1 then 1 3 1 and if\u00bc 2 14 then \u00bc 21 4 21 4 \u00bc Obviously for any column vector If

1 then 1 3 1 and if¼ 2 14 then ¼ 21 4 21 4 ¼

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1 , then ¸ = 1 3 1 . and if ¼ ¸ = 2 1 4 , then ¼ ¸ ¸ = 2 1 4 ¸ = 2 1 4 = ¼ . Obviously, for any column vector : ¸ ¸ = ¶ . If µ = 1 2 3 D E F −1 1 0 1 F 1 (3 × 4), then µ ¸ = 1 D −1 2 E 1 3 F 0 1 F 1 4 × 3 . or µ = · ¸ I F J K ¸ , µ ¸ = · I F K .
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9/11/2015 9 Transpose of ³ × ¾ matrix µ = ·· ·@ º· º@ is ¾ × ³ matrix µ ¸ = ·· º· ·@ º@ . Note that µ» ¸ = » ¸ µ ¸ . You can go through the proof of this at the end of Notes X.
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9/11/2015 10 YOU CALCULATE ONE: Write down the transpose of µ = 1 2 5 7 6 9 3 2 1 1 −1 1 0 1 0 2 . THE ZERO MATRIX A zero matrix is any matrix of which all entries are 0. Examples: 0 0 0 0 0 0 , 0 0 0 0 , 0 0 0 0 0 0 0 0 0 , 0 0 0 0 0 0 . In equations and expressions: Symbol 0. Properties: I. For any ³ × ¾ matrix µ , µ − µ = 0. Example: 1 2 5 6 7 1 2 3 4 1 2 5 6 7 1 2 3 4 = 1 − 1 2 − 2 5 − 5 6 − 6 7 − 7 1 − 1 2 − 2 3 − 3 4 − 4 = 0 0 0 0 0 0 0 0 0 = 0.
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9/11/2015 11 II. For any matrix 5 and the zero matrix with the same number of rows and columns µ + 0 = µ. Example: 1 2 5 6 7 1 2 3 4 + 0 0 0 0 0 0 0 0 0 = 1 + 0 2 + 0 5 + 0 6 + 0 7 + 0 1 + 0 2 + 0 3 + 0 4 + 0 = 1 2 5 6 7 1 2 3 4 III. Multiplying any ° × ± matrix 5 with the ± × ² zero matrix 0, results in a ° × ² zero matrix. Zero matrix acts like the number 0 in the multiplication of numbers. EXAMPLE Example 1. If µ = 1 2 3 4 1 2 and 0 = 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 , calculate µ × 0 . Solution: µ × 0 = 1 2 3 4 1 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 = 0 0 0 0 0 0 0 0 0 0 = 0.
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9/11/2015 12 THE IDENTITY MATRIX Always a square matrix : ± × ± . The diagonal from top left to bottom right always has a value 1 and the rest of the matrix is zero’s. M = 1 0 0 1 0 0 0 0 1 . Examples: 2 × 2: M = 1 0 0 1 3 × 3: M = 1 0 0 0 1 0 0 0 1 EXAMPLES Example 2. For µ = 1 2 1 3 4 2 , and M = 1 0 0 0 1 0 0 0 1 , show that µM = µ . Solution: µM = 1 2 1 3 4 2 1 0 0 0 1 0 0 0 1 = µ 1 0 0 µ 0 1 0 µ 0 0 1 = 1 + 0 + 0 0 + 2 + 0 0 + 0 + 1 3 + 0 + 0 0 + 4 + 0 0 + 0 + 2 = 1 2 1 3 4 2 = µ.
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9/11/2015 13 Example 3 . For µ = 1 2 1 3 4 2 , and M = 1 0 0 1 , show that Mµ = µ. Solution: Mµ = 1 0 0 1 1 2 1 3 4 2 = M 1 3 M 2 4 M 1 2 = 1 + 0 2 + 0 1 + 0 0 + 3 0 + 4 0 + 2 = 1 2 1 3
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