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Determinant of a Matrix Formula

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  • Last Updated : 12 May, 2022
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Matrix is one of the essential topics in mathematics. Understanding the determinant of a matrix requires some basic knowledge of matrix-like rows, columns, etc. The matrix determinant is used in various formulas like finding the inverse of a matrix and many more. It is easy to find determinants using the determinant formula. It has various properties. Let’s understand how the determinant of a matrix is determined with the help of the determinant formula. 

Determinant

The determinant of a matrix is a unique number associated with that square matrix. The determinant of a matrix can be calculated for only a square matrix. If A =[aij] is a square matrix of order n, then A’s determinant is represented by det A or |A|. The general representation of determinant of matrix A is, 

det A or |A| or \begin{vmatrix} a_{11} & a_{12} & ... & a_{1j} & ... & a_{1n}\\ a_{21} & a_{22} & ... & a_{2j} & ... & a_{2n}\\ .& . & ... &. & ... & .\\ .& . & ... &. & ... & .\\ a_{i1} & a_{i2} & ... & a_{ij} & ... & a_{in}\\ .& . & ... &. & ... & .\\ a_{n1} & a_{n2} & ... & a_{nj} & ... & a_{nn}\\ \end{vmatrix}

Determinant Formula

  • Determinant of a square matrix of order 1 

If A = [aij] is a square matrix of order 1, then the determinant of A is defined as,

|A| = a11  

  • Determinant of a square matrix of order 2

If A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} is a square matrix of order 2 then the determinant is defined as,

|A| = a11a22 – a12a21

  • Determinant of a square matrix of order 3 

If A = \begin{bmatrix} a_{11} & a_{12} &a_{13}\\ a_{21} & a_{22} &a_{23}\\a_{31} & a_{32} &a_{33}\end{bmatrix} is a square matrix of order 2 then the determinant is defined as,

|A| =  \begin{vmatrix} a_{11} & a_{12} &a_{13}\\ a_{21} & a_{22} &a_{23}\\a_{31} & a_{32} &a_{33}\end{vmatrix}

|A| = (-1)1+1 a11 \begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} +  (-1)1+2 a12  \begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{vmatrix}  + (-1)1+3 a13 \begin{vmatrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{vmatrix}

|A| = a11  \begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix}– a12   \begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{vmatrix} +a13 \begin{vmatrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{vmatrix}

|A| = a11(a22a33 – a23a32) – a12(a33a21 – a23a31) + a13(a32a21 – a22a31) ⇢ (1)

|A| = a11a22a33 + a12a23a31 + a13a32a21 – a11a23a32 – a22a13a31 – a12a21a33

Thus, the determinant of a square matrix of order 3 is the sum of the product of elements aij in ith row with (-1)i+j  times the determinant of a 2 x 2 sub-matrix obtained by leaving the ith row and jth column passing through the element. The expansion is done through the elements of ith row. Then, it is known as the expansion along the ith row. The above expansion (1) of |A| is known as the expansion along the first row.

Note 

  • Only square matrices have determinants. The matrices which are not square do not have determinants.
  • The determinant of order three can be expanded along any row and column.
  • If a row or column of a determinant is all zeros, then the determinant is zero.

Minor  

Consider a square matrix A =[aij] of order n. The minor Mij of aij in A determines the square sub-matrix of order (n – 1) obtained by leaving the ith row and jth column of A.

Cofactor

Consider a square matrix A = [aij] of order n. The cofactor Cij of aij in A is equal to (-1)i + j times the minor Mij of aij.

Cij = Cofactor of aij in A = (-1)i+j Mij    

Where Mij is the minor of aij in A.

Cij \begin{cases} \hspace{0.2cm}M_{ij}, & \text{if $n$ is even}\\ -M_{ij}, & \text{if $n$ is odd} \end{cases}

Singular Matrix

A square matrix is said to be singular if its determinant is zero. Otherwise, it is non-singular.

Properties of Determinant 

Let A be a square matrix of order n,

  1. The sum of product of elements of any row(column) with their cofactors is equal to |A|.   \sum_{j=1}^n a_{ij}C_{ij} = |A| and \sum_{i=1}^n a_{ij}C_{ij} = |A|
  2. The sum of product of elements of any row(column) with the cofactors of the corresponding elements of some other row(column) is zero.  \sum_{j=1}^n a_{ij}C_{kj} = 0 and  \sum_{i=1}^n a_{ij}C_{ik} = 0
  3. The determinant remains unchanged if its rows and columns are interchanged(transpose). |A| =|AT
  4. If two rows (columns) are interchanged, the determinant changes by a minus sign. |B| = -|A| (B is a matrix obtained by interchanging two rows (columns) of A).
  5. If two rows or columns of A are the same, then its determinant is zero. |A| = 0
  6. If B is the matrix obtained by multiplying scalar value k in each row element or column of matrix A then |B| = k|A|.
  7. If each row element (or column) is zero, then the determinant is zero.
  8. If A is a diagonal matrix, then |A| = a11 × a22 × …. ann.
  9. If A and B are square matrices of the same order, then |AB| = |A||B|.
  10. If each element of a row (column) can be expressed as the sum of two or more terms, then the determinant can be expressed as the sum of two or more determinants.
  11. If B is a matrix obtained by adding to a row (column) of A a scalar multiple of another row(column) of A, then |B| = |A|.

