Problem on using inverses to solve a 3x3 matrix equation
$\newcommand{\bfA}{\mathbf{A}}$ $\newcommand{\bfB}{\mathbf{B}}$ $\newcommand{\bfC}{\mathbf{C}}$ $\newcommand{\bfF}{\mathbf{F}}$ $\newcommand{\bfI}{\mathbf{I}}$ $\newcommand{\bfa}{\mathbf{a}}$ $\newcommand{\bfb}{\mathbf{b}}$ $\newcommand{\bfc}{\mathbf{c}}$ $\newcommand{\bfd}{\mathbf{d}}$ $\newcommand{\bfe}{\mathbf{e}}$ $\newcommand{\bfi}{\mathbf{i}}$ $\newcommand{\bfj}{\mathbf{j}}$ $\newcommand{\bfk}{\mathbf{k}}$ $\newcommand{\bfn}{\mathbf{n}}$ $\newcommand{\bfr}{\mathbf{r}}$ $\newcommand{\bfu}{\mathbf{u}}$ $\newcommand{\bfv}{\mathbf{v}}$ $\newcommand{\bfw}{\mathbf{w}}$ $\newcommand{\bfx}{\mathbf{x}}$ $\newcommand{\bfy}{\mathbf{y}}$ $\newcommand{\bfz}{\mathbf{z}}$
Find $\bfA^{-1}$ and use it to solve $\bfA \bfx = \bfb$, where $$\bfA = \begin{pmatrix}1 & -1& 1 \\ 1 & 1 & 1\\1 & 2 & 4 \end{pmatrix}, \quad \bfb = \begin{pmatrix} 1\\3\\2 \end{pmatrix}$$
Solution
Recall that
Inverse of a 3x3 matrix
To find the inverse of a $3 \times 3$ matrix,
Compute the minors of each element
Negate every other element, according to a checkerboard pattern
Take the transpose
Divide by the determinant of the original matrix
The minor of the $(i,j)$th entry of a matrix $\bfA$ is the determinant of the submatrix obtained by removing the $i$th row and the $j$th column of $\bfA$.
For example, the minor of the $1,1$ entry is $\begin{vmatrix}1 & 1 \\ 2 & 4\end{vmatrix}.$
The minor of the $1,2$ entry is $\begin{vmatrix} 1 & 1 \\ 1 & 4 \end{vmatrix}$.
Repeating this process, the minors of all entries are $$\begin{pmatrix}
\begin{vmatrix}1 & 1 \\ 2 & 4\end{vmatrix} &
\begin{vmatrix}1 & 1 \\ 1 & 4\end{vmatrix} &
\begin{vmatrix}1 & 1 \\ 1 & 2\end{vmatrix} \\
\begin{vmatrix}-1 & 1 \\ 2 & 4\end{vmatrix} &
\begin{vmatrix}1 & 1 \\ 1 & 4\end{vmatrix} &
\begin{vmatrix}1 & -1 \\ 1 & 2\end{vmatrix} \\
\begin{vmatrix}-1 & 1 \\ 1 & 1\end{vmatrix} &
\begin{vmatrix}1 & 1 \\ 1 & 1\end{vmatrix} &
\begin{vmatrix}1 & -1 \\ 1 & 1\end{vmatrix}
\end{pmatrix}$$
Recall that
Determinant of a 2x2 matrix
The determinant of a $2 \times 2$ matrix is $\begin{vmatrix}a & b \\ c & d \end{vmatrix} = ad - bc.$
Computing all the determinants, the matrix of minors is $$\begin{pmatrix} 2 & 3 & 1 \\ -6 & 3 & 3 \\ -2 &0 & 2 \end{pmatrix}.$$
Next we negate every other entry, according to the pattern $$\begin{pmatrix} + & - & + \\ - & + & - \\ + & - & +\end{pmatrix}.$$
The matrix of minors becomes $$\begin{pmatrix} 2 & -3 & 1 \\6 & 3 & -3 \\ -2 & 0 & 2\end{pmatrix}.$$
Transposing, we get $$\begin{pmatrix} 2 & 6 & -2 \\ -3 & 3 & 0 \\ 1 & -3 & 2 \end{pmatrix}.$$
In order to divide by the determinant of $\bfA$, we must first compute it.
(see details)
We can compute that $\text{det } \bfA = 6.$
Hence, $$\bfA^{-1} = \frac{1}{6}\begin{pmatrix} 2 & 6 & -2 \\ -3 & 3 & 0 \\ 1 & -3 & 2 \end{pmatrix}.$$
To solve the linear system, recall that
Linear equations
if $\bfA$ is square and invertible, then the solution to $\bfA \bfx = \bfb$ is $\bfx = \bfA^{-1} \bfb$.
Computing, we have $$\bfx = \frac{1}{6}\begin{pmatrix} 2 & 6 & -2 \\ -3 & 3 & 0 \\ 1 & -3 & 2 \end{pmatrix} \begin{pmatrix} 1\\3\\2 \end{pmatrix}$$
My matrix multiplication we get $$ \bfx = \begin{pmatrix} 8/3 \\ 1 \\ -2/3\end{pmatrix}.$$
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