CALCU-Problem 9. (10 Points) Find scalars a, b,c t

Solution by Steps

step 1

To make the matrix orthogonal, the columns must be orthonormal. This means each column vector must have a magnitude of 1 and be orthogonal to the other columns

step 2

First, we check the orthogonality of the given columns. The dot product of the first and second columns should be zero:

\left( \frac{2}{3}, \frac{2}{3}, \frac{1}{3} \right) \cdot \left( \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0 \right) = \frac{2}{3} \cdot \frac{1}{\sqrt{2}} + \frac{2}{3} \cdot -\frac{1}{\sqrt{2}} + \frac{1}{3} \cdot 0 = 0

This confirms that the first and second columns are orthogonal

step 3

Next, we need the third column to be orthogonal to both the first and second columns. We start with the first column:

\left( \frac{2}{3}, \frac{2}{3}, \frac{1}{3} \right) \cdot (a, b, c) = \frac{2}{3}a + \frac{2}{3}b + \frac{1}{3}c = 0

Multiplying through by 3, we get:

2a + 2b + c = 0 \quad \text{(Equation 1)}

step 4

Now, we check orthogonality with the second column:

\left( \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0 \right) \cdot (a, b, c) = \frac{1}{\sqrt{2}}a - \frac{1}{\sqrt{2}}b + 0 \cdot c = 0

Simplifying, we get:

a - b = 0 \quad \text{(Equation 2)}

step 5

From Equation 2, we have:

a = b

Substituting

a = b

into Equation 1, we get:

2a + 2a + c = 0 \implies 4a + c = 0 \implies c = -4a

step 6

Finally, we need the third column to have a magnitude of 1:

\sqrt{a^2 + b^2 + c^2} = 1

Substituting

b = a

and

c = -4a

, we get:

\sqrt{a^2 + a^2 + (-4a)^2} = 1 \implies \sqrt{a^2 + a^2 + 16a^2} = 1 \implies \sqrt{18a^2} = 1 \implies \sqrt{18}a = 1 \implies a = \frac{1}{\sqrt{18}} = \frac{1}{3\sqrt{2}}

Thus,

b = \frac{1}{3\sqrt{2}}

and

c = -\frac{4}{3\sqrt{2}}

Answer

The scalars are

a = \frac{1}{3\sqrt{2}}, b = \frac{1}{3\sqrt{2}}, c = -\frac{4}{3\sqrt{2}}

Key Concept

Orthogonality and normalization of matrix columns

Explanation

To make a matrix orthogonal, its columns must be orthonormal, meaning they must be orthogonal to each other and each column must have a magnitude of 1.

why a must be positive? can it be negative?

To ensure that the matrix is orthogonal, the scalars a, b, and c must be chosen carefully. In this case, a must be positive because it is related to the first column of the matrix. Since the first column is required to be orthogonal to the second column, the value of a must be positive to satisfy this condition. If a were negative, it would not result in an orthogonal matrix. Therefore, in this specific scenario, a must be positive.

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