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| Application of Matrices and Determinants |
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| Now we shall discuss the use of determinants in finding the area of a triangle and in the solution of simultaneous equations. |
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| We have already learnt in the previous class that the area of triangle whose vertices are (x1, y1), (x2, y2), (x3, y3) is given by |
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| Hence area of a triangle having vertices at (x1, y1), (x2, y2) and (x3, y3) is given by |
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| Since the area has to be a positive quantity, we always take the absolute value of the above determinant. |
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| Example: |
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| Using determinants, find the area of triangle whose vertices are |
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| (2, -7), (1, 3), (10, 8). |
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| Solution: |
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| (x1, y1) = (2, -7) |
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| (x2, y2) = (1, 3) |
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| (x3, y3) = (10, 8) |
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| Area of the triangle |
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| = -47.5 |
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| Since area has to be a positive quantity, it is given by 47.5 sq.units. |
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| We recall from our earlier classes that a system of linear equation with two variables is given by |
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| This system of linear equation may have either one solution or infinitely many solutions or no solution. |
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| Consistency of a system of linear equation |
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| If a system of linear equations has at least one solution, then the system is called consistent, otherwise it is called inconsistent. |
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| Solve the system of linear equations (1) by using method of elimination as studied earlier |
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| Multiplying the first equation by a2 and the second equation by a1, we get |
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| Similarly a1a2 x + a1a2 y = a1c2 ...(3) |
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| Subtracting (2) from (3), we get |
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| (a1b2 - a2b1)y = a1c2 - a2c1 |
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| The above discussion leads to find the solution of a system of linear equations in two variables by using Cramer's rule. |
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| Cramer's rule suggests the use of determinants to solve a system of linear equations. |
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| Let us denote a1b2 - a2b1 (Denominators of x and y in (4) and (5)) |
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| as the determinant |
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| Then the solution of the given system of linear equation (1) is given by |
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| Now consider the system of linear equations with three variable |
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| Let the determinant of the coefficients be denoted by D |
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| i.e., |
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| Multiplying both sides by x, we have |
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| Adding y times column 2 and z times column 3 to column (1), we have |
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| Similarity it can be shown that, |
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| and |
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| This method of solving a system of linear equations with three unknowns, is known as Cramer's rule. |
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| Note: |
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| In the above method note that |
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| To obtain D1, replace a1, a2, a3 by d1, d2, d3 in D |
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| To obtain D2, replace b1, b2, b3 by d1, d2, d3 in D |
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| To obtain D3, replace c1, c2, c3 by d1, d2, d3 in D |
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| Example: |
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| Solve the system of linear equations. |
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| x +2y + 3z = 6 |
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| 2x + 4y + z = 7 |
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| 3x + 2y + 9z = 14 |
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| using Cramer's rule. |
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| Suggested answer: |
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| The determinant of coefficients |
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| Note that we can also evaluate the determinant D1, D2 and D3 directly without using the properties of determinant. |
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| The solution of the system is given by |
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| It is important to mention here the consistency and inconsistency of a system of linear equations with three unknowns. |
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| Case I: |
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| Case II: |
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| If D = 0 and D1, D2 and D3 are not all zero, then the system is inconsistent, that is the system has no solution. |
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| Case III: |
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| If D = 0 and all D1, D2 and D3 are zeros, this system has either infinite solution or no solution. |
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| In this case, put x = k(y = k or z = k), in any two of the equations, find y and z in terms of k. Substitute these values of x, y and z in terms of k, in the third equation. If the third equation is satisfied by this solution, the system has infinitely many solutions. |
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| If the third equation is not satisfied, the system has no solution. |
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| Homogeneous Equations (Constant = 0) |
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| Consider the homogeneous equations |
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| a1x + b1y + c1z = 0 |
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| a2x + b2y + c2z = 0 |
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| a3x + b3y + c3z = 0 |
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| The homogenous system of equations is always consistent because x = 0, y = 0, z = 0 satisfies all the equations in the system. This solution is called the trivial solution. |
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| then the system has a non-trivial solution also. In fact, it has an infinite number of solutions and is said to be dependent. |
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| then the system has only the trivial solution x = 0, y = 0, z = 0. |
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| Non Homogenous Equations (Solution by the Matrix Method) |
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| Consider the non-homogeneous equations |
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| a1x + b1y + c1z = d1 |
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| a2x + b2y + c2z = d2 |
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| a3x + b3y + c3z = d3 |
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| This can be written as |
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| |A| may or may not be zero. |
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| Case I: |
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| Pre-multiply by A-1, |
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| \ A-1 (AX) = A-1B |
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| \ (A-1A) X = A-1B |
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| \ I X = A-1B |
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| or X = A-1B |
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| This is the matrix method to solve the equations. |
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| However, |
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| Case II: |
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| A-1 does not exist |
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| But if (adj A) B = 0, then the system is consistent with infinite number of solutions or has no solution. |
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| the system is inconsistent i.e., it has no solution. |
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| Example 1: |
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| Using matrix method solve the following systems of linear equations |
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| 2x - y + z = -3 |
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| 3x - z = - 8 |
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| 2x + 6y = 2 |
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| Suggested answer: |
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| The given equations are |
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| 2x - y + z = -3 |
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| 3x - 0.y - z = - 8 |
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| 2x + 6y + 0.z= 2 |
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| = 2(6) +1(2) + 1(18) |
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| = 12 +2 + 18 |
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| = 32 |
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The system has a unique solutions. |
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| A11 = (0 + 6) = 6, A12 = -(0 + 2) = -2, A13 = 18 |
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| A21 = 6, A22 = -2, A23 = -14 |
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| A31 = 1, A32 = 5, A33 = 3 |
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| The solution is |
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| \ x = -2, y = 1, z = 2 |
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| Example 2: |
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| Using matrix method, solve the following system of linear equations |
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| x + y + z = 6 …(1) |
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| x + 2y + 3z = 14 …(2) |
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| x + 4y + 7z = 30 …(3) |
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| Suggested answer: |
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| = (14 - 12) - (7 - 3) + (4 - 2) |
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| = 2 - 4 + 2 = 0 |
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| The system may have infinite number of solutions or no solution. |
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| Put x = k in (1) and (2) and solve |
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| y + z = 6 - k |
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| 2y + 3z = 14 - k. |
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| Solving the above two equations, we have |
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| z = k + 2 and y = 4 - 2k |
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| When x = k, substituting these values of x, y and z in (3), we get |
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| x + 4y + 7z = k + 16 - 8k + 7k +14 |
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| = 30 |
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| Since these values of x, y, z satisfy the 3rd equation, the system has infinitely many solutions x = k, y = 4 - 2k and z = k + 2, where k is any number. |
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