve but I just wanted (actually needed) to point out that using the quadratic formula when the quadratic is in factored form is ridiculous.

 

[logistic_guy]

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.

i get this $\displaystyle -\lambda^3+12\lambda^2-28\lambda+76$

i don't know how to solve this

 

[blamocur]

In your equation $\lambda \approx 9.9541159$

 

[Otis]

i get this -λ^3 + 12λ^2 - 28λ + 76

i don't know how to solve this

Hello. In general, we don't solve cubic equations by hand. There is a formula for solving ax^3+bx^2+cx+d=0, but it's too awkward to use. Mostly, we use software (i.e., computer algebra systems) to obtain decimal approximations for x.

For -λ^3 + 12λ^2 – 28λ + 76 = 0, here is what the exact solutions look like:


The first line is the Real solution, and the last two lines are the Complex solutions containing Imaginary parts.

In post #24, blamocur gives a decimal approximation of the Real solution.

 

[BigBeachBanana]

I learned a substitution method for solving cubic hands a while back. It is a little cumbersome but doable by hand. If you have done the steps correctly, you should get the results Otis has shown above.
$$−\lambda^3 +12\lambda^2 −28\lambda+76=0\\ \lambda^3 -12\lambda^2 +28\lambda-76=0\\ \text{Eliminate the quadratic term by letting }\lambda= x +4 \\ (x+4)^3 -12(x+4)^2 +28(x+4)-76=0\\ x^3-20x-92=0\\ \text{Change coordinates by letting }x= y + \frac{k}{y} \\ y^6+y^4(3k-20)-92y^3+y^2(3k^2-20k)+k^3=0\\ \text{Now choose k to eliminate the 2nd and the 4th order terms} \implies k = \frac{20}{3}\\ y^6-92y^3+\left(\frac{20}{3}\right)^3=0\\ \text{Notice this is a hidden quadratic.}\\ z^2-92z+\frac{8000}{27} =0\\ \text{From here you can solve for z and back sub starting with } z=y^3 \text{ and get 3 solutions.}$$

 

[mario99]

i get this $\displaystyle -\lambda^3+12\lambda^2-28\lambda+76$

i don't know how to solve this

You don't need to know how to solve this cubic equation. You can use Wolfram|Alpha or your calculator to solve it. The main idea is to get $\lambda_1$, $\lambda_2$, and $\lambda_3$ even if their values are just approximations.

They are:

$\lambda_1 \approx 9.9541$

$\lambda_2 \approx 1.0229 - 2.5668i$

$\lambda_3 \approx 1.0229 + 2.5668i$

Each $\lambda$ will give you three equations to solve for $k_1$, $k_2$, and $k_3$. For example $\lambda_1$ will let you find $\bold{K_1}$, so that:

$\displaystyle \bold{K_1} = \begin{bmatrix}k_1 \\ k_2 \\ k_3 \end{bmatrix}$

 

[logistic_guy]

In your equation $\lambda \approx 9.9541159$

thank you blamocur

Hello. In general, we don't solve cubic equations by hand. There is a formula for solving ax^3+bx^2+cx+d=0, but it's too awkward to use. Mostly, we use software (i.e., computer algebra systems) to obtain decimal approximations for x.

For -λ^3 + 12λ^2 – 28λ + 76 = 0, here is what the exact solutions look like:
View attachment 38134

The first line is the Real solution, and the last two lines are the Complex solutions containing Imaginary parts.

