Determinant of a Square Matrix

Examples and questions on the determinants of square matrices along with their detailed solutions are presented . The method of minors and cofactors to calculate determinants along with their properties are also discussed.

Determinant of a 2 2 Matrix - Definition

In order to explain the concept of determinant in linear algebra, we start with a 2 2 systems of equations with unknowns x and y given by
a 2 by 2 system of linear equations
Multiply all terms of the first equation in the system by the coefficient e and all the terms in the second equations by the coefficient -b to obtain a system that is equivalent to the one given above but easier to solve.
a 2 by 2 system of linear equations after multiplication
Add the left and right hand sides of the above equations and simplify to obtain an equation in one variable only.
one equation in one variable
Solve the above for x
solution for x
In a similar way, we can find y as
solution for y
We note that both x and y are expressed as ratios with the same denominator a e - b d which we define as the determinant of the square A matrix as a scalar quantity given by \[ Det (A) = |A| = \begin{vmatrix} a & b \\ d & e \end{vmatrix} = a e - bd \] Using the determinants we now write the solutions of the 2 2 system of equations as follows: \[ x = \dfrac{\begin{vmatrix} c & b \\ f & e \end{vmatrix}}{\begin{vmatrix} a & b \\ d & e \end{vmatrix}} , y = \dfrac{\begin{vmatrix} a & c \\ d & f \end{vmatrix}}{\begin{vmatrix} a & b \\ d & e \end{vmatrix}}\] In a similar way we can express the solutions of any n n systems of equations using the determinants of n n square matrices.
Determinants are scalar quantities used in solving systems of equations, in calculating the inverse of a matrix and have many other applications.
Note that the determinant of 1 by 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 n square matrix corresponding to the element \( (A)_{ij} \) is the determinant of the matrix (n-1) (n-1) matrix obtained by deleting row i and column j of matrix A.
Example
Let us find the the minors of the matrix \[ A = \begin{bmatrix} -2 & 2 & 1 \\ 3 & -3 & 4 \\ -5 & -1 & 9 \end{bmatrix} \] The minor \( M_{11} \) is equal to the determinant of the matrix obtained by deleting row 1 and column 1 from the given matrix. \[ M_{11} = \begin{vmatrix} .& . & . \\ . & -3 & 4 \\ . & -1 & 9 \end{vmatrix} = \begin{vmatrix} -3 & 4 \\ -1 & 9 \end{vmatrix} = (-3)(9) - (4)(-1) = -23\] \( M_{12} \) is the determinant of the matrix obtained by deleting row 1 and column 2 and. \[ M_{12} = \begin{vmatrix} .& . & . \\ 3 & . & 4 \\ - 5 & . & 9 \end{vmatrix} = \begin{vmatrix} 3 & 4 \\ -5 & 9 \end{vmatrix} = (3)(9) - (4)(-5) = 47\] \( M_{13} \) is calculated by deleting row 1 and column 3 \[ M_{13} = \begin{vmatrix} .& . & . \\ 3 & - 3 & . \\ - 5 & - 1 & . \end{vmatrix} = \begin{vmatrix} 3 & -3 \\ -5 & -1 \end{vmatrix} = (3)(-1) - (-3)(-5) = -18\] Find minor \( M_{21} \) by deleting row 2 and column 1 \[ M_{21} = \begin{vmatrix} .& 2 & 1 \\ . & . & . \\ . & - 1 & 9 \end{vmatrix} = \begin{vmatrix} 2 & 1 \\ -1 & 9 \end{vmatrix} = (2)(9) - (1)(-1)= 19\] \( M_{22} \) is claculated by deleting row 2 and column 2 \[ M_{22} = \begin{vmatrix} -2& . & 1 \\ . & . & . \\ - 5 & . & 9 \end{vmatrix} = \begin{vmatrix} -2 & 1 \\ -5 & 9 \end{vmatrix} = (-2)(9) - (1)(-5) = -13\] Minor \( M_{23} \) is found by deleting row 2 and column 3 \[ M_{23} = \begin{vmatrix} - 2 & 2 & . \\ . & . & . \\ -5 & -1 & . \end{vmatrix} = \begin{vmatrix} -2 & 2 \\ -5 & -1 \end{vmatrix} = (-2)(-1) - (2)(-5) = 12\] Minor \( M_{31} \) is found by deleting row 3 and column 1 \[ M_{31} = \begin{vmatrix} .& 2 & 1 \\ . & -3 & 4 \\ . & . & . \end{vmatrix} = \begin{vmatrix} 2 & 1 \\ -3 & 4 \end{vmatrix} = (2)(4) - (1)(-3) = 11\] Minor \( M_{32} \) is found by deleting row 3 and column 2 \[ M_{32} = \begin{vmatrix} -2& . & 1 \\ 3 & . & 4 \\ . & . & . \end{vmatrix} = \begin{vmatrix} -2 & 1 \\ 3 & 4 \end{vmatrix} = (-2)(4) - (1)(3) = -11\] Minor \( M_{33} \) is found by deleting row 3 and column 3 \[ M_{33} = \begin{vmatrix} -2 & 2 & . \\ 3 & -3 & . \\ . & . & . \end{vmatrix} = \begin{vmatrix} -2 & 2 \\ 3 & -3 \end{vmatrix} = (-2)(-3) - (2)(3) = 0 \] The
matrix of minors is given by \[ \begin{bmatrix} -23 & 47 & -18 \\ 19 & -13 & 12 \\ 11 & -11 & 0 \end{bmatrix} \]

