MTH405_MTB_Lect_2

# MTH405_MTB_Lect_2 - MTH405 Mid-Trimester Break Lect 2...

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Unformatted text preview: MTH405 Mid-Trimester Break Lect 2 Determinant of a Matrix We begin by dening the determinant of a 2 × 2 matrix below. Denition. Let A = of A is given by  a c  b be a 2 × 2 square matrix. Then the d a b = ad − bc. |A| = c d determinant Note that only square matrices (n × n matrices) have determinants. a11 a12 a13 Denition. Let A = a21 a22 a23 be a 3 × 3 square matrix. Then the a31 a32 a33 determinant of A is given by a11 a12 a13 a a21 a23 a22 a23 +(+1)(a13 ) 21 |A| = a21 a22 a23 = (+1)(a11 ) +(−1)(a12 ) a31 a a a a 31 33 32 33 a31 a32 a33 For the 3 × 3 matrix, notice that rst we choose the rst row (entries a11 , a12 , a13 ) and for each of these entries, we multiply by it by the determinants Remark. of the matrices as obtained below. • For the rst term, we have a11 , then we remove row 1 and column 1 from A, and nd the determinant for the rest of the matrix we get. • For the second term, we have a12 , then we remove row 1 and column 2 from A, and nd the determinant for the rest of the matrix we get. • For the third term, we have a13 , then we remove row 1 and column 3 from A, and nd the determinant for the rest of the matrix we get. • Lastly, rst term has a plus, second term has a minus and third term has a plus again. 1 a22 . a32 Exercise. Find the determinants of the following matrices.   3 −2 . 4 5 3 −2 −5 2. B = 1 4 −4 . 0 3 2 1. A = Solution. |A| = 23 and |B| = 49. Cross Product (Vector Product) Many problems in Linear Algebra, Physics and Engineering involve nding a vector orthogonal to other vectors. Remember the Flemings Left-Hand Motor Rule and Right-Hand Generator Rule; these are properly shown in the formula using cross products. Denition. Let ~u = hu1 , u2 , u3 i and ~v = hv1 , v2 , v3 i be vectors in 3-space, then the cross product ~u × ~v is the vector dened by ˆi ~u × ~v = u1 v1 ˆj u2 v2 kˆ u3 . v3 However, this is not a true determinant since the entries in an actual determinant of a matrix involve number and not vectors Theorem. If ~u, ~v and w ~ are any vectors in 1. ~u × ~v = −(~v × ~u). 2. ~u × (~v + w) ~ = (~u × ~v ) + (~u × w) ~ . 3. (~u + ~v ) × w ~ = (~u × w) ~ + (~v × w) ~ . 4. k(~u × ~v ) = (k~u) × ~v = ~u × (k~v ). 5. ~u × 0 = 0 × ~u = 0. 6. ~u × ~u = 0. 3-space and k is any scalar, then: . Exercise. Calculate the following for ~u = h1, 2, −2i and ~v = h3, 0, 1i. 1. ~u × ~v . 2. ~v × ~u. Exercise. Show that ~u × ~u = 0 for any vector ~u = hu1 , u2 , u3 i. 2...
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