Ch3-Determinants

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3.1 Introduction to Determinants
Definition
The determinant of a 2 × 2 matrix A is denoted |A| and is given by
Observe that the determinant of a 2 × 2 matrix is given by the different of the products of the two diagonals of the matrix.
The notation det(A) is also used for the determinant of A.

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Definition
Let A be a square matrix.
The minor of the element aij is denoted Mij and is the determinant of the matrix that remains after deleting row i and column j of A.
The cofactor of aij is denoted Cij and is given by
Cij = (–1)i+j Mij
Note that Cij = Mij or −Mij .

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Example 2
Solution
Determine the minors and cofactors of the elements a11 and a32 of the following matrix A.

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Definition
The determinant of a square matrix is the sum of the products of the elements of the first row and their cofactors.
These equations are called cofactor expansions of |A|.

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Example 3
Evaluate the determinant of the following matrix A.
Solution

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Theorem 3.1
The determinant of a square matrix is the sum of the products of the elements of any row or column and their cofactors.
ith row expansion:
jth column expansion:

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Example 5
Evaluate the determinant of the following 4 × 4 matrix.
Solution

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Example 6
Solve the following equation for the variable x.
Solution
There are two solutions to this equation, x = – 2 or 3.

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Computing Determinants of 2 × 2 and 3 × 3 Matrices

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Homework
Exercises will be given by the teachers of the practical classes.

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Let A be an n × n matrix and c be a nonzero scalar.
If then |B| = c|A|.
If then |B| = –|A|.
If then |B| = |A|.
3.2 Properties of Determinants
Theorem 3.2

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Example 1
Solution
Evaluate the determinant

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Example 2
If |A| = 12 is known.
Evaluate the determinants of the following matrices.
Solution
Thus |B1| = 3|A| = 36.
Thus |B2| = – |A| = –12.
Thus |B3| = |A| = 12.

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Theorem 3.3
Let A be a square matrix. A is singular if
all the elements of a row (column) are zero.
two rows (columns) are equal.
two rows (columns) are proportional. (i.e., Ri=cRj)
Proof
(c) If Ri=cRj, then , row i of B is [0 0 … 0].
⇒ |A|=|B|=0
Definition
A square matrix A is said to be singular if |A|=0.
A is nonsingular if |A|≠0.

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Example 3
Show that the following matrices are singular.
Solution
All the elements in column 2 of A are zero. Thus |A| = 0.
Row 2 and row 3 are proportional. Thus |B| = 0.

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Theorem 3.4
Let A and B be n × n matrices and c be a nonzero scalar.
|cA| = cn|A|.
|AB| = |A||B|.
|At| = |A|.
(assuming A–1 exists)
Proof

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Example 4
If A is a 2 × 2 matrix with |A| = 4, use Theorem 3.4 to compute the following determinants.
(a) |3A| (b) |A2| (c) |5AtA–1|, assuming A–1 exists
Solution
|3A| = (32)|A| = 9 × 4 = 36.
|A2| = |AA| =|A| |A|= 4 × 4 = 16.
|5AtA–1| = (52)|AtA–1| = 25|At||A–1|

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Example 6
Prove that if A and B are square matrices of the same size, with A being singular, then AB is also singular. Is the converse true?
Solution
(⇒) |A| = 0 ⇒ |AB| = |A||B| = 0
Thus the matrix AB is singular.
(⇐)
|AB| = 0 ⇒ |A||B| = 0 ⇒ |A| = 0 or |B| = 0
Thus AB being singular implies that either A or B is singular. The inverse is not true.

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Homework
Exercises will be given by the teachers of the practical classes.
Solution

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3.3 Numerical Evaluation of a Determinant
Definition
A square matrix is called an upper triangular matrix if all the elements below the main diagonal are zero.
It is called a lower triangular matrix if all the elements above the main diagonal are zero.

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Example 1
Numerical Evaluation of a Determinant

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Numerical Evaluation of a Determinant
Example 2
Evaluation the determinant.

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Example 3
Evaluation the determinant.
Solution

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Example 4
Evaluation the determinant.
Solution

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Example 5
Evaluation the determinant.
Solution

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3.4 Determinants, Matrix Inverse, and Systems of Linear Equations
Definition
Let A be an n × n matrix and Cij be the cofactor of aij.
The matrix whose (i, j)th element is Cij is called the matrix of cofactors of A.
The transpose of this matrix is called the adjoint of A and is denoted adj(A).

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Example 1
Give the matrix of cofactors and the adjoint matrix of the following matrix A.
The matrix of cofactors
of A is
The adjoint of A is

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Theorem 3.6
Let A be a square matrix with |A| ≠ 0. A is invertible with

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∴ A⋅ adj(A) = |A|In
Proof of Theorem 3.6
If i ≠ j, let
row i = row j in B
Matrices A and B have the same cofactors
Cj1, Cj2, …, Cjn.

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Theorem 3.7
A square matrix A is invertible if and only if |A| ≠ 0.
Proof
(⇒) Assume that A is invertible.
⇒ AA–1 = In.
⇒ |AA–1| = |In|.
⇒ |A||A–1| = 1
⇒ |A| ≠ 0.
(⇐) Theorem 3.6 tells us that if |A| ≠ 0, then A is invertible.
A–1 exists if and only if |A| ≠ 0.

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Example 2
Use a determinant to find out which of the following matrices are invertible.
Solution
|A| = 5 ≠ 0. A is invertible.
|B| = 0. B is singular. The inverse does not exist.
|C| = 0. C is singular. The inverse does not exist.
|D| = 2 ≠ 0. D is invertible.

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Example 3
Use the formula for the inverse of a matrix to compute the inverse of the matrix
Solution

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Homework
Exercises will be given by the teachers of the practical classes.
Exercise
Show that if A = A-1, then |A| = ±1.
Show that if At = A-1, then |A| = ±1.

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Theorem 3.8
Let AX = B be a system of n linear equations in n variables.
(1) If |A| ≠ 0, there is a unique solution.
(2) If |A| = 0, there may be many or no solutions.
Proof
If |A| ≠ 0
⇒ A–1 exists (Thm 3.7)
⇒ there is then a unique solution given by X = A–1B (Thm 2.9).
(2) If |A| = 0
⇒ since A ≈…≈ C implies that if |A|≠0 then |C|≠0 (Thm 3.2).
⇒ the reduced echelon form of A is not In.
⇒ The solution to the system AX = B is not unique.
⇒ many or no solutions.

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Example 4
Determine whether or not the following system of equations has an unique solution.

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Theorem 3.9 Cramer’s Rule
Let AX = B be a system of n linear equations in n variables such that |A| ≠ 0. The system has a unique solution given by
Where Ai is the matrix obtained by replacing column i of A with B.

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xi, the ith element of X, is given by
Proof of Cramer’s Rule

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Example 5
Solving the following system of equations using Cramer’s rule.

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Giving
Cramer’s rule now gives

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Example 6
Determine values of λ for which the following system of equations has nontrivial solutions.Find the solutions for each value of λ.
Solution
homogeneous system
⇒ x1 = 0, x2 = 0 is the trivial solution.
⇒ nontrivial solutions exist ⇒ many solutions
⇒
⇒ ⇒ ⇒
⇒ λ = – 3 or λ = 2.

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