Cofactors of a Square Matrix

The cofactor \( C_{ij} \) is given by \( C_{ij} = (-1)^{i+j} M_{ij} \)
\( C_{11} = (-1)^{1+1} M_{11} = -23 ,\qquad C_{12} = (-1)^{1+2} M_{12} = -47 ,\qquad C_{13} = (-1)^{1+3} M_{13} = -18 \)
\( C_{21} = (-1)^{2+1} M_{21} = - 19 ,\qquad C_{22} = (-1)^{2+2} M_{22} = - 13 ,\qquad C_{23} = (-1)^{2+3} M_{23} = - 12 \)
\( C_{31} = (-1)^{3+1} M_{31} = 11 ,\qquad C_{32} = (-1)^{3+2} M_{32} = 11 ,\qquad C_{33} = (-1)^{3+3} M_{33} = 0 \)
and the
matrix of cofactors is given by \[ \begin{bmatrix} -23 & -47 & -18 \\ -19 & -13 & -12 \\ 11 & 11 & 0 \end{bmatrix} \]

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 the above example, the determinant can be calculated using:
row (1)
\( \text{Det}(A) = (A)_{11} C_{11} + (A)_{12} C_{12} +(A)_{13} C_{13} = (-2)(-23) + (2)(-47) + (1)(-18) = -66 \)

or row(2)
\( \text{Det}(A) = (A)_{12} C_{12} + (A)_{22} C_{22} +(A)_{32} C_{32} = (2)(-47) + (-3)(-13) + (-1)(11) = - 66 \)

Note that \( C_{33} = 0 \) and it is therefore more efficient to calculate the determinant using row (3) or column (3)

or row(3)
\( \text{Det}(A) = (A)_{31} C_{31} + (A)_{32} C_{32} +(A)_{33} C_{33} = (-5)(11) + (-1)(11) + (9)(0) = - 66 \)

or column(3)
\( \text{Det}(A) = (A)_{13} C_{13} + (A)_{23} C_{23} +(A)_{33} C_{33} = (1)(-18) + (4)(-12) + (9)(0) = - 66 \)

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 to calculate the determinant is not very efficient for matrices with large dimensions. However another method based on
row operations to find determinants is included.
and so on.

Properties of Determinants

  1. \( \text{Det}(I_n) = 1 \) , 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. \( \text{Det}(A B) = \text{Det}(A) \text{Det}(B) \)
  5. \( \text{Det}(A^T) = \text{Det}(A) \) , where \( A^T \) is the transpose of \( A \).
  6. \( \text{Det}(A^{-1}) = \dfrac{1}{\text{Det}(A)} \) , where \( A^{-1} \) is the inverse of \( A \).
  7. \( \text{Det}( k A) = k^n \text{Det}(A) \) , where \( A \) is an n 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 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) \( \begin{vmatrix} x - 6 & - 2 \\ 1 & 3 \end{vmatrix} = \begin{vmatrix} 2 & - 2 x \\ -2 & 6 \end{vmatrix} \)

