5.3 - 1 10 TH EDITION LIAL HORNSBY SCHNEIDER COLLEGE ALGEBRA.

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Presentation transcript:

TH EDITION LIAL HORNSBY SCHNEIDER COLLEGE ALGEBRA

Determinant Solution of Linear Equations Determinants Cofactors Evaluating n  n Determinants Cramer’s Rule

Determinants Every n  n matrix A is associated with a real number called the determinant of A, written  A . The determinant of a 2  2 matrix is defined as follows.

Determinant of a 2  2 Matrix If A =

Note Matrices are enclosed with square brackets, while determinants are denoted with vertical bars. A matrix is an array of numbers, but its determinant is a single number.

Determinants The arrows in the following diagram will remind you which products to find when evaluating a 2  2 determinant.

Example 1 EVALUATING A 2  2 DETERMINANT Let A = Find  A . Use the definition with Solution a 11 a 22 a 21 a 12

Determinant of a 3  3 Matrix If A =

Evaluating The terms on the right side of the equation in the definition of  A  can be rearranged to get Each quantity in parentheses represents the determinant of a 2  2 matrix that is the part of the matrix remaining when the row and column of the multiplier are eliminated, as shown in the next slide.

Evaluating

Cofactors The determinant of each 2  2 matrix above is called the minor of the associated element in the 3  3 matrix. The symbol represents M ij, the minor that results when row i and column j are eliminated. The following table in the next slide gives some of the minors from the previous matrix.

Cofactors ElementMinorElementMinor a 11 a 22 a21a21 a 23 a31a31 a 33

Cofactors In a 4  4 matrix, the minors are determinants of matrices. Similarly, an n  n matrix has minors that are determinants of matrices. To find the determinant of a 3  3 or larger matrix, first choose any row or column. Then the minor of each element in that row or column must be multiplied by +1 or – 1, depending on whether the sum of the row number and column number is even or odd. The product of a minor and the number +1 or – 1 is called a cofactor.

Cofactor Let M ij be the minor for element a ij in an n  n matrix. The cofactor of a ij, written as A ij, is

Example 2 FINDING COFACTORS OF ELEMENTS Find the cofactor of each of the following elements of the matrix a. 6 Solution Since 6 is in the first row, first column of the matrix, i = 1 and j = 1 so The cofactor is

Example 2 FINDING COFACTORS OF ELEMENTS Find the cofactor of each of the following elements of the matrix b. 3 Solution Here i = 2 and j = 3 so, The cofactor is

Example 2 FINDING COFACTORS OF ELEMENTS Find the cofactor of each of the following elements of the matrix c. 8 Solution We have, i = 2 and j = 1 so, The cofactor is

Finding the Determinant of a Matrix Multiply each element in any row or column of the matrix by its cofactor. The sum of these products gives the value of the determinant.

Example 3 EVALUATING A 3  3 DETERMINANT Evaluate expanding by the second column. Solution Use parentheses, & keep track of all negative signs to avoid errors.

Example 3 EVALUATING A 3  3 DETERMINANT Now find the cofactor of each element of these minors.

Example 3 EVALUATING A 3  3 DETERMINANT Find the determinant by multiplying each cofactor by its corresponding element in the matrix and finding the sum of these products.

Cramer’s Rule Determinants can be used to solve a linear system in the form (1) (2) by elimination as follows. Multiply (1) by b 2. Multiply (2) by – b 1. Add.

Cramer’s Rule Multiply (1) by – a 2. Multiply (2) by a 1. Add. Similarly,

Cramer’s Rule Both numerators and the common denominator of these values for x and y can be written as determinants, since

Cramer’s Rule Using these determinants, the solutions for x and y become

Cramer’s Rule We denote the three determinants in the solution as

Note The elements of D are the four coefficients of the variables in the given system. The elements of D x are obtained by replacing the coefficients of x in D by the respective constants, and the elements of D y are obtained by replacing the coefficients of y in D by the respective constants.

Cramer’s Rule for Two Equations in Two Variables Given the system if then the system has the unique solution where

Caution As indicated in the preceding box, Cramer’s rule does not apply if D = 0. When D = 0 the system is inconsistent or has infinitely many solutions. For this reason, evaluate D first.

Example 4 APPLYING CRAMER’S RULE TO A 2  2 SYSTEM Use Cramer’s rule to solve the system Solution By Cramer’s rule, and Find D first, since if D = 0, Cramer’s rule does not apply. If D ≠ 0, then find D x and D y.

Example 4 APPLYING CRAMER’S RULE TO A 2  2 SYSTEM By Cramer’s rule, The solution set is as can be verified by substituting in the given system.

General form of Cramer’s Rule Let an n  n system have linear equations of the form Define D as the determinant of the n  n matrix of all coefficients of the variables. Define D x1 as the determinant obtained from D by replacing the entries in column 1 of D with the constants of the system. Define D xi as the determinant obtained from D by replacing the entries in column i with the constants of the system. If D  0, the unique solution of the system is

Example 5 APPLYING CRAMER’S RULE TO A 3  3 SYSTEM Use Cramer’s rule to solve the system. Solution Rewrite each equation in the form ax + by + cz +  = k.

Example 5 APPLYING CRAMER’S RULE TO A 3  3 SYSTEM Verify that the required determinants are

Example 5 APPLYING CRAMER’S RULE TO A 3  3 SYSTEM Thus, and so the solution set is

Caution As shown in Example 5, each equation in the system must be written in the form ax + by + cz +    = k before using Cramer’s rule.

Example 6 SHOWING THAT CRAMER’S RULE DOES NOT APPLY Show that Cramer’s rule does not apply to the following system. We need to show that D = 0. Expanding about column 1 gives Since D = 0, Cramer’s rule does not apply. Solution

Note When D = 0, the system is either inconsistent or has infinitely many solutions. Use the elimination method to tell which is the case. Verify that the system in Example 6 is inconsistent, so the solution set is ø.