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Determinant of a Matrix
  • Last Updated : 21 Feb, 2021

In the study of algebra and matrices, a system of linear equations is usually expressed in the form of matrices. For example, 

a1x + b1y + c1 = 0

a2x + b2x + c2 = 0 

A system like this can also be expressed as, 

\begin{bmatrix} a_{1} & b_{1}\\ a_{2} & b_{2} \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix}  =\begin{bmatrix} c_{1}\\ c_{2} \end{bmatrix}



Now, this system of equations has a unique solution or not, is determined by the number a1 b2 – a2 b1. This quantity that determines the uniqueness of the solution is called the determinant. It is used widely in the fields of computer science and electrical engineering. Determinants also give us an idea about the area or volume of space. In order to solve Determinants, we need to understand a few terms first, like Minors and Cofactors Let’s learn about them.

Minors

Minor is required to find determinant for single elements (every element) of the matrix. They are the determinants for every element obtained by eliminating the rows and columns of that element. If the matrix given is:

\begin{bmatrix}a_{11} & a_{12} &a_{13}\\a_{21} & a_{22} & a_{23}\\a_{31} & a_{32} & a_{33}\end{bmatrix}

The Minor of a12 will be the determinant:

\begin{vmatrix}a_{21} & a_{23}\\a_{31} & a_{33}\end{vmatrix}

Question: Find the Minor of 5 in the determinant \begin{vmatrix}2 & 1 & 2\\4 & 5 & 0\\2 & 0 & 1\end{vmatrix}

Answer: 

The minor of 5 will be the determinant of \begin{vmatrix}2 & 2\\2 & 1\end{vmatrix}



Calculating the determinant, the minor is obtained as: 

(2 × 1) – (2 × 2) = -2

Cofactors

Cofactors are related to minors by a small formula, for an element aij, the cofactor of this element is Cij and the minor is Mij then, cofactor can be written as: 

Cij = (-1)i+jMij

Question: Find the cofactor of the element placed in the first row and second column of the determinant:

\begin{vmatrix}2 & 1 & 2\\4 & 5 & 0\\2 & 0 & 1\end{vmatrix}

Answer: 

In order to find out the cofactor of the first row and second column element i.e the cofactor for 1. First find out the minor for 1, which will be: 

\begin{vmatrix}4 & 0\\2 & 1\end{vmatrix} \\ = (4 \times 1) - (2 \times 0) \\ = 4

M12 = 4

Now, applying the formula for cofactor: 

C12 = (-1)1 + 2M12

⇒ C12 = (-1)3 × 4

⇒ C12 = -4

Adjoint

The Adjoint of a matrix for order n can be defined as the transpose of its cofactors. For a matrix A:

Adj. A = [Cij]n×nT

Transpose of a Matrix

Transpose of a Matrix A is denoted as AT or A’. It is clear that the vertical side in the matrix is known as a column and the horizontal side is known as a row, Transposing a Matrix means replacing the Rows with columns and Vice-Versa, since the Rows and Columns are changing, the Order of the Matrix also changes.

 If a Matrix is given as A= [aij]m×n, then its Transpose will become AT or A’ = [aji]n×m

Question: What will be the transpose of the Matrix:

\begin{bmatrix}2 & 1\\3 & 0\\6 & 9\end{bmatrix}_{2\times3}

Answer: 

Interchanging Rows and Columns, AT = \begin{bmatrix}2 & 3 & 6\\1 & 0 & 9\end{bmatrix}_{3\times2}

 Determinant

A determinant is a number associated with every square matrix A = [aij] of order n. The determinant for matrix A is denoted by |A| or det(A). 

If A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}, it’s determinant is denoted by \begin{vmatrix} a & b \\ c & d \end{vmatrix}

Physical significance of Determinant

Consider a 2D matrix, each column of this matrix can be considered as a vector on the x-y plane.  So, the determinant between two vectors on a 2d plane gives us the area enclosed between them. If we extend this concept, in 3D the determinant will give us the volume enclosed between two vectors. 

