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Solve the following equations using Cramer's rule (determinant) a) x-3y=-3, 2x+5y=8 b) x−2y−z=3, 3x−6y−5z=3, 2X−y+z=0 c) x+2y+3z=17, 3x+2y+z=11, x-5y+3z=-5

Mathematics

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Here's the solution for question 9:

To determine if vectors are orthogonal, we check if their dot product is zero. If they are orthogonal, we then check if they are orthonormal by verifying if both vectors are unit vectors (i.e., their magnitudes are 1).

i) Given vectors a=(4,2,−1)a = (4, 2, -1) and b=(1,−3,−2)b = (1, -3, -2).

  1. Calculate the dot product a⋅ba \cdot b.

    a⋅b=(4)(1)+(2)(−3)+(−1)(−2)a \cdot b = (4)(1) + (2)(-3) + (-1)(-2) a⋅b=4−6+2a \cdot b = 4 - 6 + 2 a⋅b=0a \cdot b = 0
  2. Determine if the vectors are orthogonal.

    Since a⋅b=0a \cdot b = 0, the vectors aa and bb are orthogonal.

  3. Calculate the magnitudes of aa and bb.

    ∣∣a∣∣=42+22+(−1)2=16+4+1=21||a|| = \sqrt{4^2 + 2^2 + (-1)^2} = \sqrt{16 + 4 + 1} = \sqrt{21} ∣∣b∣∣=12+(−3)2+(−2)2=1+9+4=14||b|| = \sqrt{1^2 + (-3)^2 + (-2)^2} = \sqrt{1 + 9 + 4} = \sqrt{14}
  4. Determine if the vectors are orthonormal.

    Since ∣∣a∣∣=21≠1||a|| = \sqrt{21} \neq 1 and ∣∣b∣∣=14≠1||b|| = \sqrt{14} \neq 1, the vectors are not unit vectors.
    Therefore, the vectors aa and bb are not orthonormal.

    ii) Given vectors a=(5,−1,3)a = (5, -1, 3) and b=(4,7,−2)b = (4, 7, -2).

  5. Calculate the dot product a⋅ba \cdot b.

    a⋅b=(5)(4)+(−1)(7)+(3)(−2)a \cdot b = (5)(4) + (-1)(7) + (3)(-2) a⋅b=20−7−6a \cdot b = 20 - 7 - 6 a⋅b=13−6a \cdot b = 13 - 6 a⋅b=7a \cdot b = 7
  6. Determine if the vectors are orthogonal.

    Since a⋅b=7≠0a \cdot b = 7 \neq 0, the vectors aa and bb are not orthogonal.

  7. Determine if the vectors are orthonormal.

    Since the vectors are not orthogonal, they cannot be orthonormal.

    Final Answer:

    i) The vectors a=(4,2,−1)a = (4, 2, -1) and b=(1,−3,−2)b = (1, -3, -2) are orthogonal but not orthonormal.

    ii) The vectors a=(5,−1,3)a = (5, -1, 3) and b=(4,7,−2)b = (4, 7, -2) are not orthogonal and therefore not orthonormal.

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