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Determinant of a Matrix

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  • Last Updated : 20 Jan, 2023
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Determinant of a Matrix is defined as the function that gives the unique output (real number) for every input value of the matrix. the scalar value computed for a given square matrix. Determinant of the matrix is considered the scaling factor that is used for the transformation of a matrix. Determinant of a matrix is useful for finding the solution of a system of linear equations, the inverse of the square matrix, and others. The determinant of only square matrices exists.

Definition of Determinant of Matrix

Determinant of a Matrix is defined as the sum of products of the elements of any row or column along with their corresponding co-factors. Determinant is defined only for square matrices.

Determinant of any square matrix of order 2×2, 3×3, 4×4, or n × n, where n is the number of rows or the number of columns. (For a square matrix number of rows and columns are equal). Determinant can also be defined as the function which maps every matrix with the real numbers. 

For any set S of all square matrices, and R the set of all numbers the function f, f: S → R is defined as f (x) = y, where x ∈ S and y ∈ R, then f (x) is called the determinant of the input matrix.

Symbol of Determinant

Let’s take any square matrix A, then the determinant of A is denoted as det A (or) |A|. Determinant is also denoted by the symbol Δ.

Minor of Element of Matrix

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}

Minor of a12 will be the determinant:

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

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

Answer:

The minor of element 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 of Element of Matrix

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, the 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 of a Matrix

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 A =

\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 of a 1×1 Matrix

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

Determinant of a 2×2 Matrix

The determinant of any 2×2 square matrix A = \begin{bmatrix}a & b\\c & d\end{bmatrix}_{2\times2}
is calculated using the formula |A| = ad – bc

Example: Find the Determinant of A = 

Solution:

Determinant of A = \begin{bmatrix}3 & 2\\2 & 3\end{bmatrix}_{2\times2}    is calculated as,

| A | = \begin{vmatrix}3 & 2\\2 & 3\end{vmatrix}

| A | = 3×3 – 2×2 
       = 9 – 4 
       = 5

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. 

Physical Significance of Determinant

Area enclosed between two vectors in 2D

Determinant of a 3×3 Matrix

Determinant of a 3×3 Matrix is 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 a  matrix A of order 3×3 

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

For calculating the Determinant of 3×3 Matrix use the following step:

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 the second element of the first rowR1,  with the determinant obtained after deleting the 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 the 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 the 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. 

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

Solution: 

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

Laplace Formula for Determinant

Laplace’s formula, is used to expressed the determinant of a matrix in terms of the minors of the matrix.

If  An×n is the given suare matrix and Cij is the cofactor of Aij the solving for any row i or column j

det (A) = \sum_{i =1}^{n}A_{ij}C_{ij}

Properties of Determinants

Various Properties of the Determinants of the square matrix are discussed below:

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

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

Solution:

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}

Also, Check

Solved Examples on Determinant of Matrix

Example 1: If x, y, and 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. 

Solution: 

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

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

Solution: 

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{}

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

Solution: 

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{}

FAQs on Determinant of Matrix

Question 1: What is meant by the determinant formula?

Answer:

For any 3×3 matrix A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \\   the shortcut formula for computing its determinant is: 

|A| = a (ei − fh) − b (di − fg) + c (dh − eg)

Question 2: Can determinant of any matrix be negative?

Answer:

Yes, the determinant of any matrix can be negative.

Question 3: Can determinant of any matrix ever be equal to 0?

Answer:

Yes, the determinant of any matrix can be zero if any one row or column of the matrix has all the zero values. It can also be zero if any two rows or columns of the matrix are equal.

Question 4: How to find the Determinant of Matrix?

Answer:

The determinant of any matrix can be found by using the following steps:

Step 1: Select any row or column of our choice.

Step 2: Calculate the cofactors of all the elements of the selected row or column

Step 3: The product of the elements of the row or column by their corresponding cofactors is found. The calculated product is added with alternate negative sign.


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