High School
SAT
ACT
University
MCAT
LSAT
DAT
Wize High School Algebra II Textbook (Common Core) > Matrices (Extension Topic)

Basics of Determinants

Solving Matrix EquationsPrevious Section
Basics of Systems of Linear EquationsNext Section
Popular Courses
Find My Course
0:00 / 0:00

Basics of Determinants

Determinant Definition

Every square matrix has an associated special scalar number called the determinant of the matrix.
The determinant has many uses, including calculating inverse matrices and finding solutions to SLEs.
Notation
Given a square matrix A=[a11a12…a1na21a22…a2n⋮⋮⋮an1an2…ann]A = \begin{bmatrix} a_{11} & a_{12} & \dots & a_{1n}\\ a_{21} & a_{22} & \dots & a_{2n}\\ \vdots & \vdots & & \vdots\\ a_{n1} & a_{n2} & \dots & a_{nn}\\ \end{bmatrix} , the determinant of AA is denoted with vertical bars:
det(A)=∣a11a12…a1na21a22…a2n⋮⋮⋮an1an2…ann∣\det{A} = \left \vert \begin{array}{cccc} a_{11} & a_{12} & \dots & a_{1n}\\ a_{21} & a_{22} & \dots & a_{2n}\\ \vdots & \vdots & & \vdots\\ a_{n1} & a_{n2} & \dots & a_{nn}\\ \end{array} \right\vert
Calculating Determinants of Small Matrices
  • 1×11\times1 matrix: det([ a ])=a\det{[\ a\ ]}=a
  • 2×22\times2 matrix: det[abcd] = ∣abcd∣ = ad−bc{\rm det} \begin{bmatrix} a&b\\ c&d \end{bmatrix} \ =\ \left \vert \begin{array}{rr} a&b\\ c&d \end{array} \right \vert \ =\ ad-bc
PAGE BREAK

Determinant Properties

Let An×nA_{n \times n} and  Bn×n\ B_{n \times n} be square matrices.
  • If AA has a zero row or zero column, then det(A)=0\det{A}=0
  • If two rows or two columns are scalar multiples of one another, then det(A)=0\det{A}=0
  • det(A)=det(AT)\det{A}=\det{A^T}
  • det(A−1)=1det(A)\det{A^{-1}}=\dfrac{1}{\det{A}}
  • det(cA)=cn det(A)\det{cA}=c^n\ \det{A}
  • det(AB)=det(A)⋅det(B)\det{AB} = \det{A} \cdot \det{B}
Note: In general, det(A+B)≠det(A)+det(B)\det{A+B} \colorTwo{\neq} \det{A} + \det{B}.
PAGE BREAK

Triangular Matrices

  • Upper triangular matrix: all entries under the main diagonal are 0s. Example:[abc0de00f]\begin{bmatrix} \colorTwo a & \colorTwo b & \colorTwo c\\ 0 & \colorTwo d & \colorTwo e\\ 0 & 0 & \colorTwo f\\ \end{bmatrix}
  • Lower triangular matrix: all entries above the main diagonal are 0s. Example: [a00bc0def]\begin{bmatrix} \colorTwo a & 0& 0 \\ \colorTwo b & \colorTwo c &0\\ \colorTwo d & \colorTwo e & \colorTwo f\\ \end{bmatrix}
  • Diagonal matrix: both upper and lower triangular. All entries that are not on the main diagonal are 0s. Example: [a000b000c]\begin{bmatrix} \colorTwo a & 0 & 0\\ 0 & \colorTwo b &0\\ 0 & 0 & \colorTwo c\\ \end{bmatrix}
Wize Tip
If a square matrix is triangular then the determinant is the product of the entries on the main diagonal:
det(A)=a11×a22×...×ann\det{A}=a_{11}\times a_{22}\times ...\times a_{nn}

0:00 / 0:00

Example: Basics of Determinants

Part A)