Evaluation of Determinant

Evaluation of the determinant of a matrix of order 2 is easy. Still, for a higher-order matrix, the evaluation of the determinant can be made more accessible by performing row or column operation to make the whole row or column zero or identical or to make more elements zero so that the calculation in the evaluation of the determinant becomes easier.

Sample Questions 

Question 1: Evaluate: 

  1.  \begin{vmatrix} -2 & 6 \\ 4 & 3 \end{vmatrix}  
  2.  \begin{vmatrix} sin\theta & cos\theta \\ -cos\theta & sin\theta \end{vmatrix}

Solution: 

  1. \begin{vmatrix} -2 & 6 \\ 4 & 3 \end{vmatrix} = (-2) × 3 – 6 × 4 = -30
  2.  \begin{vmatrix} sin\theta & cos\theta \\ -cos\theta & sin\theta \end{vmatrix}   = sin2θ – (-cos2θ) = sin2θ + cos2θ = 1

Question 2: Evaluate:  

  • D = \begin{vmatrix} -3 & 2 &4\\ 6& 9&5\\1 & 2 &3\end{vmatrix}   expand along first row.
  • D= \begin{vmatrix} -1 & 2 &0\\ 3&5&4\\1 & -2 &1\end{vmatrix}   expand along second column.

Solution:

  • D= \begin{vmatrix} -3 & 2 &4\\ 6& 9&5\\1 & 2 &3\end{vmatrix}

= (-1)1 + 1(-3)\begin{vmatrix} 9 & 5 \\ 2& 3 \end{vmatrix}   +  (-1)1+2(2)\begin{vmatrix} 6 & 5 \\ 1 & 3 \end{vmatrix}   + (-1)1+3(4) \begin{vmatrix} 6 & 9 \\ 1 & 2 \end{vmatrix}

= -3\begin{vmatrix} 9 & 5 \\ 2& 3 \end{vmatrix}   – 2\begin{vmatrix} 6 & 5 \\ 1 & 3 \end{vmatrix}   + 4\begin{vmatrix} 6 & 9 \\ 1 & 2 \end{vmatrix}

= -3[9 × 3 – 5 × 10 ] – 2[6 × 3 – 5 × 1] + 4[6 × 2 – 9 × 1] = -3[27 -10] – 2[18 -5] + 4[12 -9] = -3 × 17 – 2 × 13 + 4 × 3 = -51 -26 +12 = -65

  • D =\begin{vmatrix} -1 & 2 &0\\ 3&5&4\\1 & -2 &1\end{vmatrix}

= (-1)1 + 2 (2) \begin{vmatrix} 3 & 4 \\ 1& 1 \end{vmatrix}+  (-1)2+2(5)\begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix}   + (-1)3+2(-2)\begin{vmatrix} -1 & 2 \\ 3 & 5 \end{vmatrix}

= -2\begin{vmatrix} 3 & 4 \\ 1& 1 \end{vmatrix} + 5\begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix}   + 2\begin{vmatrix} -1 & 2 \\ 3 & 5 \end{vmatrix}

= -2[3 × 1 – 4 × 1] + 5[1 × 1 – 1 × 0] + 2[(-1) × 5 – 2 × 3] = -2[3 – 4] + 5[1 – 0] + 2[-5 – 6] = (-2) × (-1) + 5 × 1 + 2 × (-11) = (-2) × (-1) + 5 × 1 + 2 × (-11)

= 2 + 5 – 22 = -15

Question 3: Evaluate: D = \begin{vmatrix}1 & 0 &2 &0\\ 3& -1 & 8 & 6\\0 & -4 &-2& 3\\2&6&7&1 \\\end{vmatrix}   expand along first row.

Solution: 

D = \begin{vmatrix}1 & 0 &2 &0\\ 3& -1 & 8 & 6\\0 & -4 &-2& 3\\2&6&7&1 \\\end{vmatrix}

D = (-1)1+1(1)\begin{vmatrix}\\ -1 & 8 & 6\\ -4 &-2& 3\\6&7&1 \\\end{vmatrix}  + (-1)1+2(0) \begin{vmatrix}\\ 3&  8 & 6\\0 & -2& 3\\2&7&1 \\\end{vmatrix}  + (-1)1+3(2) \begin{vmatrix}\\ 3& -1 &  6\\0 & -4 & 3\\2&6&1 \\\end{vmatrix} + (-1)1+4(0)\begin{vmatrix}\\ 3& -1 & 8 \\0 & -4 &-2\\2&6&7 \\\end{vmatrix}

