Given a system of equations, solve with matrix inverses using a calculator.
Save the coefficient matrix and the constant matrix as matrix variables
$\text{\hspace{0.17em}}\left[A\right]\text{\hspace{0.17em}}$ and
$\text{\hspace{0.17em}}\left[B\right].$
Enter the multiplication into the calculator, calling up each matrix variable as needed.
If the coefficient matrix is invertible, the calculator will present the solution matrix; if the coefficient matrix is not invertible, the calculator will present an error message.
Using a calculator to solve a system of equations with matrix inverses
Solve the system of equations with matrix inverses using a calculator
On the matrix page of the calculator, enter the
coefficient matrix as the matrix variable
$\text{\hspace{0.17em}}\left[A\right],\text{\hspace{0.17em}}$ and enter the constant matrix as the matrix variable
$\text{\hspace{0.17em}}\left[B\right].$
On the home screen of the calculator, type in the multiplication to solve for
$\text{\hspace{0.17em}}X,\text{\hspace{0.17em}}$ calling up each matrix variable as needed.
An identity matrix has the property
$\text{\hspace{0.17em}}AI=IA=A.\text{\hspace{0.17em}}$ See
[link] .
An invertible matrix has the property
$\text{\hspace{0.17em}}A{A}^{\mathrm{-1}}={A}^{\mathrm{-1}}A=I.\text{\hspace{0.17em}}$ See
[link] .
Use matrix multiplication and the identity to find the inverse of a
$\text{\hspace{0.17em}}2\times 2\text{\hspace{0.17em}}$ matrix. See
[link] .
The multiplicative inverse can be found using a formula. See
[link] .
Another method of finding the inverse is by augmenting with the identity. See
[link] .
We can augment a
$\text{\hspace{0.17em}}3\times 3\text{\hspace{0.17em}}$ matrix with the identity on the right and use row operations to turn the original matrix into the identity, and the matrix on the right becomes the inverse. See
[link] .
Write the system of equations as
$\text{\hspace{0.17em}}AX=B,\text{\hspace{0.17em}}$ and multiply both sides by the inverse of
$\text{\hspace{0.17em}}A:{A}^{\mathrm{-1}}AX={A}^{\mathrm{-1}}B.\text{\hspace{0.17em}}$ See
[link] and
[link] .
We can also use a calculator to solve a system of equations with matrix inverses. See
[link] .
Section exercises
Verbal
In a previous section, we showed that matrix multiplication is not commutative, that is,
$\text{\hspace{0.17em}}AB\ne BA\text{\hspace{0.17em}}$ in most cases. Can you explain why matrix multiplication is commutative for matrix inverses, that is,
$\text{\hspace{0.17em}}{A}^{\mathrm{-1}}A=A{A}^{\mathrm{-1}}?$
If
$\text{\hspace{0.17em}}{A}^{\mathrm{-1}}\text{\hspace{0.17em}}$ is the inverse of
$\text{\hspace{0.17em}}A,\text{\hspace{0.17em}}$ then
$\text{\hspace{0.17em}}A{A}^{\mathrm{-1}}=I,\text{\hspace{0.17em}}$ the identity matrix. Since
$\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ is also the inverse of
$\text{\hspace{0.17em}}{A}^{\mathrm{-1}},{A}^{\mathrm{-1}}A=I.\text{\hspace{0.17em}}$ You can also check by proving this for a
$\text{\hspace{0.17em}}2\times 2\text{\hspace{0.17em}}$ matrix.
Does every
$\text{\hspace{0.17em}}2\times 2\text{\hspace{0.17em}}$ matrix have an inverse? Explain why or why not. Explain what condition is necessary for an inverse to exist.
Can you explain whether a
$\text{\hspace{0.17em}}2\times 2\text{\hspace{0.17em}}$ matrix with an entire row of zeros can have an inverse?
No, because
$\text{\hspace{0.17em}}ad\text{\hspace{0.17em}}$ and
$\text{\hspace{0.17em}}bc\text{\hspace{0.17em}}$ are both 0, so
$\text{\hspace{0.17em}}ad-bc=0,\text{\hspace{0.17em}}$ which requires us to divide by 0 in the formula.
