Skip to main content
Logo image

Discover Linear Algebra: A first course in linear algebra

Jeremy Sylvestre

Section 9.3 Examples

Subsection 9.3.1 Determinants by row reduction

As discussed in Warning 8.3.3, determinants by cofactor expansions are extremely inefficient for matrices larger than 3×3. Here we provide an example of using the row reduction method to compute a determinant (Procedure 9.2.2).
🔗

Example 9.3.1. Using row reduction to compute a determinant.

Let’s recompute the determinant of
🔗
A=[1−1212011010−31−2−10],
the same matrix from Example 8.4.8.
🔗
🔗
We will carry out the first two steps of Procedure 9.2.2 simultaneously, displaying the determinant of each matrix below it. For the purposes of describing how the sequence of determinant values progresses afterward, we’ll label our matrices as we go. As in the procedure, let d represent the unknown determinant value of A.
🔗
(i) detA=d⟶A=[1−1212011010−31−2−10]R2−2R1R4−R1(i) detA1=d⟶A1=[1−12102−3−1010−30−1−3−1]XR2↔R3X(ii) detA2=−d⟶A2=[1−121010−302−3−10−1−3−1]R3−2R2R4+R2(iii) detA3=−d⟶A3=[1−121010−300−3500−3−4]XX−13R3X(iv) detA4=d3⟶A4=[1−121010−3001−5300−3−4]xxxR4+3R3(v) detA5=d3⟶A5=[1−121010−3001−53000−9]
We would need one more operation to get to REF, but we are already at upper triangular so we don’t need to bother. And notice that we didn’t bother clearing entries above leading ones, since our goal was to get to an upper triangular matrix, which only requires entries below leading ones to be cleared.
🔗
🔗
Here is how we arrived at our sequence of determinant formulas:
🔗
  1. Matrix A1 was produced from A by a pair of row combination operations, but that type of operation does not change the determinant. So detA1=detA=d.
  2. Matrix A2 was produced from A1 by swapping rows, so the determinants of those two matrices have opposite signs: detA2=−detA1=−d.
  3. Matrix A3 was produced from A2 by another pair of row combination operations, so the determinants of those two matrices are equal: detA3=detA2=−d.
  4. Matrix A4 was produced from A3 by multiplying a row, and so the determinants are related by the same multiplication factor: detA4=−13detA3=d3.
  5. Matrix A5 was produced from A4 by another row combination operations, so detA5=detA4=d3.
But since A5 is an upper triangular matrix, we easily calculate its determinant as the product of its diagonal entries:
🔗
detA5=1⋅1⋅1⋅(−9)=−9.
Equating this to our formula for detA5 from our row operation analysis, we obtain
🔗
d3=−9d=−27.
Therefore, detA=−27, which agrees with the result of the calculation of detA by cofactor expansion in Example 8.4.8.
🔗
🔗
🔗

Example 9.3.2. A determinant of a matrix whose entries involve a variable.

Suppose we would like to know the values of the variable x that make the determinant of the matrix below equal to 0.
🔗
C=[x+634−72x+5−4−13−3−3x−17660x−14]
Let d represent the determinant of this matrix. We will reduce to upper triangular, but we will not strictly follow Procedure 2.3.2 to do so. In particular, we will not worry too much about obtaining leading ones, but will focus more on obtaining zeros. The reasons for this are:
🔗
  • A matrix with entries that contain a variable are complicated enough, so we will avoid introducing fractions into our matrix. To do this, we will multiply other rows to match a leading value instead of dividing to obtain a leading one.
  • We should not attempt to obtain a leading one by dividing a row by an expression involving x, since there may be values of x for which that operation would actually be dividing by zero.
We will not explain each choice of row operation or the progression of determinant values — see if you can fill in those details for yourself. We will also simplify/factor entries as we go to make it easier to see potential simplifying row operations.
🔗
detC=d⟶C=[x+634−72x+5−4−13−3−3x−17660x−14]XR1↔R2detC1=−d⟶C1=[2x+5−4−13x+634−7−3−3x−17660x−14]2R22R3XdetC2=−4d⟶C2=[2x+5−4−132(x+6)68−14−6−62(x−1)7660x−14]R2−(x+6)R1R3+3R1R4−3R1detC3=−4d⟶C3=[2x+5−4−130−(x+8)(x+3)4(x+8)13x+6403(x+3)2(x−7)−250−3(x+3)12x+25]3R2XXdetC4=−12d⟶C4=[2x+5−4−130−3(x+8)(x+3)12(x+8)39x+19203(x+3)2(x−7)−250−3(x+3)12x+25]R2+(x+8)R3R4+R3detC5=−12d⟶C5=[2x+5−4−13002(x−1)(x+8)14x−803(x+3)2(x−7)−25002(x−1)x]XR2↔R3XdetC6=12d⟶C6=[2x+5−4−1303(x+3)2(x−7)−25002(x−1)(x+8)14x−8002(x−1)x]R3↔R4detC7=−12d⟶C7=[2x+5−4−1303(x+3)2(x−7)−25002(x−1)x002(x−1)(x+8)14x−8]R4−(x+8)R3detC8=−12d⟶C8=[2x+5−4−1303(x+3)2(x−7)−25002(x−1)x000−(x−2)(x−4)]
We can now calculate the determinant of upper triangular matrix C6 as the product of its diagonal entries:
🔗
detC8=2⋅3(x+3)⋅2(x−1)⋅(−(x−2)(x−4))=−12(x+3)(x−1)(x−2)(x−4).
Since we also know detC8=−12d, where d represents detC, we must have
🔗
detC=(x+3)(x−1)(x−2)(x−4),
and so we have detC=0 precisely when x is one of the values −3,1,2,4.
🔗
🔗
🔗
🔗

Subsection 9.3.2 Matrices of determinant zero

Example 9.3.3. Recognizing detA=0.

Here are a few examples of recognizing matrices that have determinant 0.
🔗
  1. [1−121010−3010−31234]
  2. [1−1210155710−3−22−4−2]
  3. [1−101410−3−110−31204]
  4. [1−12−101517101−22−42]
The first matrix has two identical rows, the second matrix has two proportional rows (R4=−2R1), the third matrix has a column of zeros, and the fourth matrix has two identical columns. So the determinant of each of these matrices is 0.
🔗
🔗
🔗
🔗
🔗
PrevTopNext