Determine whether A is invertible, and if so, ﬁnd the inverse. (3x3) 
In Exercises 37-38, determine whether $A$ is invertible, and if so, find the inverse. [Hint: Solve $AX = I$ for $X$ by equating corresponding entries on the two sides.
37. $A = \begin{bmatrix} 1&0&1 \\ 1&1&0 \\ 0&1&1 \end{bmatrix}$

How the heck am I supposed to find an inverse of a 3x3? If I'm supposed to solve for $X$ I'd normally multiply both sides by $A^{-1}$ but this book gave no mention of a formula for more than 4 elements in a matrix... so if I equate $AX$ to $I$, how do I solve for $X$?
 A: There are a couple of things you can do that do not involve finding the actual inverse:
1) Do Gaussian elimination. Then if you are left with a matrix with all zeros in a row, your matrix is not invertible.
You do this by adding multiples of the first row as the "pivot row" to other rows, so that you get rid of the leading entries; in your matrix, start by adding (-1)(first row) to the second row (note that this is one of the three basic operations that does not change the solutions of your system)
2) Compute the determinant of your matrix and use the fact that a matrix is invertible iff its determinant is nonzero. 
Given$$A = \left[\begin{array}{ccc}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31}& a_{32} & a_{33} \end{array} \right]$$
You can find the determinant by "cofactor expansion": start with the term $a_{11}$. Now, delete the first column and first row and multiply $a_{11}$ by the determinant of the remaining matrix. Then take $a_{12}$, delete the first row and the second column and multiply $-a_{12}$ by the remaining matrix. Do the same with $a_{13}$, by first deleting the first row of $A$ and the third column and multiplying $a_{13}$ by the determinant of the remaining matrix. Add these three terms and you have found the determinant of A. If this determinant is $0$, your matrix is not invertible; otherwise, it is invertible.
A: It looks like they are wanting you to consider an arbitrary matrix $X$, say:
$$X = \left[\begin{array}{ccc}
a & b & c \\
d & e & f \\
g & h & i \end{array} \right]$$
Calculate $AX$, then set the diagonal entries equal to $1$, and the rest equal to $0$.  Can you find values for $a, b, c, ...$ so that everything works out?  If so, then the matrix must be invertible.
There are FAR easier ways to determine whether a matrix is invertible, however.  If you have learned these methods, then here are two:


*

*Put the matrix into echelon form.  Does the matrix have full rank?  If so, it is invertible.

*Calculate $\det(A)$.  Is $\det(A) \neq 0$?  If so, the matrix is invertible.

A: This can be done by inspection. Just ask yourself: what linear combination of the columns gives me $$(1): [1,0,0]^T$$ The answer is $c_1+c_3-c_2$ ("$c_i$" is "column $i$") divided by $2$.
So the first column of the matrix $A^{-1}$ is just ${1\over 2}[1,-1,1]^T$. Do the same for the other two columns.
