Step 1
To find the inverse of a 3 x 3 matrix, we start by augmenting the original matrix to the 3 x 3 identity matrix.
$\begin{bmatrix} 5 & 2 & 1 \\ 5 & 6 & 7 \\ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
We now perform elementary row operations until the matrix on the left is transformed into the identity matrix. This will transform the identity matrix on the right into the
inverse of the original matrix.
To get a 1 in row 1, column 1, we can simply switch row 3 with row 1.
$\begin{bmatrix} 5 & 2 & 1 \\ 5 & 6 & 7 \\ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 1 & 1 \\ 5 & 6 & 7 \\ 5 & 2 & 1 \end{bmatrix} \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}$
Step 2
To get a zero in row 2, column 1 we can add -5 times row 1 to row 2 and put the result in row 2. To get a zero in row 3, column 1 we can add -5 times row 1 to
the result in row 3.
$\begin{bmatrix} 1 & 1 & 1 \\ 5 & 6 & 7 \\ 5 & 2 & 1 \end{bmatrix} \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & -3 & -4 \end{bmatrix} \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & -5 \\ 1 & 0 & -5 \end{bmatrix}$Enter a number.
