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:
$\text{\hspace{0.17em}}{A}^{\mathrm{-1}}=\frac{1}{0(0)\mathrm{-1}(1)}\left[\begin{array}{cc}0& \mathrm{-1}\\ \mathrm{-1}& 0\end{array}\right]=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right].$
No because a negative times a negative is a positive. No matter what you do you can never multiply the same number by itself and end with a negative
lurverkitten
Actually you can. you get what's called an Imaginary number denoted by i which is represented on the complex plane. The reply above would be correct if we were still confined to the "real" number line.
Liam
Suppose P= {-3,1,3} Q={-3,-2-1} and R= {-2,2,3}.what is the intersection