How do you find the inverse of A=((4, 9, 15), (15, 17, 6), (24, 0, 17))?

Nov 2, 2017

$\frac{1}{5963} \left(\begin{matrix}- 289 & 153 & 201 \\ 111 & 292 & - 201 \\ 408 & - 216 & 67\end{matrix}\right)$

Explanation:

To find the inverse of the matrix (if it exists), create an augmented matrix by appending to the right the identity matrix of the same size (3x3 in this case), like so:

$\left(\begin{matrix}4 & 9 & 15 & | & 1 & 0 & 0 \\ 15 & 17 & 6 & | & 0 & 1 & 0 \\ 24 & 0 & 17 & | & 0 & 0 & 1\end{matrix}\right)$

Now, using elementary row operations, manipulate the left hand side of this augmented matrix until it becomes the identity matrix, being sure to carry your operations throughout the entire matrix. Given the numbers involved in this problem, a scientific calculator may be helpful to reduce fractions as you go along. I will use the notation ${R}_{n}$ to represent Row n in this problem:

$\left(\begin{matrix}4 & 9 & 15 & | & 1 & 0 & 0 \\ 15 & 17 & 6 & | & 0 & 1 & 0 \\ 24 & 0 & 17 & | & 0 & 0 & 1\end{matrix}\right) \left.\begin{matrix}\div 4 \\ \div 15 \to \\ \div 24\end{matrix}\right.$

$\left(\begin{matrix}1 & \frac{9}{4} & \frac{15}{4} & | & \frac{1}{4} & 0 & 0 \\ 1 & \frac{17}{15} & \frac{2}{5} & | & 0 & \frac{1}{15} & 0 \\ 1 & 0 & \frac{17}{24} & | & 0 & 0 & \frac{1}{24}\end{matrix}\right) \left.\begin{matrix}\null \\ - {R}_{1} \to \\ - {R}_{1}\end{matrix}\right.$

$\left(\begin{matrix}1 & \frac{9}{4} & \frac{15}{4} & | & \frac{1}{4} & 0 & 0 \\ 0 & - \frac{67}{60} & - \frac{67}{20} & | & - \frac{1}{4} & \frac{1}{15} & 0 \\ 0 & - \frac{9}{4} & - \frac{73}{24} & | & - \frac{1}{4} & 0 & \frac{1}{24}\end{matrix}\right) \left.\begin{matrix}\null \\ \times - \frac{60}{67} \to \\ \times - \frac{4}{9}\end{matrix}\right.$

$\left(\begin{matrix}1 & \frac{9}{4} & \frac{15}{4} & | & \frac{1}{4} & 0 & 0 \\ 0 & 1 & 3 & | & \frac{15}{67} & - \frac{4}{67} & 0 \\ 0 & 1 & \frac{73}{54} & | & \frac{1}{9} & 0 & - \frac{1}{54}\end{matrix}\right) \left.\begin{matrix}- \frac{9}{4} {R}_{2} \\ \to \\ - {R}_{2}\end{matrix}\right.$

$\left(\begin{matrix}1 & 0 & - 3 & | & - \frac{17}{67} & \frac{9}{67} & 0 \\ 0 & 1 & 3 & | & \frac{15}{67} & - \frac{4}{67} & 0 \\ 0 & 0 & - \frac{89}{54} & | & - \frac{68}{603} & \frac{4}{67} & - \frac{1}{54}\end{matrix}\right) \left.\begin{matrix}\null \\ \to \\ \times - \frac{54}{89}\end{matrix}\right.$

$\left(\begin{matrix}1 & 0 & - 3 & | & - \frac{17}{67} & \frac{9}{67} & 0 \\ 0 & 1 & 3 & | & \frac{15}{67} & - \frac{4}{67} & 0 \\ 0 & 0 & 1 & | & \frac{408}{5963} & - \frac{216}{5963} & \frac{1}{89}\end{matrix}\right) \left.\begin{matrix}+ 3 {R}_{3} \\ - 3 {R}_{3} \to \\ \null\end{matrix}\right.$

$\left(\begin{matrix}1 & 0 & 0 & | & - \frac{289}{5963} & \frac{153}{5963} & \frac{3}{89} \\ 0 & 1 & 0 & | & \frac{111}{5963} & \frac{292}{5963} & - \frac{3}{89} \\ 0 & 0 & 1 & | & \frac{408}{5963} & - \frac{216}{5963} & \frac{1}{89}\end{matrix}\right)$

The final result on the right hand side, where the identity matrix first was placed, is the inverse ${A}^{-} 1$:

$\left(\begin{matrix}- \frac{289}{5963} & \frac{153}{5963} & \frac{3}{89} \\ \frac{111}{5963} & \frac{292}{5963} & - \frac{3}{89} \\ \frac{408}{5963} & - \frac{216}{5963} & \frac{1}{89}\end{matrix}\right)$

If the fractions bother you, the denominator of 5963 can be factored out and the inverse could be written as this:

$\frac{1}{5963} \left(\begin{matrix}- 289 & 153 & 201 \\ 111 & 292 & - 201 \\ 408 & - 216 & 67\end{matrix}\right)$