Inverse of a $4 \times 4$ matrix with variables

Question

I missed my class on the inverses of matrices. I'm catching up well, but there's a problem in the book that got me stumped.

It's a $4 \times 4$ matrix that is almost an identity matrix, but whose bottom row is $a,b,c,d$ instead of $0,0,0,1$.

$$\begin{pmatrix} 1 &0 &0 &0 \\ 0 &1 &0 &0 \\ 0& 0 & 1 &0 \\ a &b & c & d \end{pmatrix}$$

Any pointers?

Answer 1

Recall that one way to compute an inverse is by forming the augmented matrix

$$ (A \vert I) $$ and then using Gaussian elimination to completely row-reduce $A$. The final result will be of the form $$ (I \vert A^{-1}). $$

So, in this case, you would write $$ \begin{bmatrix} 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 \\ a & b & c & d & 0 & 0 & 0 & 1 \end{bmatrix} $$

and the first step would be to add $-a \times$ (first row) to (last row), i.e. $$ \begin{bmatrix} 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 \\ 0 & b & c & d & -a & 0 & 0 & 1 \end{bmatrix} $$

etc.

Answer 2

The systematic way to compute an inverse to matrix $A$ is as follows.

1. Adjoin the identity to $A$, i.e. $[A|I]$, to form a matrix with $n$ rows and $2n$ columns.

2. Perform row reduction to turn $A$ into the identity.

3a. If you can't, i.e. there's a row of all zeroes, then $A$ is not invertible.

3b. If you can, the result will be $[I|B]$, for some matrix $B$. This $B$ is the inverse of $A$.