Section 5.3 Systems of Linear Equations: Determinants

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1 Section 5. Systems of Linear Equations: Determinants In this section, we will explore another technique for solving systems called Cramer's Rule. Cramer's rule can only be used if the number of equations matches the number of variables. Also, if the problem consists of a system of linear dependent equations, then Cramer's Rule only tells us that our equations are dependent. Cramer's Rule works by using what are called determinants. Objective : Evaluate x determinants. If a matrix has the same number of rows and columns, it is called a square matrix. Every square matrix has a real number associated with it called its determinant. The determinant is defined as follows for matrices: Determinants The determinant for matrix is denoted as a b a b and is defined as: a b a b = a b a b. It is almost like cross multiplication except we are subtracting instead of setting the products equal. Evaluate the following: Ex. 5 4 Ex. 5 = 5() 4( ) = 5 + = 7. 4 = ( ) () = 9 4 =.

2 Objective : Cramer s Rule for Solving a System of Two Linear Equations in Two Variables Cramer s Rule for x Systems For the system: ax + by = s (the equations must be in standard form) cx + dy = t We can first calculate the following determinants: D = a b D x = s b D y = a s c d t d c t Then, the solution to the system is x = D x and y = D y D D If D =, and if either D x or D y or both are not zero, then system is inconsistent. If D, D x, and D y are all zero, then the equations are dependent. We must then use another technique to solve the system. To find D x, replace the coefficients for x in the determinant by the constant terms. To find D y, replace the coefficients for y in the determinant by the constant terms. To see where Cramer s Rule comes from, let us solve the following system using the elimination method. Solve using elimination: Ex. ax + by = s cx + dy = t Let s first eliminate x. c eqn. # + a eqn. #: c ) acx cby = cs a ) acx + ady = at ady cby = at cs (factor out y) (ad cb)y = at cs (solve for y) y = at cs ad cb But at cs = a c s t = D y and ad cb = a b c d = D.So, y = D y D. Now, let s eliminate y. d eqn. # + b eqn. #: d ) adx + bdy = ds b ) cbx bdy = bt

3 adx cbx = ds bt (factor out x) (ad cb)x = ds bt (solve for x) x = ds bt ad cb But ds bt = s b t d = D x and ad cb = a b c d = D. So, x = D x Solve using Cramer s Rule: Ex. 4 5x y = 4x + y = 7 D = 5 = 5() 4( ) = 5 + = 7. 4 D x = = () 7( ) = + = 4. 7 D y = 5 = 5(7) 4() = 5 5 = x = D x D = 4 7 = and y = D y D So, the solution is (, ). = 7 7 =. Ex. 5 6x 4y = 9x + 6y = D = = 6(6) ( 9)( 4) = 6 6 = D x = = (6) ( 8)( 4) = 7 7 = D y = = 6( 8) ( 9)() = = 9 8 The system is dependent. Solve for y: 6x 4y = 4y = 6x + y =.5x So, the solution is { (x, y) y =.5x, x is any real number}. D.

4 4 Objective : Understanding and evaluating determinants. The determinant will have the form: a a a a a a a a a Before we can find the determinant for an n n matrix with n being or higher, we need to define the minor and cofactors of an entry of the determinant. Minors For any n n determinant, the minor of M ij of entry a ij is the determinant of the matrix with the i th row and j th column deleted. Hence, minor M can be found be crossing out the second row and third column where as the minor M can be found by crossing out the third row and first column. M = a a a a a a = a a a a a a a M = a a a a a a = a a a a a a a Cofactor For any n n determinant, the cofactor of entry a ij, denoted by A ij, is given by: A ij = ( ) i + j M ij where M ij is the minor of a ij. Find the following cofactors of the given matrix on the right: Ex. 6a A 4 Ex. 6b A a) A = ( ) b) A = ( ) = ( ) 4 7 = ( ) = ( + 8) = 6 = ( 4 ) = 4

5 Evaluating an n n determinant To calculate the value of an n n determinant, multiply each entry in any column or row by its corresponding cofactor and add the results. The process is called expanding across a column or row. Evaluate the following: Ex. 7 Ex. 8 4 Expand across column #: = ( ) 4 + ( ) 4 + ( )4 4 = [( ) 4( )] [( )( ) 4()] + [( )( ) ()] = [ ] [ ] + [ ] = 4 + = 6. 4 Expand across row #: 4 = ( ) 4 + ( )4 + ( 4 )5 = [4() ()] + [()() ()] [()() (4)] = [ ] + [] [] = + =. Objective 4: Cramer s Rule for solving a system of three linear equations in three variables. We will now state Cramer s Rule for solving a system of three linear equations in three variables. The proof of Cramer s Rule is quite a bit more complicated than the proof of Cramer s Rule for x systems and it will be omitted. The proof does follow the same spirit as for the x systems. 5

