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Chapter 4: Problem 333

Construct an orthogonal matrix using the skew-symmetric matrix (a) \(\left|\begin{array}{cl}-(3 / 5) & -(4 / 5) \\ (4 / 5) & -(3 / 5)\end{array}\right|\) (c) \(\left|\begin{array}{ll}(4 / 5) & (3 / 5) \\ (3 / 5) & (4 / 5)\end{array}\right|\) $$ \text { (d) }\left|\begin{array}{ll} -(4 / 5) & -(3 / 5) \\ -(3 / 5) & -(4 / 5) \end{array}\right| $$

Short Answer

Expert verified
The orthogonal matrices constructed using the given skew-symmetric matrices are: (a) \(B =\) \(\left|\begin{array}{cc} \frac{3}{5} & \frac{4}{5} \\ -\frac{4}{5} & \frac{3}{5} \end{array}\right|\) (c) \(B =\) \(\left|\begin{array}{cc} \frac{4}{5} & -\frac{3}{5} \\ -\frac{3}{5} & -\frac{4}{5} \end{array}\right|\) (d) \(B =\) \(\left|\begin{array}{cc} \frac{4}{5} & \frac{-3}{5} \\ \frac{3}{5} & \frac{4}{5} \end{array}\right|\)

Step by step solution

01

(Step 1): Checking whether the given matrices are skew-symmetric

(For each of the given matrices, transpose it and check if it is equal to the negative of the original matrix to confirm they are skew-symmetric.)
02

(Step 2): Constructing the orthogonal matrix using skew-symmetric matrices

(Since the matrix A satisfies the skew-symmetric condition, we can construct the orthogonal matrix B by applying the condition \(B^T B = I\). We have \(B^T = -A\), so \((-A) B = I\). We can get the orthogonal matrix B by multiplying A and the inverse of \(-A\).)
03

(Step 3): Calculate the inverse of \(-A\)

(To calculate the inverse of a 2x2 matrix, we swap the elements in the main diagonal, change the signs of the other elements, and divide by the determinant. Apply this method to \(-A\) for each skew-symmetric matrix.)
04

(Step 4): Multiply A with the inverse of \(-A\)

(Multiply each skew-symmetric matrix A by the inverse of \(-A\) calculated in step 3 to obtain the orthogonal matrix B.) Applying these steps, we have the following orthogonal matrices: (a) \(B =\) \(\left|\begin{array}{cc} \frac{3}{5} & \frac{4}{5} \\ -\frac{4}{5} & \frac{3}{5} \end{array}\right|\) (c) \(B =\) \(\left|\begin{array}{cc} \frac{4}{5} & -\frac{3}{5} \\ -\frac{3}{5} & -\frac{4}{5} \end{array}\right|\) (d) \(B =\) \(\left|\begin{array}{cc} \frac{4}{5} & \frac{-3}{5} \\ \frac{3}{5} & \frac{4}{5} \end{array}\right|\)

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