  4. Part 4
    Find the determinant of the matrices A and B given below.
    a) \( A = \begin{bmatrix} 4 & 0 \\ 2 & 5 \end{bmatrix} \begin{bmatrix} 0 & 4 \\ 2 & 5 \end{bmatrix} ,\qquad \) 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
    The determinant of the matrix
    Given that \( \begin{vmatrix} a & b & c\\ d & e & f \\ g & h & i \end{vmatrix} = K\)
    Find the following:
    a) \( \begin{vmatrix} a & b & c\\ d & e & f \\ - 4 g & - 4 h & - 4i \end{vmatrix} \) , b) \( \begin{vmatrix} d & e & f \\ a & b & c\\ g & h & i \end{vmatrix} \) , c) \( \begin{vmatrix} 3 d & 3 e & 3 f \\ - 4 a & - 4 b & - 4 c\\ 7 g & 7 h & 7 i \end{vmatrix} \), d) \( \begin{vmatrix} a & b & c\\ d - 8a & e - 8 b & f - 8c \\ g & h & i \end{vmatrix} \)

Solutions to the Above Questions

  1. Part 1
    a)Given matrix \( A = \begin{bmatrix} -1 & 23 \\ 2 & -2 \end{bmatrix} \), we first calculate the matrix of minors

    \( M_{11} = Det \begin{bmatrix} . & . \\. & -2 \end{bmatrix} = -2 \) , \( \quad M_{12} = Det \begin{bmatrix} . & . \\ 2 & . \end{bmatrix} = 2 \)

    \( M_{21} = Det \begin{bmatrix} . & 23 \\ . & . \end{bmatrix} = 23 \) , \( \quad M_{22} = Det \begin{bmatrix} -1 & . \\ . & . \end{bmatrix} = -1 \)
    Hence the matrix M of minors is given by
    \[ M = \begin{bmatrix} -2 & 2 \\ 23 & -1 \end{bmatrix} \] Cofactors are given by \[ C_{ij} = (-1)^{i+j}M_{ij} \] Hence the matrix C of cofactors is given by
    \[ C = \begin{bmatrix} (-1)^{1+1}(-2) & (-1)^{1+2}(2) \\ (-1)^{2+1}(23) & (-1)^{2+2}(-1) \end{bmatrix} = \begin{bmatrix} -2 & -2 \\ -23 & -1 \end{bmatrix} \]

    b)
    Given matrix \( B = \begin{bmatrix} -1 & 0 & 3\\ 2 & - 4 & -2 \\ -5 & 2 & -9 \end{bmatrix} \), we first calculate the matrix of minors

    \( M_{11} = Det \begin{bmatrix} . & . & .\\ . & - 4 & -2 \\ . & 2 & -9 \end{bmatrix} = (-4)(-9) - (-2)(2)= 40 \)

    \( M_{12} = Det \begin{bmatrix} . & . & .\\ 2 & . & -2 \\ -5 & . & -9 \end{bmatrix} = (2)(-9) - (-2)(-5)= -28 \)

    \( M_{13} = Det \begin{bmatrix} . & . & .\\ 2 & - 4 & . \\ -5 & 2 & . \end{bmatrix} = (2)(2) - (-4)(-5)= -16 \)

    \( M_{21} = Det \begin{bmatrix} . & 0 & 3\\ . & . & . \\ . & 2 & -9 \end{bmatrix} = (0)(-9) - (3)(2)= -6 \)

    \( M_{22} = Det \begin{bmatrix} -1 & . & 3\\ . & . & . \\ -5 & . & -9 \end{bmatrix} = (-1)(-9) - (3)(-5)= 24 \)

    \( M_{23} = Det \begin{bmatrix} -1 & 0 & .\\ . & . & . \\ -5 & 2 & . \end{bmatrix} = (-1)(2) - (0)(-5)= -2 \)

    \( M_{31} = Det \begin{bmatrix} . & 0 & 3\\ . & - 4 & -2 \\ . & . & . \end{bmatrix} = (0)(-2) - (3)(-4)= 12 \)

    \( M_{32} = Det \begin{bmatrix} -1 & . & 3\\ 2 & . & -2 \\ . & . & . \end{bmatrix} = (-1)(-2) - (3)(2)= -4 \)

    \( M_{33} = Det \begin{bmatrix} -1 & 0 & .\\ 2 & - 4 & . \\ . & . & . \end{bmatrix} = (-1)(-4) - (0)(2)= 4 \)

