Question

$A = \begin{bmatrix} 0 & 2b & c \\ a & b & -c \\ a & -b & c \end{bmatrix}$ is orthogonal matrix, then 36|abc| =

• A. 4
• B. 6
• C. 9
• D. 1

Answer 1

Answer: (B)

6

Answer 2

The problem states that the given matrix A is an orthogonal matrix. An orthogonal matrix A satisfies the condition $A A^T = I$ (where I is the identity matrix) and its determinant, $\det(A)$, is either +1 or -1.

Given matrix: $A = \begin{bmatrix} 0 & 2b & c \\ a & b & -c \\ a & -b & c \end{bmatrix}$

We will use the property that for an orthogonal matrix, $|\det(A)| = 1$.

First, calculate the determinant of matrix A: $\det(A) = 0 \cdot \det \begin{pmatrix} b & -c \\ -b & c \end{pmatrix} - 2b \cdot \det \begin{pmatrix} a & -c \\ a & c \end{pmatrix} + c \cdot \det \begin{pmatrix} a & b \\ a & -b \end{pmatrix}$

$\det(A) = 0 - 2b(a \cdot c - (-c) \cdot a) + c(a \cdot (-b) - b \cdot a)$

$\det(A) = -2b(ac + ac) + c(-ab - ab)$

$\det(A) = -2b(2ac) + c(-2ab)$

$\det(A) = -4abc - 2abc$

$\det(A) = -6abc$

Since A is an orthogonal matrix, we know that $|\det(A)| = 1$. Therefore, $|-6abc| = 1$. Using the property $|xy| = |x||y|$, we get: $|-6| \cdot |abc| = 1$

$6 \cdot |abc| = 1$

$|abc| = \frac{1}{6}$

The question asks for the value of $36|abc|$.

$36|abc| = 36 \cdot \left(\frac{1}{6}\right)$

$36|abc| = 6$