Let *A* = [ *a _{ij} *] be a square matrix. The transpose of the matrix whose (

*i, j*) entry is the

*a*cofactor is called the classical

_{ij}**adjoint**of

*A*:

**Example 1**: Find the adjoint of the matrix

The first step is to evaluate the cofactor of every entry:

Therefore,

Why form the adjoint matrix? First, verify the following calculation where the matrix *A* above is multiplied by its adjoint:

Now, since a Laplace expansion by the first column of *A* gives

This result gives the following equation for the inverse of *A*:

By generalizing these calculations to an arbitrary *n* by *n* matrix, the following theorem can be proved:

**Theorem H**. A square matrix *A* is invertible if and only if its determinant is not zero, and its inverse is obtained by multiplying the adjoint of *A* by (det *A*) ^{−1}. [Note: A matrix whose determinant is 0 is said to be **singular**; therefore, a matrix is invertible if and only if it is nonsingular.]

**Example 2**: Determine the inverse of the following matrix by first computing its adjoint:

First, evaluate the cofactor of each entry in *A*:

These computations imply that

Now, since Laplace expansion along the first row gives

*A* is

*AA* ^{−1} = *A* ^{−1} *A* = *I*.

**Example 3**: If *A* is an invertible *n* by *n* matrix, compute the determinant of Adj *A* in terms of det *A*.

Because *A* is invertible, the equation *A* ^{−1} = Adj *A*/det *A* implies

Recall that if *B* is *n* x *n* and *k* is a scalar, then det( *kB*) = *k ^{n} *det

*B*. Applying this formula with

*k*= det

*A*and

*B*=

*A*

^{−1}gives

Thus,

**Example 4**: Show that the adjoint of the adjoint of *A* is guaranteed to equal *A* if *A* is an invertible 2 by 2 matrix, but not if *A* is an invertible square matrix of higher order.

First, the equation *A* · Adj *A* = (det *A*) *I* can be rewritten

Next, the equation *A* · Adj *A* = (det *A*) *I* also implies

This expression, along with the result of Example 3, transforms (*) into

*n* is the size of the square matrix *A*. If *n* = 2, then (det *A*) ^{n−2 }= (det *A*) ^{0} = 1—since det *A* ≠ 0—which implies Adj (Adj *A*) = *A*, as desired. However, if *n* > 2, then (det *A*) ^{n−2 }will not equal 1 for every nonzero value of det *A*, so Adj (Adj *A*) will not necessarily equal *A*. Yet this proof does show that whatever the size of the matrix, Adj (Adj *A*) will equal *A* if det *A* = 1.

**Example 5**: Consider the vector space *C* ^{2}( *a, b*) of functions which have a continuous second derivative on the interval ( *a, b*) ⊂ **R**. If *f, g*, and *h* are functions in this space, then the following determinant,

**Wronskian** of *f, g*, and *h*. What does the value of the Wronskian say about the linear independence of the functions *f, g*, and *h*?

The functions *f, g*, and *h* are linearly independent if the only scalars *c* _{1}, *c* _{2}, and *c* _{3} which satisfy the equation *c* _{1} = *c* _{2} = *c* _{3} = 0. One way to obtain three equations to solve for the three unknowns *c* _{1}, *c* _{2}, and *c* _{3} is to differentiate (*) and then to differentiate it again. The result is the system

**c** = ( *c* _{1}, *c* _{2}, *c* _{3}) ^{T}. A homogeneous square system—such as this one—has only the trivial solution if and only if the determinant of the coefficient matrix is nonzero. But if **c** = **0** is the only solution to (**), then *c* _{1} = *c* _{2} = *c* _{3} = 0 is the only solution to (*), and the functions *f, g*, and *h* are linearly independent. Therefore,

To illustrate this result, consider the functions *f, g*, and *h* defined by the equations

Since the Wronskian of these functions is

Here's another illustration. Consider the functions *f, g*, and *h* in the space *C* ^{2}(1/2, ∞) defined by the equations

By a Laplace expansion along the second column, the Wronskian of these functions is

Since this function is not identically zero on the interval (1/2, ∞)—for example, when *x* = 1, *W*( *x*) = *W*(1) = *e* ≠ 0—the functions *f, g*, and *h* are linearly independent.