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A Determinant, in Linear Algebra, is the unique numerical value determined from a square matrix. So, to every square matrix A = [aij] of order n, we can associate a number (real or complex) called determinant of the square matrix A. It is denoted by det A or |A|.
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Key Terms: Determinant, Linear Algebra, Matrix, Repetition property, Switching property, Scalar multiple property, Sum property, Invariance property, Factor property, Triangle property, Cofactor matrix property, Square matrix
Also Check: Identity Matrix
Properties of Determinants
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Determinants have some properties which are very useful as they permit us to generate the same results with different and simpler configurations of entries (elements). These properties of determinants simplify its evaluation by obtaining a maximum number of zeros in a row or a column. These properties are true for determinants of any order.
There are mainly ten properties of determinants namely Reflection property, All-zero property, Proportionality or Repetition property, Switching property, Scalar multiple property, Sum property, Invariance property, Factor property, Triangle property, and Cofactor matrix property.
Given below are the properties of determinants:
1. Reflection Property
The value of the determinant remains unchanged if its rows and columns are interchanged.
\(\Delta=\left|\begin{array}{lll} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{array}\right|=\left|\begin{array}{lll} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{array}\right|\)
2. Switching Property
If any two rows (or columns) of a determinant are interchanged, then the sign of determinant changes.
\(\Delta=\left|\begin{array}{lll} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{array}\right|=-\left|\begin{array}{ccc} a_{1} & a_{2} & a_{3} \\ c_{1} & c_{2} & c_{3} \\ b_{1} & b_{2} & b_{3} \end{array}\right|\)
3. All-Zero Property
If any two rows (or columns) of a determinant are identical (all corresponding elements are same), then the value of the determinant is zero.
Proof:
If we interchange the identical rows (or columns) of the determinant Δ, then Δ does not change. However, by Property 2, it follows that Δ has changed its sign
Therefore Δ = – Δ
Or Δ = 0
Alternatively, if we exchange the 1st and 2nd rows, Δ stays the same, but by the previous property, it should be −Δ, so
Δ = −Δ
Δ = 0
4. Proportionality (Repetition) Property
If the all elements of a row (or column) are proportional or identical to the elements of some other row (or column), then the determinant is zero.
5. Scalar Multiple Property
If all the elements of a row (or column) of a determinant are multiplied by a non-zero constant, then the determinant gets multiplied by the same constant.
Example:
\(\Delta=\left|\begin{array}{ccc} \lambda a_{1} & \lambda a_{2} & \lambda a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{array}\right|=\lambda\left|\begin{array}{ccc} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{array}\right|\)
Note carefully that λ is multiplied with elements of just one row and not of the entire determinant.
This property is trivial and can be proved easily by expansion.
6. Sum Property
If some or all elements of a row or column of a determinant are expressed as sum of two (or more) terms, then the determinant can be expressed as sum of two (or more) determinants.
\(\left|\begin{array}{lll} a_{1}+b_{1} & c_{1} & d_{1} \\ a_{2}+b_{2} & c_{2} & d_{2} \\ a_{3}+b_{3} & c_{3} & d_{3} \end{array}\right|=\left|\begin{array}{lll} a_{1} & c_{1} & d_{1} \\ a_{2} & c_{2} & d_{2} \\ a_{3} & c_{3} & d_{3} \end{array}\right|+\left|\begin{array}{lll} b_{1} & c_{1} & d_{1} \\ b_{2} & c_{2} & d_{2} \\ b_{3} & c_{3} & d_{3} \end{array}\right|\)
7. Property of Invariance
If, to each element of any row or column of a determinant, the equimultiples of corresponding elements of other row (or column) are added, then value of determinant remains the same, i.e., the value of determinant remain same if we apply the operation Ri → Ri + kRj or Ci → Ci + kCj
\(\left|\begin{array}{lll} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{array}\right|=\left|\begin{array}{lll} a_{1}+\alpha b_{1}+\beta c_{1} & b_{1} & c_{1} \\ a_{2}+\alpha b_{2}+\beta c_{2} & b_{2} & c_{2} \\ a_{3}+\alpha b_{3}+\beta c_{3} & b_{3} & c_{3} \end{array}\right|\)
8. Factor Property
If a determinant Δ becomes zero when we put x=α, then (x−α) is a factor of Δ.