Area enclosed between two vectors in 2D

Determinant of a matrix of Order One

Let X = [a] be the matrix of order one, then it’s determinant is given by det(X) = a. 

Determinant of a matrix of Order Two

Let X = \begin{bmatrix} a & b \\ c & d \end{bmatrix}, it’s determinant is given by cross multiplication of it’s elements. 

det(X) = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc

Question: Evaluate A = \begin{vmatrix} 2 & 4 \\ 3 & 1 \end{vmatrix}

Answer: 

A = \begin{vmatrix} 2 & 4 \\ 3 & 1 \end{vmatrix} \\ = 2(1) - 4(3) \\ = 2 - 12 \\ = -10

Determinant of a matrix of Order 3×3

It can be determined by expressing it in terms of 2nd order determinants. It can be expanded either along rows(R1, R2 or R3) or column(C1, C2 or C3). Consider an order 3 matrix A. 

A = \begin{bmatrix} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{bmatrix}  

Step 1: Multiply the first element a11 of row R1 with (-1)(1 + 1)[(-1)sum of suffixes in a11] and with the second order determinant obtained by deleting the elements of row R1 and C1 of A as a11 lies in R1 and C1. (-1)^{1 + 1}a_{1}\begin{vmatrix} b_{2} & b_{3} \\ c_{2} & c_{3} \end{vmatrix}

Step 2: Similarly, multiply second element of the first rowR1,  with the determinant obtained after deleting first row and second column. (-1)^{1 + 2}a_{12}\begin{vmatrix} b_{1} & b_{3}   \\ c_{1} & c_{3} \end{vmatrix}

Step 3: Multiply the third element of row R1 with the determinant obtained after deleting first row and third column. (-1)^{1 + 3}a_{3}\begin{vmatrix} b_{1} & b_{2} \\ c_{1} & c_{2} \end{vmatrix}

Step 4: Now the expansion of determinant of A, that is |A| can be written as |A| =  (-1)^{1 + 1}a_{1}\begin{vmatrix} b_{2} & b_{3} \\ c_{2} & c_{3} \end{vmatrix} + (-1)^{1 + 2}a_{12}\begin{vmatrix} b_{1} & b_{3}   \\ c_{1} & c_{3} \end{vmatrix} + (-1)^{1 + 3}a_{3}\begin{vmatrix} b_{1} & b_{2} \\ c_{1} & c_{2} \end{vmatrix}

Similarly, in this way, we can expand it along any row and any column. 

Question: Evaluate the determinant det(A) = \begin{vmatrix} 1 & 3 & 0 \\ 4 & 1 & 0 \\ 2 & 0 & 1 \end{vmatrix}

Answer: 

We see that the third column has most number of zeros, so it will be easier to expand along that column. 

det(A) = (-1)^{1 + 3}0\begin{vmatrix}4 & 1 \\ 2 & 0 \end{vmatrix} + (-1)^{2 + 3}0\begin{vmatrix}1 & 3 \\ 2 & 0 \end{vmatrix}  + (-1)^{1 + 3}1\begin{vmatrix}1 & 3 \\ 4 & 1 \end{vmatrix} \\ = -11

Properties of Determinants

  • Reflection Property: Value of the determinant remains unchanged even after rows and columns are interchanged. That determinant of a matrix, and it’s transpose remains the same.
  • Switching Property: If any two rows or columns of a determinant are interchanged, then the sign of determinant changes.