D = 1[(-1)1+1(-1) \begin{vmatrix}\\-2& 3\\7&1 \\\end{vmatrix} + (-1)1+2(8)\begin{vmatrix}\\-4 & 3\\6&1 \\\end{vmatrix} +(-1)1+3(6) \begin{vmatrix}\\ -4 &-2\\6&7 \\\end{vmatrix} ] – 0 +2[(-1)1+1(3)\begin{vmatrix}\\-4 & 3\\6&1 \\\end{vmatrix}  + (-1)1+2(-1) \begin{vmatrix}\\0 & 3\\2&1 \\\end{vmatrix}  +(-1)1+3(6)\begin{vmatrix}\\0 & -4\\2&6 \\\end{vmatrix} ] – 0

D = 1[(-1){(-2} × 1 – 3 × 7} – 8{(-4) × 1 – 6 × 3} + 6{(-4) × 7 – 6 × (-2)}] + 2[3{(-4} × 1 – 6 × 3} + 1{0 × 1 – 3 × 2} + 6{0 × 6 -(-4) × 2}]

D = [-1 × (-2 – 21) – 8 × (-4 – 18) + 6(-28 + 12)] + 2[3 × (-4 – 18) + (0 – 6) + 6 × (0 + 8)]

D = [-1 × (-23) – 8 × (-22) + 6 × (-16)] + 2[3 × (-22) -6 + 48]

D = [23 + 176 – 96] + 2[-66 – 6 + 48]

D = 103 + 2 × (-24) = 103 – 48

D = 55 

Question 4: Find the value of x for which matrix A = \begin{bmatrix} 1 & -2 &3\\ 1&2 & 1\\x& 2& -3 \end{bmatrix} is singular. 

Solution: 

Matrix A is singular, if |A| = 0

\begin{vmatrix} 1 & -2 &3\\ 1&2 & 1\\x& 2& -3 \end{vmatrix} = 0 

⇒ 1\begin{vmatrix} 2 & 1\\ 2&-3  \end{vmatrix} + 2\begin{vmatrix} 1 & 1\\ x&-3  \end{vmatrix} + 3\begin{vmatrix} 1 & 2\\ x&2  \end{vmatrix} = 0

⇒ (-6 – 2) + 2(-3 – x) + 3(2 – 2x) = 0

⇒ -8 – 6 – 2x + 6 – 6x = 0

⇒ -8x – 8 = 0

⇒ x = -1     

Question 5: Show that \begin{vmatrix} 1 & a &b+c\\ 1&b & a+c\\1& c& a+b \end{vmatrix}=0.

Solution: 

D = \begin{vmatrix} 1 & a &b+c\\ 1&b & a+c\\1& c& a+b \end{vmatrix}    

Apply C2 ⇢ C2 + C3

D = \begin{vmatrix} 1 & a+b+c &b+c\\ 1&b+a+c & a+c\\1& c+a+b& a+b \end{vmatrix}

D = (a+ b+ c)\begin{vmatrix} 1 & 1 &b+c\\ 1&1& a+c\\1& 1& a+b \end{vmatrix}

D = (a+ b+ c) x 0 = 0 [Since, C1 and C2 are identical so determinant = 0 (Property of determinant)]

Question 6: Evaluate : \begin{vmatrix} 1 & 2 &3\\ 2&4 & 6\\5&9&7\end{vmatrix}

Solution: 

D = \begin{vmatrix} 1 & 2 &3\\ 2&4& 6\\5&9&7\end{vmatrix}

R2 ⇢ R2 – 2R1

D = \begin{vmatrix} 1 & 2 &3\\ 0&0&0\\5&9&7\end{vmatrix}

[Since, all the elements of R2 are 0 so, D = 0]

D = 0

Question 7: Prove that :  \begin{vmatrix}a+b & a & a\\ 5a+4b& 4a & 2a \\10a+8b & 8a &3a \\\end{vmatrix}  = a

Solution: 

Let  D = \begin{vmatrix}a+b & a & a\\ 5a+4b& 4a & 2a \\10a+8b & 8a &3a \\\end{vmatrix}

Since, C2 contains sum of two elements it can be represented as sum of two determinants.

D = \begin{vmatrix}a & a & a\\ 5a& 4a & 2a \\10a & 8a &3a \\\end{vmatrix} + \begin{vmatrix}b & a & a\\ 4b& 4a & 2a \\8b & 8a &3a \\\end{vmatrix}  

Taking a and b common from rows and columns.           

D = a3 \begin{vmatrix}1 & 1 & 1\\ 5& 4 & 2 \\10 & 8 &3 \\\end{vmatrix}   +ba2  \begin{vmatrix}1& 1 & 1\\ 4& 4 & 2 \\8 & 8 &3\\\end{vmatrix}

Since C1 and C2 in the second determinant are the same, its determinant is zero.

D = a3  \begin{vmatrix}1 & 1 & 1\\ 5& 4 & 2 \\10 & 8 &3 \\\end{vmatrix} + ba2 × 0   

Applying C1 ⇢ C1 – C3, C2 ⇢ C2 – C

D = a \begin{vmatrix}0 & 0 & 1\\ 3& 2 & 2 \\7 & 5 &3 \\\end{vmatrix}

D = a3× 1 × \begin{vmatrix} 3& 2  \\7 & 5  \\\end{vmatrix}

D = a3 (15 – 14) = a3


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