First Principles AI

Accedmychevron_right11thchevron_rightmathchevron_rightMatrices And Determinantschevron_rightExercise 4.3

Solution

To find $A^{-1}$ , we row-reduce $[A\mid I]$ to $[I\mid A^{-1}]$ .

(i) $A=\begin{bmatrix}2&6&-3\\0&-2&0\\-2&5&6\end{bmatrix}$

The inverse is:

A^{-1}= \begin{bmatrix} 1&\frac{17}{4}&\frac{1}{2}\\ 0&-\frac{1}{2}&0\\ \frac{1}{3}&\frac{11}{6}&\frac{1}{3} \end{bmatrix}

(ii) $A=\begin{bmatrix}1&2&-1\\0&-2&8\\1&0&2\end{bmatrix}$

A^{-1}= \begin{bmatrix} -\frac{2}{5}&-\frac{2}{5}&\frac{7}{5}\\ \frac{4}{5}&\frac{3}{10}&-\frac{4}{5}\\ \frac{1}{5}&\frac{1}{5}&-\frac{1}{5} \end{bmatrix}

(iii) $A=\begin{bmatrix}1&6&2\\2&13&0\\0&-1&1\end{bmatrix}$

A^{-1}= \begin{bmatrix} -\frac{13}{3}&\frac{8}{3}&\frac{26}{3}\\ \frac{2}{3}&-\frac{1}{3}&-\frac{4}{3}\\ \frac{2}{3}&-\frac{1}{3}&-\frac{1}{3} \end{bmatrix}