Determinant of a Square Matrix

Determinant of a 2 × 2 Matrix - Definition

In order to explain the concept of determinant in linear algebra, we start with a 2 × 2 system of equations with unknowns $x$ and $y$ given by:

\[ \begin{cases} ax + by = c \\ dx + ey = f \end{cases} \]

Multiply all terms of the first equation in the system by the coefficient $e$ and all the terms in the second equation by the coefficient $-b$ to obtain an equivalent system:

\[ \begin{cases} aex + bey = ce \\ -dbx - eby = -fb \end{cases} \]

Add the left and right-hand sides of the above equations and simplify to obtain an equation in one variable only:

\[ aex - dbx = ce - fb \]

Solve for $x$:

\[ x = \frac{ce - bf}{ae - bd} \]

In a similar way, we can find $y$ as:

\[ y = \frac{af - cd}{ae - bd} \]

We note that both $x$ and $y$ are expressed as ratios with the same denominator $ae - bd$, which we define as the determinant of the square matrix $A$ as a scalar quantity given by:

\[ \det(A) = |A| = \begin{vmatrix} a & b \\ d & e \end{vmatrix} = ae - bd \]

Using determinants, we now write the solutions of the 2 × 2 system of equations as follows:

\[ x = \frac{\begin{vmatrix} c & b \\ f & e \end{vmatrix}}{\begin{vmatrix} a & b \\ d & e \end{vmatrix}} , \quad y = \frac{\begin{vmatrix} a & c \\ d & f \end{vmatrix}}{\begin{vmatrix} a & b \\ d & e \end{vmatrix}} \]

Similarly, we can express the solutions of any $n \times n$ system of equations using determinants of $n \times n$ square matrices. Determinants are scalar quantities used in solving systems of equations, calculating the inverse of a matrix, and many other applications.

Note that the determinant of a $1 \times 1$ matrix is equal to the entry in that matrix:

\[ \det[-2] = -2 \]

Determinant of a 3 × 3 Matrix or Higher - Minors and Cofactors

One possibility to calculate the determinant of a matrix is to use minors and cofactors of a square matrix.

Minors of a Square Matrix

The minor $M_{ij}$ of an $n \times n$ square matrix corresponding to the element $(A)_{ij}$ is the determinant of the $(n-1) \times (n-1)$ matrix obtained by deleting row $i$ and column $j$ of matrix $A$.

Cofactors of a Square Matrix

The cofactor $C_{ij}$ is given by $C_{ij} = (-1)^{i+j} M_{ij}$.

• $C_{11} = (-1)^{1+1} M_{11} = -23, \quad C_{12} = (-1)^{1+2} M_{12} = -47, \quad C_{13} = (-1)^{1+3} M_{13} = -18$
• $C_{21} = (-1)^{2+1} M_{21} = -19, \quad C_{22} = (-1)^{2+2} M_{22} = -13, \quad C_{23} = (-1)^{2+3} M_{23} = -12$
• $C_{31} = (-1)^{3+1} M_{31} = 11, \quad C_{32} = (-1)^{3+2} M_{32} = 11, \quad C_{33} = (-1)^{3+3} M_{33} = 0$

And the matrix of cofactors is given by:

\[ C = \begin{bmatrix} -23 & -47 & -18 \\ -19 & -13 & -12 \\ 11 & 11 & 0 \end{bmatrix} \]

Find Determinant Using Cofactors

Knowing the cofactors, the determinant of a square matrix can be calculated by adding the product of entries and their corresponding cofactors in any row or column.

Using row 1:

\[ \det(A) = (A)_{11} C_{11} + (A)_{12} C_{12} + (A)_{13} C_{13} = (-2)(-23) + (2)(-47) + (1)(-18) = -66 \]

Using row 2:

\[ \det(A) = (A)_{21} C_{21} + (A)_{22} C_{22} + (A)_{23} C_{23} = (3)(-19) + (-3)(-13) + (4)(-12) = -66 \]

Note that $C_{33} = 0$, so it is more efficient to calculate using row 3 or column 3:

Using row 3:

\[ \det(A) = (A)_{31} C_{31} + (A)_{32} C_{32} + (A)_{33} C_{33} = (-5)(11) + (-1)(11) + (9)(0) = -66 \]

Using column 3:

\[ \det(A) = (A)_{13} C_{13} + (A)_{23} C_{23} + (A)_{33} C_{33} = (1)(-18) + (4)(-12) + (9)(0) = -66 \]

Notes:

1. When computing the determinant of a square matrix, it is more efficient to use a column or a row with the highest number of zeros.
2. The method of cofactors is not very efficient for matrices with large dimensions. Another method based on row operations to find determinants is also available.

