math111_logo Properties of Determinant

2. Invertibility and determinant

If A is an invertible matrix. Let B be the inverse of A. Then by the properties in this section, we have

detA detB = detAB = detI = 1.

In particular, we just proved the ⇒ direction of the following property.

A square matrix A invertible ⇔ detA ≠ 0.

To prove the ⇐ direction, we need to use the assumption detA ≠ 0 to construct an inverse of A. We illustrate the idea by a 4 by 4 matrix.

A = [ a11 a12 a13 a14 ].
a21 a22 a23 a24
a31 a32 a33 a34
a41 a42 a43 a44

Recall the cofactor expansion along the first row

a11C11 + a12C12 + a13C13 + a14C14 = det[ a11 a12 a13 a14 ].
a21 a22 a23 a24
a31 a32 a33 a34
a41 a42 a43 a44

If we replace the terms a11, a12, a13, a14 in the formula by a21, a22, a23, a24, then we have

a21C11 + a22C12 + a23C13 + a24C14 = det[ a21 a22 a23 a24 ].
a21 a22 a23 a24
a31 a32 a33 a34
a41 a42 a43 a44

The determinant vanishes by the following result.

A has a zero row (column) ⇒ detA = 0.

A row (column) in A is a scalar multiple of another row (column) ⇒ detA = 0.

Proof If A has a zero row, then by cofactor expansion along the zero row, we get detA = 0. If [row i] is r multiple of [row j], then apply (-r)[row i] + [row j] to A to get B. By this property, we have detA = detB. Moreover, the j-th row of B is zero, so that detB = 0.

The argument we had clearly generalizes to the following result.

ai1Cj1 + ai2Cj2 + ai3Cj3 + ai4Cj4 = { detA i = j
0 ij

In other words, if we define the adjugate

adjA = [ C11 C21 C31 C41 ],
C12 C22 C32 C42
C13 C23 C33 C43
C14 C24 C34 C44

then we have

A adjA = (detA)I.

Since A is a square matrix, by this property, we conclude

detA ≠ 0 ⇒ A is invertible, with A-1 = (detA)-1 adjA.

We had the 2 by 2 case of this result before.

Example In an earlier example, we used row operations to find the inverse of

A = [ 3 1 -1 ]
1 -1 1
2 2 1

Now we use our new formula. First,

detA = det[ 3 4 -4 ]
= -det[ 4 -4 ] = -12.
4 -1
1 0 0
2 4 -1

Then

C11 = (-1)1+1det[ -1 1 ] = -3, C21 = (-1)2+1det[ 1 -1 ] = -3, C31 = (-1)3+1det[ 1 -1 ] = 0,
2 1 2 1 -1 1
 
C12 = (-1)1+2det[ 1 1 ] = 1, C22 = (-1)2+2det[ 3 -1 ] = 5, C32 = (-1)3+2det[ 3 -1 ] = -4,
2 1 2 1 1 1
 
C13 = (-1)1+3det[ 1 -1 ] = 4, C23 = (-1)2+3det[ 3 1 ] = -4, C33 = (-1)3+3det[ 3 1 ] = -4,
2 2 2 2 1 -1

and

A-1 = (-1/12)[ -3 -3 0 ] = [ 1/4 1/4 0 ].
1 5 -4 -1/12 -5/12 1/3
4 -4 -4 -1/3 1/3 1/3

One find the result to be the same as the earlier computation. However, the computation here is much more complicated. It is even more complicated if you try the inverse of matrices of bigger size. Therefore row operation is still the practical way of computing the inverse.


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