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Find the inverse of the following matrix:
\\[A = \\left(\\begin{array}{rrr} \\var{a11} & \\var{a12} & \\var{a13}\\\\ \\var{a21} & \\var{a22} & \\var{a23}\\\\ \\var{a31} & \\var{a32} & \\var{a33}\\\\ \\end{array}\\right)\\]

Form the $3 \\times 6$ augmented matrix $B$ by placing $I_3$ to the right of $A$ as below:
\\[B = \\left(\\begin{array}{rrr|ccc} \\var{a11} & \\var{a12} & \\var{a13} &\\var{1}&\\var{0}&\\var{0}\\\\ \\var{a21} & \\var{a22} & \\var{a23}&\\var{0}&\\var{1}&\\var{0}\\\\ \\var{a31} & \\var{a32} & \\var{a33}&\\var{0}&\\var{0}&\\var{1}\\\\ \\end{array}\\right)\\]

In subsequent parts work with this matrix using row operations and introduce the identity matrix on the left hand side with the inverse of A eventually appearing on the right hand side.

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$A$ a $3 \\times 3$ matrix. Using row operations on the augmented matrix $\\left(A | I_3\\right)$ reduce to $\\left(I_3 | A^{-1}\\right)$.

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We begin with the augmented matrix

\\[B = \\left(\\begin{array}{rrr|ccc} \\var{a11} & \\var{a12} & \\var{a13} &\\var{1}&\\var{0}&\\var{0}\\\\ \\var{a21} & \\var{a22} & \\var{a23}&\\var{0}&\\var{1}&\\var{0}\\\\ \\var{a31} & \\var{a32} & \\var{a33}&\\var{0}&\\var{0}&\\var{1}\\\\ \\end{array}\\right)\\]

Adding $\\var{-a21}$ lots of row 1 to row 2 gives:

\\[B = \\left(\\begin{array}{rrr|ccc} \\var{a11} & \\var{a12} & \\var{a13} &\\var{1}&\\var{0}&\\var{0}\\\\ 0 & \\var{a22-a21*a12} & \\var{a23-a21*a13}&\\var{-a21}&\\var{1}&\\var{0}\\\\ \\var{a31} & \\var{a32} & \\var{a33}&\\var{0}&\\var{0}&\\var{1}\\\\ \\end{array}\\right)\\]

Adding $\\var{-a31}$ lots of row 1 to row 3 gives:

\\[B = \\left(\\begin{array}{rrr|ccc} \\var{a11} & \\var{a12} & \\var{a13} &\\var{1}&\\var{0}&\\var{0}\\\\ 0 & \\var{a22-a21*a12} & \\var{a23-a21*a13}&\\var{-a21}&\\var{1}&\\var{0}\\\\ 0 & \\var{a32-a31*a12} & \\var{a33-a31*a13}&\\var{-a31}&\\var{0}&\\var{1}\\\\ \\end{array}\\right)\\]

We make the leading element of row 2 equal 1 by dividing row 2 by $-1$

\\[B = \\left(\\begin{array}{rrr|ccc} \\var{a11} & \\var{a12} & \\var{a13} &\\var{1}&\\var{0}&\\var{0}\\\\ 0 & 1 & \\var{-(a23-a21*a13)}&\\var{a21}&\\var{-1}&\\var{0}\\\\ 0 & \\var{a32-a31*a12} & \\var{a33-a31*a13}&\\var{-a31}&\\var{0}&\\var{1}\\\\ \\end{array}\\right)\\]

We can obtain a 0 at the bottom of column 2 by adding $\\var{-(a32-a31*a12)}$ lots of row 2 to row 3:

\\[B = \\left(\\begin{array}{rrr|ccc} \\var{a11} & \\var{a12} & \\var{a13} &\\var{1}&\\var{0}&\\var{0}\\\\ 0 & 1 & \\var{-(a23-a21*a13)}&\\var{a21}&\\var{-1}&\\var{0}\\\\ 0 & 0 & \\var{1}&\\var{-a31-(a32-a31*a12)*a21}&\\var{-(a32-a31*a12)*-1}&\\var{1}\\\\ \\end{array}\\right)\\]

We can create a zero in column 3, row 1 by adding $\\var{-a13}$ lots of row 3 to row 1

\\[B = \\left(\\begin{array}{rrr|ccc} \\var{a11} & \\var{a12} & \\var{0}&\\var{1-a13*(-a31-(a32-a31*a12)*a21)} &\\var{a13*(-a32+a31*a12)}&\\var{-a13}\\\\ 0 & 1 & \\var{-(a23-a21*a13)}&\\var{a21}&\\var{-1}&\\var{0}\\\\ 0 & 0 & \\var{1}&\\var{-a31-(a32-a31*a12)*a21}&\\var{-(a32-a31*a12)*-1}&\\var{1}\\\\ \\end{array}\\right)\\]

