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:
$\text{\hspace{0.17em}}{A}^{\mathrm{-1}}=\frac{1}{0(0)\mathrm{-1}(1)}\left[\begin{array}{cc}0& \mathrm{-1}\\ \mathrm{-1}& 0\end{array}\right]=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right].$
show that the set of natural numberdoes not from agroup with addition or multiplication butit forms aseni group with respect toaaddition as well as multiplication
how we can draw three triangles of distinctly different shapes. All the angles will be cutt off each triangle and placed side by side with vertices touching