Find the determinant of A=[−12−34]A=\begin{bmatrix} -1&2\\ -3&4 \end{bmatrix}.
det(A) = ad−bc = (−1)(4)−(2)(−3) = −4−(−6) = 2{\rm det}(A) \ =\ ad - bc \ =\ \left(−1\right)\left(4\right)−\left(2\right)\left(-3\right) \ =\ −4−\left(−6\right) \ =\ 2

PAGE BREAK

Part B)

Find the determinant of B=[3007−5062−1]B= \left[\begin{array}{rrr} 3&0&0\\ 7&-5&0\\ 6&2&-1 \end{array}\right].
Notice that BB is lower triangular (0s above the main diagonal).
Then its determinant is the product of the diagonal entries.
det(B) = (3)(−5)(−1) = 15\det{B} \ =\ (3)(-5)(-1) \ =\ 15


PAGE BREAK

Part C)

Given C=[23−11]C= \left[ \begin{array}{rr} 2&3\\ -1&1 \end{array} \right] and D=[−331−1]D= \left[ \begin{array}{rr} -3&3\\ 1&-1 \end{array} \right], find det(3CD)\det{3CD} without using matrix multiplication.

Using determinant properties:
  • det(3CD)=32⋅det(CD)\det{3CD} = 3^2 \cdot\det{CD}
  • Since the product CDCD is a 2×22 \times 2 matrix, we must raise the scalar to the power of 22.
  • det(CD)=det(C) det(D)\det{CD} = \det{C} \ \det{D}
So we can compute:
det(3CD)=32⋅det(C)⋅det(D)=9⋅∣23−11∣⋅∣−331−1∣=9⋅[(2)(1)−(3)(−1)]⋅[(−3)(−1)−(3)(1)]=9⋅5⋅0=0\begin{aligned} \det{3CD} &= 3^2\cdot \det{C}\cdot\det{D}\\[0.5em] &=9 \cdot \left\vert \begin{array}{rr} 2&3\\ -1&1 \end{array} \right\vert \cdot \left\vert \begin{array}{rr} -3&3\\ 1&-1 \end{array} \right\vert \\[1em] &=9 \cdot \big[(2)(1)-(3)(-1)\big]\cdot \big[(-3)(-1)-(3)(1)\big] \\[0.5em] &=9 \cdot 5 \cdot 0 \\[0.5em] &= 0 \end{aligned}
Note: det(D)=0\det{D}=0 since it contains rows (and columns) that are scalar multiples of one another.
Exam Tip
Before computing determinants, inspect whether any rows/columns are scalar multiples of one another.
If so, you know the determinant is 0.


Determinant and Area


A determinant is an important piece when finding the inverse of a matrix, but that's not all we can do with it.

The absolute value of the determinant of a 2x2 matrix gives us the area of a parallelogram if we interpret the sides as the column vectors.

Use this idea to find the area of the follow parallelogram.
Hint, first find the vectors that represent the sides. Put them together to make a 2x2 matrix.



ANSWER:
From the graph we can see two vectors that make up the sides


Using these as column vectors we have the matrix A=[−1324]A = \left[ \begin{array}{cc} -1 & 3 \\ 2 & 4 \end{array} \right]
From this we can calculate the absolute value of the determinant.

∣det(A)∣=∣(−1)(4)−(2)(3)∣=∣−4−6∣=∣−10∣=10\begin{aligned} |\det A| &= |(-1)(4) - (2)(3)| \\ &= |-4 - 6| \\ &= |-10| \\ &=10 \end{aligned}
So the area of the parallelogram is 10 units.
Let CC be an invertible matrix with det(C)=3\det C = 3, and let B=[−2−1001000−5]B= \left[ \begin{array}{rrr} -2 & -1 & 0\\ 0 & 1 & 0\\ 0 & 0 & -5\\ \end{array} \right].
If A=C−1BCA = C^{-1}BC, calculate det(A)\det A.

Answer the following true/false questions.

True or False?
Given A4×4A_{4 \times 4} and det(A)=−3\det A = -3, then det(AT)=1det(A−1)\det{A^T}=\dfrac{1}{\det{A^{-1}}}.