We can create a zero in column 3, row 2 by adding $\\var{(a23-a21*a13)}$ lots of row 3 to row 2

\\[B = \\left(\\begin{array}{rrr|ccc} \\var{a11} & \\var{a12} & \\var{0}&\\var{1-a13*(-a31-(a32-a31*a12)*a21)} &\\var{a13*(-a32+a31*a12)}&\\var{-a13}\\\\ 0 & 1 & 0&\\var{a21+(a23-a21*a13)*(-a31-(a32-a31*a12)*a21)}&\\var{-1+(a23-a21*a13)*(a32-a31*a12)}&\\var{(a23-a21*a13)}\\\\ 0 & 0 & \\var{1}&\\var{(-a31-(a32-a31*a12)*a21)}&\\var{(a32-a31*a12)}&\\var{1}\\\\ \\end{array}\\right)\\]

Finally we can create a zero in column 2, row 1 by adding $\\var{-a12}$ lots of row 2 to row 1

\\[B = \\left(\\begin{array}{rrr|ccc} \\var{a11} & 0 & \\var{0}&\\var{1-a13*(-a31-(a32-a31*a12)*a21)-a12*(a21+(a23-a21*a13)*(-a31-(a32-a31*a12)*a21))} &\\var{a13*(-a32+a31*a12)-a12*(-1+(a23-a21*a13)*(a32-a31*a12))}&\\var{-a13-a12*(a23-a21*a13)}\\\\ 0 & 1 & 0&\\var{a21+(a23-a21*a13)*(-a31-(a32-a31*a12)*a21)}&\\var{-1+(a23-a21*a13)*(a32-a31*a12)}&\\var{(a23-a21*a13)}\\\\ 0 & 0 & \\var{1}&\\var{(-a31-(a32-a31*a12)*a21)}&\\var{(a32-a31*a12)}&\\var{1}\\\\ \\end{array}\\right)\\]

Then the inverse $A^{-1}$ is the $3 \\times 3$ matrix on the right hand side, namely:

\\[A^{-1} = \\left(\\begin{array} &\\var{1-a13*(-a31-(a32-a31*a12)*a21)-a12*(a21+(a23-a21*a13)*(-a31-(a32-a31*a12)*a21))} &\\var{a13*(-a32+a31*a12)-a12*(-1+(a23-a21*a13)*(a32-a31*a12))}&\\var{-a13-a12*(a23-a21*a13)}\\\\ \\var{a21+(a23-a21*a13)*(-a31-(a32-a31*a12)*a21)}&\\var{-1+(a23-a21*a13)*(a32-a31*a12)}&\\var{(a23-a21*a13)}\\\\ \\var{(-a31-(a32-a31*a12)*a21)}&\\var{(a32-a31*a12)}&\\var{1}\\\\ \\end{array}\\right)\\]

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Part 1.

Introduce zeros in the first column below the first entry by adding suitable multiples of the first row to rows 2 and 3.

Input all numbers as fractions or integers and not as decimals.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

\\[\\left( \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\end{matrix} \\right.\\]	$\\var{a11}$	$\\var{a12}$	$\\var{a13}$		$1$	$0$	$0$	\\[\\left) \\begin{matrix} \\phantom{.} \\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\end{matrix} \\right.\\]
$0$	[[0]]	[[1]]		[[2]]	$1$	$0$
$0$	[[3]]	[[4]]		[[5]]	$0$	$1$

Now, if necessary, multiply the second row by a suitable number so that the second entry in the second row is 1.

Part 2.

Now using this matrix, introduce a zero in the second column below the second entry of the second column by:

Adding [[0]] times the second row to the third row to get the matrix:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

\\[\\left( \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\end{matrix} \\right.\\]	$\\var{a11}$	$\\var{a12}$	$\\var{a13}$	$1$	$0$	$0$	\\[\\left) \\begin{matrix} \\phantom{.} \\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\end{matrix} \\right.\\]
$0$	$1$	[[1]]	[[2]]	[[3]]	$0$
$0$	$0$	[[4]]	[[5]]	[[6]]	$1$

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Part 3.

Now, if necessary, multiply the third row by a suitable constant so that the third entry in the third column is 1.

With this matrix, use the third row to introduce zeros into the second and first entries in the third column by adding suitable multiples of the third row to the second and first rows.

Multiply third row by [[0]]and add to the second row.

Multiply third row by [[1]]and add to the first row.

Using this new matrix there is one more operation needed.

Multiplying the second row by $\\var{-a12}$ and adding to the first row to obtain the inverse matrix appearing on the right hand side.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

\\[\\left( \\begin{matrix} \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\end{matrix} \\right.\\]	$1$	$0$	$0$	[[2]]	[[3]]	[[4]]	\\[\\left) \\begin{matrix} \\phantom{.} \\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\end{matrix} \\right.\\]
$0$	$1$	$0$	[[5]]	[[6]]	[[7]]
$0$	$0$	$1$	[[8]]	[[9]]	[[10]]

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