Can a matrix with zeros on the diagonal have an inverse? If so, find an example. If not, prove why not. For simplicity, assume a
$\text{\hspace{0.17em}}2\times 2\text{\hspace{0.17em}}$ matrix.
Yes. Consider the matrix
$\text{\hspace{0.17em}}\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right].\text{\hspace{0.17em}}$ The inverse is found with the following calculation:
$\text{\hspace{0.17em}}{A}^{\mathrm{-1}}=\frac{1}{0(0)\mathrm{-1}(1)}\left[\begin{array}{cc}0& \mathrm{-1}\\ \mathrm{-1}& 0\end{array}\right]=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right].$
I've run into this:
x = r*cos(angle1 + angle2)
Which expands to:
x = r(cos(angle1)*cos(angle2) - sin(angle1)*sin(angle2))
The r value confuses me here, because distributing it makes:
(r*cos(angle2))(cos(angle1) - (r*sin(angle2))(sin(angle1))
How does this make sense? Why does the r distribute once
this is an identity when 2 adding two angles within a cosine. it's called the cosine sum formula. there is also a different formula when cosine has an angle minus another angle it's called the sum and difference formulas and they are under any list of trig identities
Brad
How can you tell what type of parent function a graph is ?
generally by how the graph looks and understanding what the base parent functions look like and perform on a graph
William
if you have a graphed line, you can have an idea by how the directions of the line turns, i.e. negative, positive, zero
William
y=x will obviously be a straight line with a zero slope
William
y=x^2 will have a parabolic line opening to positive infinity on both sides of the y axis
vice versa with y=-x^2 you'll have both ends of the parabolic line pointing downward heading to negative infinity on both sides of the y axis
William
y=x will be a straight line, but it will have a slope of one. Remember, if y=1 then x=1, so for every unit you rise you move over positively one unit. To get a straight line with a slope of 0, set y=1 or any integer.
Aaron
yes, correction on my end, I meant slope of 1 instead of slope of 0
Typically a function 'f' will take 'x' as input, and produce 'y' as output. As
'f(x)=y'.
According to Google,
"The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain."
Thomas
Sorry, I don't know where the "Â"s came from. They shouldn't be there. Just ignore them. :-)
Thomas
GREAT ANSWER THOUGH!!!
Darius
Thanks.
Thomas
Â
Thomas
It is the Â that should not be there. It doesn't seem to show if encloses in quotation marks.
"Â" or 'Â' ... Â
I've been struggling so much through all of this. my final is in four weeks 😭
Tiffany
this book is an excellent resource! have you guys ever looked at the online tutoring? there's one that is called "That Tutor Guy" and he goes over a lot of the concepts
Darius
thank you I have heard of him. I should check him out.
Tiffany
is there any question in particular?
Joe
I have always struggled with math. I get lost really easy, if you have any advice for that, it would help tremendously.
Tiffany
Sure, are you in high school or college?
Darius
Hi, apologies for the delayed response. I'm in college.
The center is at (3,4) a focus is at (3,-1) and the lenght of the major axis is 26 what will be the answer?
Rima
I done know
Joe
What kind of answer is that😑?
Rima
I had just woken up when i got this message
Joe
Can you please help me. Tomorrow is the deadline of my assignment then I don't know how to solve that
Rima
i have a question.
Abdul
how do you find the real and complex roots of a polynomial?
Abdul
@abdul with delta maybe which is b(square)-4ac=result then the 1st root -b-radical delta over 2a and the 2nd root -b+radical delta over 2a. I am not sure if this was your question but check it up
Nare
This is the actual question: Find all roots(real and complex) of the polynomial f(x)=6x^3 + x^2 - 4x + 1
Abdul
@Nare please let me know if you can solve it.
Abdul
I have a question
juweeriya
hello guys I'm new here? will you happy with me
mustapha
The average annual population increase of a pack of wolves is 25.
Period =2π
if there is a coefficient (b), just divide the coefficient by 2π to get the new period
Am
if not then how would I find it from a graph
Imani
by looking at the graph, find the distance between two consecutive maximum points (the highest points of the wave). so if the top of one wave is at point A (1,2) and the next top of the wave is at point B (6,2), then the period is 5, the difference of the x-coordinates.
Am
you could also do it with two consecutive minimum points or x-intercepts