In post #24, blamocur gives a decimal approximation of the Real solution.

thank you Otis. that's an scary solution. do i have to check it to make sure it's equal to blamocur's answer. i'll try

$\displaystyle \frac{1}{3}(1242+6\sqrt{36849})^{\frac{1}{3}} + 20\frac{1}{(1242+6\sqrt{36849})^{\frac{1}{3}}}+4 = 9.9541159$

it seems correct

I learned a substitution method for solving cubic hands a while back. It is a little cumbersome but doable by hand. If you have done the steps correctly, you should get the results Otis has shown above.
$$−\lambda^3 +12\lambda^2 −28\lambda+76=0\\ \lambda^3 -12\lambda^2 +28\lambda-76=0\\ \text{Eliminate the quadratic term by letting }\lambda= x +4 \\ (x+4)^3 -12(x+4)^2 +28(x+4)-76=0\\ x^3-20x-92=0\\ \text{Change coordinates by letting }x= y + \frac{k}{y} \\ y^6+y^4(3k-20)-92y^3+y^2(3k^2-20k)+k^3=0\\ \text{Now choose k to eliminate the 2nd and the 4th order terms} \implies k = \frac{20}{3}\\ y^6-92y^3+\left(\frac{20}{3}\right)^3=0\\ \text{Notice this is a hidden quadratic.}\\ z^2-92z+\frac{8000}{27} =0\\ \text{From here you can solve for z and back sub starting with } z=y^3 \text{ and get 3 solutions.}$$

thank you BigBeachBanana. that's a nice way to solve. do this consider like quadratic formula for $\displaystyle x^3$? it's complicated but i'll try to learn it

You don't need to know how to solve this cubic equation. You can use Wolfram|Alpha or your calculator to solve it. The main idea is to get $\lambda_1$, $\lambda_2$, and $\lambda_3$ even if their values are just approximations.

They are:

$\lambda_1 \approx 9.9541$

$\lambda_2 \approx 1.0229 - 2.5668i$

$\lambda_3 \approx 1.0229 + 2.5668i$

Each $\lambda$ will give you three equations to solve for $k_1$, $k_2$, and $k_3$. For example $\lambda_1$ will let you find $\bold{K_1}$, so that:

$\displaystyle \bold{K_1} = \begin{bmatrix}k_1 \\ k_2 \\ k_3 \end{bmatrix}$

thank you Mario99. i think i can do the matrix now

$\displaystyle \begin{bmatrix}6 - 9.9541 & 1 & 1\\ 8 & 7 - 9.9541 & -1 \\ 2 & 9 & -1 - 9.9541 \end{bmatrix}$

$\displaystyle (6 - 9.9541)k_1 + 1k_2 + 1k_3 = 0$
$\displaystyle 8k_1 + (7 - 9.9541)k_2 + -1k_3 = 0$
$\displaystyle 2k_1 + 9k_2 + (-1 - 9.9541)k_3 = 0$

i know how to solve $\displaystyle 2$ equations, not $\displaystyle 3$

 

[mario99]

i know how to solve $\displaystyle 2$ equations, not $\displaystyle 3$

It does not matter you have 2, 3, or 10 equations. You will not do a single calculation (at least me). Just enter this website, reduce the default matrix to 4 columns, enter your values there, and you will get $x_1$, $x_2$, and $x_3$. You can change the variables to $k_1$, $k_2$, and $k_3$ if you want.

System of linear equations calculator

You can solve systems of linear equations using Gauss-Jordan elimination, Cramer's rule, inverse matrix, and other methods. Also, you can analyze the compatibility.

matrixcalc.org

Bonus. You can also calculate the complex matrices there for $\lambda_2$ and $\lambda_3$.

-Mario

 

[logistic_guy]

when i solve matrix through website, i get zero for all $\displaystyle k$. what's wrong?

 

[mario99]

when i solve matrix through website, i get zero for all $\displaystyle k$. what's wrong?

Are you sure that you have posted a correct problem in the original post? This is the first time in my life to see a system that does not have a complementary solution!

Maybe they intended to make your life difficult. And they want you to find the particular solution. Go back to post #1 or better #3 done by Dan, and you will see this matrix:

$\begin{bmatrix}t \\10t \\6t \end{bmatrix}$

This matrix makes the system nonhomogeneous. And you have to make a guess what the particular solution will look like. Start guessing like when you guess a particular solution for a normal nonhomogeneous differential equation. What will your guess be?