9. Triangle Property
If all the elements of a determinant above or below the main diagonal consist of zeros, then the determinant is equal to the product of diagonal elements. That is,
\(\left|\begin{array}{ccc} a_{1} & a_{2} & a_{3} \\ 0 & b_{2} & b_{3} \\ 0 & 0 & c_{3} \end{array}\right|=\left|\begin{array}{ccc} a_{1} & 0 & 0 \\ a_{2} & b_{2} & 0 \\ a_{3} & b_{3} & c_{3} \end{array}\right|=a_{1} b_{2} c_{3}\)
10. Cofactor Matrix Property
\(\Delta=\left|\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right| \text { then } \Delta_{1}=\left|\begin{array}{lll} C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23} \\ C_{31} & C_{32} & C_{33} \end{array}\right|=\Delta^{2}\)
where Cij denotes the cofactor of the element aij in Δ.
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Determinants Detailed Video Explanation:
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Sample Questions
Ques. What do you mean by the value of a Determinant? (1 Mark)
Ans. The determinant happens to be a scalar value that one can compute from the square matrix’s elements. Furthermore, it encodes certain properties that belong to the linear transformation as described by the matrix. The denotation of the determinant of a matrix A is as det(A), det A, or |A|.
Ques. What does the determinant of a matrix mean? (2 Marks)
Ans. A Matrix is an array of many numbers. For a square matrix, i.e., a matrix with the same number of rows and columns, one can capture important information about the matrix in a just single number, called the determinant. The determinant is useful for solving linear equations, capturing how linear transformation Change area or volume, and Changing variables in integrals.
Ques. When does it mean to have a Determinant of a Matrix equivalent to zero? (2 Marks)
Ans. When the determinant of a matrix is zero, the system of equations associated with it is linearly dependent; that is, if the determinant of a matrix is zero, at least one row of such a matrix is a scalar multiple of another.
Ques. What is the difference between Matrices and Determinants? (2 Marks)
Ans. A matrix or matrices is a rectangular grid of numbers or symbols that is represented in a row and column format. A determinant is a component of a square matrix and it cannot be found in any other type of matrix. Matrices and determinants are important concepts in linear mathematics. These concepts play a huge part in linear equations and are also applicable to solving real-life problems in physics, mechanics, optics, etc.
Ques. If the determinant of a matrix is zero, what is the matrix called? (2 Marks)
Ans. If the determinant of a matrix is zero,then the matrix is called a singular matrix. Inverse of such a matrix can’t be found out as Inverse of a matrix=Adjoint of the matrix/Determinant of the matrix.
Previous Year Questions
Ques. Write the value of \(\begin{vmatrix} 2&7&65\\ 3&8&75\\ 5&9&86 \end{vmatrix}\). (All India 2014C) (3 Marks)
Ans. Given,
\(\begin{vmatrix} 2&7&65\\ 3&8&75\\ 5&9&86 \end{vmatrix}\)
= \(2 \begin{vmatrix} 8&75\\ 9&86 \end{vmatrix} -7 \begin{vmatrix} 3&75\\ 5&86 \end{vmatrix} +65 \begin{vmatrix} 3&8\\ 5&9 \end{vmatrix}\) [expanding the determinant along R1]
= 2 (688 – 675) – 7 (258 – 375) + 65 (27 – 40)
= 26 + 819 – 845
= 845 – 845
= 0
Ques. Prove the following, using properties of determinants. (Delhi 2014) (5 Marks)
\(\begin{vmatrix} a+b+2c & a & b \\ c & b+c+2a & b \\ c & a & c+a+2b \end{vmatrix} = 2(a+b+c)^3\)
Ans.
