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Solve the system using the inverse of the coefficient matrix.
$$X=\left[\begin{array}{c}4\\ 38\\ 58\end{array}\right]$$
Given a system of equations, solve with matrix inverses using a calculator.
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.
Evaluate the expression.
Access these online resources for additional instruction and practice with solving systems with inverses.
Identity matrix for a $2\text{}\times \text{}2$ matrix | ${I}_{2}=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ |
Identity matrix for a $\text{3}\text{}\times \text{}3$ matrix | ${I}_{3}=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$ |
Multiplicative inverse of a $2\text{}\times \text{}2$ matrix | ${A}^{\mathrm{-1}}=\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right],\text{where}ad-bc\ne 0$ |
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 an entire column of zeros have an inverse? Explain why or why not.
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: $\text{\hspace{0.17em}}{A}^{\mathrm{-1}}=\frac{1}{0(0)\mathrm{-1}(1)}\left[\begin{array}{cc}0& \mathrm{-1}\\ \mathrm{-1}& 0\end{array}\right]=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right].$
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