Properties of Determinants

1. $\det(I_n) = 1$, the determinant of the identity matrix of any order is equal to 1.
2. The determinant of a square matrix with one row or one column of zeros is equal to zero.
3. The determinant of any triangular matrix is equal to the product of the entries in the main diagonal (top left to bottom right).
4. $\det(A B) = \det(A) \det(B)$
5. $\det(A^T) = \det(A)$, where $A^T$ is the transpose of $A$.
6. $\det(A^{-1}) = \dfrac{1}{\det(A)}$, where $A^{-1}$ is the inverse of $A$.
7. $\det(k A) = k^n \det(A)$, where $A$ is an $n \times n$ matrix and $k$ is a constant.
8. If matrix $B$ is obtained from matrix $A$ by multiplying a row or a column of $A$ by a constant $k$, then $\det(B) = k \det(A)$.
9. If two rows or two columns of a square matrix are the same, the determinant of that matrix is equal to zero.
10. If two rows or two columns of a square matrix are interchanged, the determinant of that matrix changes sign.
11. If you add a multiple of one row to another row, the determinant does not change.

Questions on Determinants

1. Part 1
   Calculate the matrix of minors and the matrix of cofactors of each of the following matrices: \[ a) \, A = \begin{bmatrix} -1 & 23 \\ 2 & -2 \end{bmatrix}, \quad b) \, B = \begin{bmatrix} -1 & 0 & 3 \\ 2 & -4 & -2 \\ -5 & 2 & -9 \end{bmatrix} \]
2. Part 2
   Find, if it is defined, the determinant of each of the following matrices: \[ a) \, A = \begin{bmatrix} -1 & 23 & 10 \\ 0 & -2 & -11 \end{bmatrix}, \quad b) \, B = \begin{bmatrix} 2 & 0 & 1 \\ 3 & -3 & -1 \\ -5 & -1 & 2 \end{bmatrix}, \quad c) \, C = \begin{bmatrix} -3 \end{bmatrix}, \quad d) \, D = \begin{bmatrix} -6 & -4 & 2 \\ 0 & -3 & 23 \\ 0 & 4 & 9 \end{bmatrix}, \quad e) \, E = \begin{bmatrix} -3 & -3 & -4 \\ 0 & 4 & 0 \\ 0 & 0 & 5 \end{bmatrix} \]
3. Part 3
   Solve for $x$ each of the following equations: \[ a) \, \begin{vmatrix} x & 8 \\ 2 & x \end{vmatrix} = 0, \quad b) \, \begin{vmatrix} x - 6 & -2 \\ 1 & 3 \end{vmatrix} = -3, \quad c) \, \begin{vmatrix} x - 6 & -2 \\ 1 & 3 \end{vmatrix} = \begin{vmatrix} 2 & -2x \\ -2 & 6 \end{vmatrix} \]
4. Part 4
   Find the determinant of matrices $A$ and $B$ given below: \[ a) \, A = \begin{bmatrix} 4 & 0 \\ 2 & 5 \end{bmatrix} \begin{bmatrix} 0 & 4 \\ 2 & 5 \end{bmatrix}, \quad b) \, B = \begin{bmatrix} 4 & 0 & 0 \\ 2 & -1 & 7 \\ -4 & 4 & 8 \end{bmatrix} \begin{bmatrix} 5 & 9 & 0 \\ -3 & 1 & 0 \\ 7 & 11 & -6 \end{bmatrix} \begin{bmatrix} -7 & -1 & 2 \\ 0 & 1 & 0 \\ 7 & 2 & -6 \end{bmatrix} \]
5. Part 5
   Given that $\begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} = K$, find the following:
   a) $\begin{vmatrix} a & b & c \\ d & e & f \\ -4g & -4h & -4i \end{vmatrix}$, \quad b) $\begin{vmatrix} d & e & f \\ a & b & c \\ g & h & i \end{vmatrix}$, \quad c) $\begin{vmatrix} 3d & 3e & 3f \\ -4a & -4b & -4c \\ 7g & 7h & 7i \end{vmatrix}$, \quad d) $\begin{vmatrix} a & b & c \\ d - 8a & e - 8b & f - 8c \\ g & h & i \end{vmatrix}$