For Example: \begin{vmatrix} 3 & 3 & 0 \\ 2 & 1 & 1 \\ 5 & 0 & 1 \end{vmatrix}

det. A = [3×{(1×1)-(0×1)}]-[3×{(2×1)-(5×1)}]+[0×{(2×0)-(5×1)}]

= {3×(1-0)}-{3×(2-5)+0

= [3-{3(-3)}+0]

= (3+9)

=12

Now, Interchanging Row 1 with Row 2, determinant will be:

\begin{vmatrix} 2 & 1 & 1 \\ 3 & 3 & 0 \\ 5 & 0 & 1 \end{vmatrix}

det. A = [2×{(3×1)-(0×0)}]-[1×{(3×1)-(5×0)}]+[1×{(3×0)-(5×3)}]

= (6-3-15)

= -12

  • Repetition Property/Proportionality Property: If any two rows or any two columns of a determinant are identical, then the value of the determinant becomes zero.
  • Scalar Multiple Property: If each element of a row (or a column) of a determinant is multiplied by a constant k, then its value gets multiplied by k

 \begin{vmatrix} ka & kb \\ c & d \end{vmatrix} = k\begin{vmatrix} a & b \\ c & d \end{vmatrix}

  • Sum Property: If some or all elements of a row or column can be expressed as the sum of two or more terms, then the determinant can also be expressed as the sum of two or more determinants.

 \begin{vmatrix} a_{1} + \lambda_{1} & a_{2} + \lambda_{2} & a_{3} + \lambda_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{vmatrix} = \begin{vmatrix} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{vmatrix} + \begin{vmatrix} \lambda_{1} & \lambda_{2} & \lambda_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{vmatrix}

Examples of Properties

Question 1: If x, y, z are different. and A = \begin{vmatrix} x & x^{2} & 1 + x^{3} \\ y & y^{2} & 1 + y^{3} \\ z & z^{2} & 1 + z^{3} \end{vmatrix} = 0, then show that 1 + xyz = 0. 

Answer: 

Using Sum Property:

 \begin{vmatrix} x & x^{2} & 1 + x^{3} \\ y & y^{2} & 1 + y^{3} \\ z & z^{2} & 1 + z^{3} \end{vmatrix} = \begin{vmatrix} x & x^{2} & 1 \\ y & y^{2} & 1\\ z & z^{2} & 1 \end{vmatrix}  + \begin{vmatrix} x & x^{2} & x^{3} \\ y & y^{2} & y^{3} \\ z & z^{2} & z^{3} \end{vmatrix} \text{} \\ = (-1)^{2}\begin{vmatrix} 1 & x & x^{2} \\ 1 & y & y^{2}\\ 1 & z & z^{2} \end{vmatrix} + xyz\begin{vmatrix} 1 & x & x^{2} \\ 1 & y & y^{2}\\ 1 & z & z^{2} \end{vmatrix} \\ = (1 + xyz) \begin{vmatrix} 1 & x & x^{2} \\ 1 & y & y^{2}\\ 1 & z & z^{2} \end{vmatrix} = 0 

On solving this determinant and expanding it, 

A = (1 + xyz)(y- x)(z-y)(z-x) 

Since it’s given in the question, that all x, y and z have different values and A =0. So the only term that can be zero is 1 + xyz. 

Hence, 1 + xyz = 0

Question 2: Evaluate \begin{vmatrix} 102 & 18 & 36 \\ 1 & 3 & 4 \\ 17 & 3 & 6 \\ \end{vmatrix}

Answer: 

Using Scalar Multiple Property and Repetition Property: 

 \begin{vmatrix} 102 & 18 & 36 \\ 1 & 3 & 4 \\ 17 & 3 & 6 \\ \end{vmatrix} = \begin{vmatrix} 17(6) & 6(3) & 6(6) \\ 1 & 3 & 4 \\ 17 & 3 & 6 \\ \end{vmatrix} \\ = 6\begin{vmatrix} 17 & 3 & 6 \\ 1 & 3 & 4 \\ 17 & 3 & 6 \\ \end{vmatrix} = 0  \text{}

Question 3: Evaluate the determinant A. A = \begin{vmatrix} 2 & 3 & 1 \\ 1 & 0 & 5 \\ 2 & 3 & 1 \end{vmatrix} \\

Answer: 

Using Proportionality Property:

Two of the rows of the matrix are identical. 

So, A = \begin{vmatrix} 2 & 3 & 1 \\ 1 & 0 & 5 \\ 2 & 3 & 1 \end{vmatrix} \\ = 0 \text{}

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