6 6 Cramer s Rule for x Systems For the system: a x + a y + a z = c a x + a y + a z = c a x + a y + a z = c We can first calculate the following determinants: a a a c a a D = a a a D x = c a a a a a c a a D y = a c a a c a D z = a c a Then, the solution to the system is a a c a a c a a c x = D x D, y = D y D, and z = D z D If D, D x, D y, and D z are all zero, then the system is dependent and there are an infinite number of solutions. If D is zero and if at least one of D x, D y, and D z is not zero, then the system inconsistent and has no solution. To find D x, replace the coefficients for x in the determinant by the constant term. To find D y, replace the coefficients for y in the determinant by the constant term. To find D z, replace the coefficients for z in the determinant by the constant term. Solve the following systems using Cramer s Rule: Ex. 9 x + y z = 6 x + y + z = 4 x 5y + 6z = 9 Expand across column #: D = = ( ) 5 6 ( ) ( )4 5 6 = (8 ( 5)) + ( 5) + ( ( )) = = 86.

7 7 Expand across column #: 6 D X = 4 = 6( ) ( ) ( ) = 6(8 ( 5)) 4( 5) + 9( ( )) = = 55. Expand across column #: 6 D y = 4 = ( ) ( 6 ) ( 6 ) = (4 9) + (6 ( 9)) + (6 ( 4)) = =. Expand across column #: 6 D z = 4 = ( ) ( 6 ) ( 6 ) = (7 ( )) + (8 ( )) + (8 8) = = 69. Thus, x = D X D = 55 86, y = D y D = 86 = 55 4, z = D z D = So, the solution is ( 86, 55 4, ). Ex. y + 6z = 7 x y + z = x + y + z = Expand across column #: D = 6 = ( ) ( 6 ) + ( 6 )4 = ( 4) + ( ) + (4 + 6) = + = Expand across row #: 7 6 D X = = ( ) 6 ( 7 6 )4 + ( 7 )5 = ( ) (7 + ) (4 + 4) = 9 6 = 55. Since D =, but D x, then the system is inconsistent. There is no solution.

8 Ex. x + y + z = x + y + z = 5 x y = 4 Expand across row #: D = = ( ) 4 ( )5 + ( )6 = ( ) + ( + ) + ( + ) = + + = Expand across column #: D X = 5 = ( ) ( )5 4 + ( )6 5 4 = ( 5 + 8) ( + 4) + ( 5) = + = 8 Expand across column #: D y = 5 = ( ) ( )5 4 + ( )6 5 4 = (4 ) ( 4 ) + (5 + ) = = Expand across column #: D z = 5 = ( ) 5 4 ( ) 4 + ( )4 5 4 = ( 8 + 5) + ( 4 + ) + (5 ) = + ( ) + 6 = Since D, D x, D y, D z are all zero, the equations are dependent. We will have to use one of our other techniques to find the solution. Write as an augmented matrix: 5 4 Use the elementary row operations to get in row echelon form:

9 R = r + r 6 6 R = r R = r r Thus, our system is: x + z = x = z y + z = y = z + Hence, the solution is: {(x, y, z) x = z, y = z +, z is any real number}. Objective 5: Properties of Determinants. Property #: The value of the determinant changes signs if you interchange any two rows or interchange any two columns. To illustrate, let's interchange column # and # and compare the determinants: D = 5 4 = 5() 4( ) = 5 + = 7. D = 5 4 = (4) (5) = 5 = 7. R = r + r R = r r

10 Property #: If all the entries of any row or column are zeros, the determinant is zero. To prove this, expand the determinant across the row or column that contains all zeros for entries. Thus, all the cofactor will be multiplied by which will yield a determinant of. 4 Property #: If any two rows or any two columns of a determinant have corresponding entries that are equal, the value of the determinant is zero. To illustrate this, let row # and row # be the same. 5 = ( ) 5 5 ( ) 5 ( )4 5 5 = ( ) + (5 ) (5 ) = + = Property #4: If any row or column of a determinant is multiplied by a nonzero number k, the value of the determinant is changed by a factor of k. To illustrate this, multiply every entry in row # by and compare to the original determinant: D = 5 = 5() 4( ) = 5 + = 7. 4 D = 5 = 5() 8( ) = + 4 = 4 = 7. 8 Property #5: If the entries of any row or any column of a determinant are multiplied by a nonzero constant k and added to the corresponding entries of another row or column, the value of the determinant is unchanged. To illustrate this, add r to row # and compare to the original determinant: D = 5 = 5() 4( ) = 5 + = D = = 5( 5) 4( ) = =