Solutions to the Above Questions

1. Part 1 Solution
   a) Given matrix $A = \begin{bmatrix} -1 & 23 \\ 2 & -2 \end{bmatrix}$, we first calculate the matrix of minors:
   $M_{11} = -2, \quad M_{12} = 2, \quad M_{21} = 23, \quad M_{22} = -1$
   The matrix $M$ of minors is: \[ M = \begin{bmatrix} -2 & 2 \\ 23 & -1 \end{bmatrix} \] Using $C_{ij} = (-1)^{i+j} M_{ij}$, the matrix of cofactors is: \[ C = \begin{bmatrix} -2 & -2 \\ -23 & -1 \end{bmatrix} \]
   b) Given matrix $B = \begin{bmatrix} -1 & 0 & 3 \\ 2 & -4 & -2 \\ -5 & 2 & -9 \end{bmatrix}$:
   $M_{11} = 40, \quad M_{12} = -28, \quad M_{13} = -16$
   $M_{21} = -6, \quad M_{22} = 24, \quad M_{23} = -2$
   $M_{31} = 12, \quad M_{32} = -4, \quad M_{33} = 4$
   The matrix $M$ of minors is: \[ M = \begin{bmatrix} 40 & -28 & -16 \\ -6 & 24 & -2 \\ 12 & -4 & 4 \end{bmatrix} \] The matrix $C$ of cofactors is: \[ C = \begin{bmatrix} 40 & 28 & -16 \\ 6 & 24 & 2 \\ 12 & 4 & 4 \end{bmatrix} \]
2. Part 2 Solution
   a) Matrix $A$ is not square, so its determinant is undefined.
   b) $\det(B) = 2 \begin{vmatrix} -3 & -1 \\ -1 & 2 \end{vmatrix} + 1 \begin{vmatrix} 3 & -3 \\ -5 & -1 \end{vmatrix} = 2(-6 - 1) + 1(-3 - 15) = -32$.
   c) Matrix $C$ has one entry, so $\det(C) = -3$.
   d) Using column 1: $\det(D) = -6 \left( (-3)(9) - (23)(4) \right) = -6(-27 - 92) = 714$.
   e) Matrix $E$ is triangular: $\det(E) = (-3)(4)(5) = -60$.
3. Part 3 Solution
   a) $\begin{vmatrix} x & 8 \\ 2 & x \end{vmatrix} = x^2 - 16 = 0 \implies x = 4, x = -4$.
   b) $\begin{vmatrix} x - 6 & -2 \\ 1 & 3 \end{vmatrix} = 3(x-6) - (-2)(1) = 3x - 16 = -3 \implies x = 13/3$.
   c) $3x - 16 = 12 - 4x \implies 7x = 28 \implies x = 4$.
4. Part 4 Solution
   a) $\det(A) = \det\begin{bmatrix} 4 & 0 \\ 2 & 5 \end{bmatrix} \det\begin{bmatrix} 0 & 4 \\ 2 & 5 \end{bmatrix} = (20)(-8) = -160$.
   b) $\det(B) = (4)((-1)(8) - (7)(4)) \times (-6)((5)(1) - (9)(-3)) \times (1)((-7)(-6) - (2)(7)) = (4)(-36)(-6)(32)(1)(28) = 774144$.
5. Part 5 Solution
   a) $-4K$
   b) $-K$
   c) $(3)(-4)(7)K = -84K$
   d) $K$

More References and Links

• Row operations to find determinants
• Online calculator for the Determinant of a 3 by 3 Matrix
• Step-by-Step Solver to Calculate the Determinant of a 3 by 3 Matrix
• Determinant