The document Area of a Triangle Using Determinants Mathematics Notes | EduRev is a part of the Mathematics Course Algebra for IIT JAM Mathematics.

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**Determinants**

A determinant of a square matrix [a_{ij}] of order n where a_{ij} = (i, j)^{th} element of A is a number (real or complex) associated with it. The number of rows is the same as the number of the columns in a square matrix. If A is a square matrix such as

Then determinant of A is |A| = Î” = a_{11} [(a_{22} Ã— a_{33}) â€“ (a_{23 }Ã— a_{32})] â€“ a_{12} [(a_{21 }Ã— a_{33}) â€“ (a_{31} Ã— a_{23})] + a_{13} [(a_{21} Ã— a_{32}) â€“ (a_{31} Ã— a_{22})].

**Determinants and Matrices as Equation Solver**

Since now we are familiar with the way of calculating the determinant of a square matrix. Here, we will discuss the way to solve a system of linear equations in two or three variables. With the help of the determinant, we can also check for the consistency of linear equations.

â€¢** Consistent System:** If one or more solution(s) exists for a system of equations then it is a consistent system

â€¢** Inconsistent System:** A system of equations with no solution is an inconsistent system.

**The Solution of System of Linear Equations**

A solution for a system of linear Equations can be found by using the inverse of a matrix. Suppose we have the following system of equations

a_{11} x + a_{12} y + a_{13} z = b_{1}

a_{21} x + a_{22} y + a_{23} z = b_{2}

a_{31} x + a_{32} y + a_{33 }z = b_{3}

where, x, y, and z are the variables and a_{11}, a_{12}, â€¦ , a_{33} are the respective coefficients of the variables and b_{1}, b_{2}, and b_{3} are the constants. We need to find the solution for the values of the variables in this system of equations.

**Determinant as an Equation Solver**

The above system of equations can be represented in the form of a square matrix as

i.e., AX = B or,

Here arise two cases

**Case1**

If A is a non-singular matrix i.e., |A| â‰ 0, then its inverse exists.

We have A X = B

or, A^{â€“1} (A X) = A^{â€“1} B (pre-multiplying by A^{â€“1})

or, (A^{â€“1} A) X = A^{â€“1} B

and, IX = A^{â€“ 1} B (I is the identity matrix)

or, X = A^{â€“1} B where, A^{â€“1} = (adj A)/|A|

This matrix equation provides a unique solution and is known as the Matrix Method.

**Case2**

If A is a singular matrix, then |A| = 0 then we calculate (adj A) B. If (adj A) B â‰ 0 (zero matrix), then the solution does not exist. The system of equations is inconsistent. Else, if (adj A) B = 0 then the system will either have infinitely many solutions (consistent system) or no solution (inconsistent system).

**Solved Example for You****Problem:** Suppose you have three numbers. The sum of the two numbers and the twice of the second equals 2. The sum of the second and third when subtracted from the twice of first gives 1. The difference of thrice of first and five times the third gives 5. Rewrite the statement in form of the system of equations. Solve it using Matrix Method as an equation solver.**Solution:** Assume that x, y, and z are the three numbers. Rewriting the above statement we have the following system of equations

x + 2y + z = 2

2x â€“ y â€“ z = 1

3x â€“ 5y = 5

In matrix notation, we have

Here, the determinant of A = |A| = 1(5 â€“ 0) â€“ 2(â€“10 + 3) + 1(0 + 3) = 22 â‰ 0. Hence there exists a unique solution for X.

Calculating adj (A), we have A_{ij} = (â€“1)^{(i + j)} M_{ij} , where M_{ij} is the co-factor of a_{ij}

â€¢** **A_{11} = 1(5 â€“ 0) = 5, A_{12} = â€“1(â€“10 + 3) = 7, A_{13} = 1(0 + 3) = 3,

â€¢** **A_{21} = â€“1(â€“10 â€“0) = 10, A_{22} = 1(â€“5 â€“ 3) = â€“8, A_{23} = â€“1(0 â€“ 6) = 6,

â€¢ A_{31} = 1(â€“2 + 1) = â€“1, A_{32} = â€“1(â€“1 â€“ 2) = 3, A_{33} = 1(â€“1 â€“ 4) = â€“5

The inverse of the matrix A is A^{âˆ’1}.

