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Lecture 31.pdf

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Department
Mathematics
Course
MATH 136
Professor
Robert Sproule
Semester
Winter

Description
Monday, March 24 − Lecture 31 : Determinant applications. Concepts: 1. Applying Cramers rule to find to solve a system of linear equations. 2. Define the adjoint of a matrix. 3. Find the inverse of a matrix by using the adjoint of the matrix. 4. Area of a parallelogram and volume of a parallelepipedon. Cramer’s rule for solving systems of linear equations using 31.1 Theorem − Cramer’s rule. If A is an invertible n × n matrix and b is a vector in ℝ , n th then A(i) denotes the matrix obtained by replacing the i column of A by b. The unique solution x of the system Ax = b is given by 31.1.1 Example: Solve the system by using Cramer’s rule. Solution: We have the coefficient matrix A: Expanding along row 1 we get det A = 6. Thus Adjoint (or adjugate) of a matrix. 31.2 Definition − Let Abe an n × n matrix. The adjoint of A, adj(A), (also called the adjugate of A) is the transpose of the matrix whose entries are the cofactors c of A. That ij is - Recall that the (i, j) cofactor of A , c , isijhe number (−1) i + jm ijere m is tij th number called the (i, j) minor (the number you obtaithby taking ththdeterminant of the "minor matrix" you get by deleting the i row and j column). th 31.2.1 Example − Find the (1, 2) entry of the adjoint, adj(A), of the matrix Answer: - Let D = [d ij = adj(A). We want the value of d . 12 T - Since D = adj(A) = [ c ] tijn d 12c 21 1 + 2 = (−1) m 21 = (−1) det M 21 = (−1)( −6 + 3) = 3. Once computed the complete matrix D= adj(A) is 31.3 Theorem − Adjoint method of finding the inverse of A. (Also called the Cofactor method). For a square n × n matrix A, 31.3.1 Example − Find the inverse of the matrix Answer: We easily compute det A= −2. We have from above 31.3 Area of a parallelogram as a determinant We show to ways of computing the area of a parallelogram: Method 1: To compute the area P of a parallelogram, we assume the parallelogram is in quadrant I with vertices (0, 0), v = (a, c), u = (b, d) and u + v = (a + c, c + d) as in the diagram below, where a,b, c and d are positive. The area of the rectangle A = (a + b) (c + d) can be expressed as the sum ofthe area P of the parallelogram plus the area of four rectangles. Two of these rectangles areso that the length of the base is c+ d and the length of the height is d. The other two of rectangles are so that the length of the base is a + b and the length of the height is c. Then, Note that, given any two vectors u = (a, c) and v = (b, d), det[ u v ] = − det[ v u ]. Then depending on the order in which we place the column vectors of the matrix, the number det[ u v ] may not represent the area of the parallelogram with vertices (0, 0), u and v. However taking the absolute value of det[ u v ] will produce the desired area: Method 2: We know that the area of a parallelogram is normally computed by multiplying the height h of the parallelogramby the length of its base. The parallelogram can be drawn in the Cartesian plane so that one of its vertices is at the origin. If the two vertices closest to the origin are u = (u , 1 ) 2nd v = (v , v1) t2en the vertex which is the furthest from the origin must be u + v = (u + v ,1u + 1 ).2Supp2se we project the vector v onto the vector u to obtain proj v. Then perp v = v – proj v. From this we can obtain u u u the height of the parallelogram when the directed line segment for the vector u represents its base. Then the area of the parallelogram is A = ||u|||| perp u ||.
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