    Hence the matrix M of minors is given by
    \[ M = \begin{bmatrix} 40 & -28 & -16\\ -6 & 24 & -2 \\ 12 & -4 & 4 \end{bmatrix} \]
    Cofactors are given by
    \[ C_{ij} = (-1)^{i+j}M_{ij} \]
    Hence the matrix C of cofactors is given by
    \[ C = \begin{bmatrix} (-1)^{1+1}(40) & (-1)^{1+2}(-28) & (-1)^{1+3}(-16) \\ (-1)^{2+1}(-6) & (-1)^{2+2}(24) & (-1)^{2+3}(-2) \\ (-1)^{3+1}(12) & (-1)^{3+2}(-4) & (-1)^{3+3}(4) \end{bmatrix} = \begin{bmatrix} 40 & 28 & -16 \\ 6 & 24 & 2 \\ 12 & 4 & 4 \end{bmatrix} \]

  2. Part 2
    a) Matrix A is not square and therefore its determinant is undefined.

    b) The determinant of matrix B is efficiently calculated if the first row (or the second column) is used since one of its entries is equal to zero. Using the first row, we have
    \( Det(B) = 2 \begin{vmatrix} -3 & - 1 \\ -1 & 2 \end{vmatrix} + 1 \begin{vmatrix} 3 & - 3 \\ -5 & -1 \end{vmatrix} = 2(-6 - 1) + (-3 - 15) = - 32 \)

    c) Matrix C has one entry and its determinant is equal to that entry; hence
    \( Det(C) = - 3 \)

    d) Column 1 of matrix D has two zeros and may therefore be efficiently used to evaluate the determinant as follows:
    \( Det(D) = -6( (-3)(9) - (32)(4)) = 714 \)

    e) Matrix E is a triangular matrix and its determinant is given by the product of the elements in the main diagonal; hence
    \( Det(E) = (-3)(4)(5) = -60 \)

  3. Part 3
    a) \( \begin{vmatrix} x & 8 \\ 2 & x \end{vmatrix} = (x)(x)-(8)(2) = x^2 - 16 ,\quad\)
    Solve \( x^2 - 16 = 0 \)
    The solutions are \( x = 4 \) and \( x = -4\)

    b) Evaluate the determinant: \( \begin{vmatrix} x - 6 & - 2 \\ 1 & 3 \end{vmatrix} = (x-6)(3) -(-2)(1) = 3x - 16 ,\quad\)
    Solve \( 3x - 16 = -3 \)
    The solution is \( x = 13/3 \)

    c) Evaluate the determinants: \( \begin{vmatrix} x - 6 & - 2 \\ 1 & 3 \end{vmatrix} = (x-6)(3) - (-2)(1) = 3x - 16 \)
    and
    \( \begin{vmatrix} 2 & - 2 x \\ -2 & 6 \end{vmatrix} = (2)(6)-(-2x)(-2) = 12 - 4x\)
    Solve \( 3x - 16 = 12 - 4x \)
    The solution is \( x = 4 \).

  4. Part 4
    Use property (4) above that gives the determinant of the product of matrices.
    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) Use property (4) above that gives the determinant of the product of matrices taking. The matrix on the left has 2 zeros in row (1) and this will be used to find its determinant. The matrix in the middle 2 zeros in column (3) and this will be used to evaluate its determinant. The matrix on the right has 2 zeros in the second row which will be used to evaluate its determinant. Hence \( Det (B) = Det\begin{bmatrix} 4 & 0 & 0\\ 2 & -1 & 7 \\ -4 & 4 & 8 \end{bmatrix} Det\begin{bmatrix} 5 & 9 & 0\\ -3 & 1 & 0 \\ 7 & 11 & -6 \end{bmatrix} Det\begin{bmatrix} -7 & -1 & 2\\ 0 & 1 & 0 \\ 7 & 2 & -6 \end{bmatrix} \\\\ = (4) ( (-1)(8) - (7)(4) ) \times (-6) ( (5)(1) - (9)(-3) ) \times (1)( (-7)(-6) - (2)(7) ) = 774144\)
  5. Part 5
    a) The last row is multiplied by - 4, using property 8, the answer is - 4 K.
    b) Two rows are exchanged, using property 10, the answer - K.
    c) The first row is multiplied by 3, the second row by - 4 and the third row by 7, using property 8 three times, the answer is
    (3)(-4)(7)K = - 84 K
    d) A multiple of the first row is added to the second row, use property 11, the answer is K.

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