LHS = \(\begin{vmatrix} a+b+2c & a & b \\ c & b+c+2a & b \\ c & a & c+a+2b \end{vmatrix}\)
On applying C1 → C1 + C2 + C3, we get
LHS = \(\begin{vmatrix} 2(a+b+c) & a & b \\ 2(a+b+c) & b+c+2a & b \\ 2(a+b+c) & a & c+a+2b \end{vmatrix}\)
On taking 2(a+b+c) common from C1, we get
LHS = \(2(a+b+c) \begin{vmatrix} 1 & a & b \\ 1 & b+c+2a & b \\ 1 & a & c+a+2b \end{vmatrix}\)
On applying R2 → R2 – R1 and R3 → R3 – R1, we get
LHS = \(2(a+b+c) \begin{vmatrix} 1 & a & b \\ 0 & b+c+a & 0 \\ 0 & 0 & c+a+b \end{vmatrix}\)
On taking (a+b+c) common from R2 and R3, we get
LHS = \(2(a+b+c)^3 \begin{vmatrix} 1 & a & b \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix}\)
On expanding along R3, we get
LHS = 2(a + b + c)3 [(1)(1–0)]
= 2(a + b + c)3
= RHS
Hence Proved.
Ques. Using properties of determinants, prove that: (Delhi 2014) (5 Marks)
\(\begin{vmatrix} x^2+1 & xy & xz \\ xy & y^2+1 & yz \\ xz & yz & z^2+1 \end{vmatrix} = 1 + x^2 + y^2 + z^2\)
Ans. On taking common factors x, y and z from R1, R2, and R3, we get
LHS = \(xyz \begin{vmatrix} x+\frac{1}{x} & y & z \\ x & y+\frac{1}{y} & z \\ x & y & z+\frac{1}{z} \end{vmatrix}\)
On applying R2 → R2 – R1 and R3 → R3 – R1, we get
LHS = \(xyz \begin{vmatrix} x+\frac{1}{x} & y & z \\ -\frac{1}{x} & \frac{1}{y} & 0 \\ -\frac{1}{x} & 0 & \frac{1}{z} \end{vmatrix}\)
On multiplying and dividing C1 by x, C2 by y and C3 by z and taking common \(\frac{1}{x}\), \(\frac{1}{y}\), \(\frac{1}{z}\) from C1, C2 and C3, we get
LHS = \(xyz \times \frac{1}{xyz} \begin{vmatrix} x^2+1 & y^2 & z^2 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{vmatrix}\)
= \(\begin{vmatrix} x^2+1 & y^2 & z^2 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{vmatrix}\)
On expanding along R3, we get
LHS = -1 × (-z2) + 1[1(x2 + 1) + 1(y2)]
= 1 + x2 + y2 + z2
= RHS
Hence Proved.
Ques. Using properties of determinants, prove that: (All India 2014, 2010C; Delhi 2012, 2010, 2009C) (5 Marks)
\(\begin{vmatrix} b+c & c+a & a+b \\ q+r & r+p & p+q \\ y+z & z+x & x+y \end{vmatrix} =2 \begin{vmatrix} a&b&c\\ p&q&r\\ x&y&z \end{vmatrix}\)
Ans.