Since X = A^{â€“ 1} B

Thus, x = 15/22, y = 21/22, and z = â€“13/22.

**Definition of Adjoint of a Matrix**

The adjoint of a square matrix A = [a_{ij}]_{n x n} is defined as the transpose of the matrix [A_{ij}]n x n, where Aij is the cofactor of the element a_{ij}. Adjoing of the matrix A is denoted by **adj A**.

**Example 1: **Find the adjoint of the matrix:**Solution: **We will first evaluate the cofactor of every element,

Therefore,

**The Relation between Adjoint and Inverse of a Matrix**

To find the inverse of a matrix A, i.e A-1 we shall first define the adjoint of a matrix. Let A be an n x n matrix. The (i,j) cofactor of A is defined to be

A_{ij} = (-1)^{ij} det(M_{ij}),

where M_{ij} is the (i,j)^{th} minor matrix obtained from A after removing the ith row and jth column. Letâ€™s consider the n x n matrix A = (Aij) and define the n x n matrix Adj(A) = A^{T}. The matrix Adj(A) is called the adjoint of matrix A. When A is invertible, then its inverse can be obtained by the formula given below.

The inverse is defined only for non-singular square matrices. The following relationship holds between a matrix and its inverse:

AA^{-1} = A^{-1}A = I, where I is the identity matrix.

**Example 2: **Determine whether the matrix given below is invertible and if so, then find the invertible matrix using the above formula.**Solution:** Since A is an upper triangular matrix, the determinant of A is the product of its diagonal entries. This, we have det(A) = -1, which is a non-zero value and hence, A is invertible. To find the inverse using the formula, we will first determine the cofactors A_{ij} of A. We have,

Then the adjoint matrix of A is

Using the formula, we will obtain the inverse matrix as

**Theorems on Adjoint and Inverse of a Matrix**

Theorem 1

If A be any given square matrix of order n, then A adj(A) = adj(A) A = |A|I, where I is the identitiy matrix of order n.

Proof: Let

Since the sum of the product of elements of a row (or a column) with corresponding cofactors is equal to |A| and otherwise zero, we have

Similarly, we can show that adj(A) A = |A| I

Hence, A adj(A) = adj(A) A

Theorem 2

If A and B are non-singular matrices of the same order, then AB and BA are also non-singular matrices of the same order.

Theorem 3

The determinant of the product matrices is equal to the product of their respective determinants, that is, |AB| = |A||B|, where A and B are square matrices of the same order.**Remark:** We know that

Writing determinants of matrices on both sides, we have

i.e,

i.e, |adj(A)| |A| = |A|^{3}

or |adj(A)| = |A|^{2}

In general, if A is a quare matrix of order n, then |adj(A)| = |A|^{n-1}.

Theorem 4

A square matrix A is invertible if and only if A is a non-singular matrix.

Proof: Let A be an invertible matrix of order n and I be the identity matrix of the same order. Then there exists a square matrix B of order n such that AB = BA = I. Now, AB = I.

So |A| |B| = |I| = 1 (since |I| = 1 and |AB| = |A| |B|). This gives |A| to be a non-zero value. Hence A is a non-singular matrix. Conversely, let A be a non-singular matrix, then |A| is non-zero. Now A adj(A) = adj(A) A = |A| I (Theorem 1). Or,

or AB = BA = I, where

Thus A is invertible and

**Solved Examples for You****Question: **With 1, Ï‰, Ï‰2 as cube roots of units, inverse of which of the following matrices exist?**Solution:** Option D. Inverse of a matrix exists if its determinant if not equal to 0. For option A, B and C, the determinants are equal to zero, hence the inverse does not exist for any of the given matrices.

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