Find the inverse matrix using Gauss-Jordan method

So far our knowledge about inverse matrices is restricted to

2 \times 2

matrices. The following method will allow us to find the inverse of any dimensional matrix - if it exists - using Gauss-Jordan method.

We are given a matrix and the identity of the same dimentions:

A = (\begin{array}{ccc} a_{11} & \dots & a_{1 n} \\ ⋮ & ⋱ & ⋮ \\ a_{n 1} & \dots & a_{n n} \end{array}) and I_{n} = (\begin{array}{ccc} 1 & \dots & 0 \\ ⋮ & ⋱ & ⋮ \\ 0 & \dots & 1 \end{array})

If we define the matrix

(\begin{array}{cccccc} a_{11} & \dots & a_{1 n} & 1 & \dots & 0 \\ ⋮ & ⋱ & ⋮ & ⋮ & ⋱ & ⋮ \\ a_{n 1} & \dots & a_{n n} & 0 & \dots & 1 \end{array}) = (A ∣ I)

And we apply now Gauss-Jordan so that matrix

A

is transformed into

I

, then our initial right part of

(A ∣ I)

will be transformed into

A^{- 1}

finalising the process as

(\begin{array}{cccccc} 1 & \dots & 0 & a_{11}^{- 1} & \dots & a_{1 n}^{- 1} \\ ⋮ & ⋱ & ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & \dots & 1 & a_{n 1}^{- 1} & \dots & a_{n n}^{- 1} \end{array}) = (I ∣ A^{- 1})

Find the inverse of

A = (\begin{array}{rrr} 8 & 4 & 4 \\ - 3 & - 9 & - 6 \\ 1 & - 7 & - 5 \end{array})

Recommended for you