Consider \(\Delta = \begin{vmatrix} b+c & c+a & a+b \\ q+r & r+p & p+q \\ y+z & z+x & x+y \end{vmatrix}\)
On splitting Δ along C1, we get
\(\Delta = \begin{vmatrix} b & c+a & a+b \\ q & r+p & p+q \\ y & z+x & x+y \end{vmatrix} + \begin{vmatrix} c & c+a & a+b \\ r & r+p & p+q \\ z & z+x & x+y \end{vmatrix}\)
Again, on splitting both above determinants along their respective second columns, we get
\(\Delta = \begin{vmatrix} b & c & a+b \\ q & r & p+q \\ y & z & x+y \end{vmatrix} + \begin{vmatrix} b & a & a+b \\ q & p & p+q \\ y & x & x+y \end{vmatrix} + \begin{vmatrix} c & c & a+b \\ r & r & p+q \\ z & z & x+y \end{vmatrix} + \begin{vmatrix} c & a & a+b \\ r & p & p+q \\ z & x & x+y \end{vmatrix}\)
\(= \begin{vmatrix} b & c & a+b \\ q & r & p+q \\ y & z & x+y \end{vmatrix} + \begin{vmatrix} b & a & a+b \\ q & p & p+q \\ y & x & x+y \end{vmatrix} + \begin{vmatrix} c & a & a+b \\ r & p & p+q \\ z & x & x+y \end{vmatrix}\)
\(\left[ \because \begin{vmatrix} c & c & a+b \\ r & r & p+q \\ z & z & x+y \end{vmatrix} = 0, \text{as C}_1\ \text{is identical to C}_2 \right]\)
Similarly, on splitting all above determinants together, we get
\(\Delta = \begin{vmatrix} b & c & a \\ q & r & p \\ y & z & x \end{vmatrix} + \begin{vmatrix} b & c & b \\ q & r & q \\ y & z & y \end{vmatrix} + \begin{vmatrix} b & a & a \\ q & p & p \\ y & x & x \end{vmatrix} + \begin{vmatrix} b & a & b \\ q & p & q \\ y & x & y \end{vmatrix} + \begin{vmatrix} c & a & a \\ r & p & p \\ z & x & x \end{vmatrix} + \begin{vmatrix} c & a & b \\ r & p & q \\ z & x & y \end{vmatrix}\)
\(= \begin{vmatrix} b & c & a \\ q & r & p \\ y & z & x \end{vmatrix} + \begin{vmatrix} c & a & b \\ r & p & q \\ z & x & y \end{vmatrix}\)
[∵ all other determinants have their two columns identical, so their value is 0]
On applying C1 ↔ C3 in first and C1 ↔ C2 in second determinant, we get
\(\Delta = -\begin{vmatrix} a & c & b \\ p & r & q \\ x & z & y \end{vmatrix} - \begin{vmatrix} a & c & b \\ p & r & q \\ x & z & y \end{vmatrix}\)
[∵ when any two columns or rows of a determinant are interchanged, its value becomes negative]
On applying C2 ↔ C3 in both determinants, we get
\(\Delta = \begin{vmatrix} a & b & c \\ p & q & r \\ x & y & z \end{vmatrix} + \begin{vmatrix} a & b & c \\ p & q & r \\ x & y & z \end{vmatrix} \)
\(= 2 \begin{vmatrix} a & b & c \\ p & q & r \\ x & y & z \end{vmatrix} \)
Hence Proved.
Ques. Using properties of determinants, prove that: (All India 2014, 2009) (5 Marks)
\(\begin{vmatrix} x+y & x & x \\ 5x+4y & 4x & 2x \\ 10x+8y & 8x & 3x \end{vmatrix} = x^3\)
Ans.
To Prove: \(\begin{vmatrix} x+y & x & x \\ 5x+4y & 4x & 2x \\ 10x+8y & 8x & 3x \end{vmatrix} = x^3\)
LHS = \(\begin{vmatrix} x+y & x & x \\ 5x+4y & 4x & 2x \\ 10x+8y & 8x & 3x \end{vmatrix}\)
On applying R2 → R2 – 2R1 and R3 → R3 – 3R1, we get
LHS = \(\begin{vmatrix} x+y & x & x \\ 3x+2y & 2x & 0 \\ 7x+5y & 5x & 0 \end{vmatrix}\)
On expanding along C3, we get
LHS = \(x\begin{vmatrix} 3x+2y & 2x \\ 7x+5y & 5x \end{vmatrix}\)
= x [5x(3x + 2y) – 2x(7x + 5y)]
= x [15x2 + 10xy – (14x2 + 10xy)]
= x3
= RHS
Hence Proved.
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