Question

If $\alpha ,\beta ,\gamma$ are three real numbers and A = \left[ {\begin{array}{*{20}{c}} 1&{\cos \left( {\alpha - \beta } \right)}&{\cos \left( {\alpha - \gamma } \right)} \\\ {\cos \left( {\beta - \alpha } \right)}&1&{\cos \left( {\beta - \gamma } \right)} \\\ {\cos \left( {\gamma - \alpha } \right)}&{\cos \left( {\gamma - \beta } \right)}&1 \end{array}} \right], then which of the following is not true.
a. $A$ is singular
b. $A$ is symmetric
c. $A$ is orthogonal
d. $A$ is not invertible

Answer 1

We will rearrange the elements of the matrix by using the property $\cos \left( { - x} \right) = \cos x$. Then, compare the elements of the matrix with that of a symmetric matrix. Here, we will see if ${a_{12}} = {a_{21}}$, ${a_{23}} = {a_{32}}$ and ${a_{13}} = {a_{31}}$. If all such elements are equal, then the matrix is a symmetric matrix.

Complete step by step solution:
We are given a matrix A = \left[ {\begin{array}{*{20}{c}} 1&{\cos \left( {\alpha - \beta } \right)}&{\cos \left( {\alpha - \gamma } \right)} \\\ {\cos \left( {\beta - \alpha } \right)}&1&{\cos \left( {\beta - \gamma } \right)} \\\ {\cos \left( {\gamma - \alpha } \right)}&{\cos \left( {\gamma - \beta } \right)}&1 \end{array}} \right]
Here, we know that $\cos \left( { - x} \right) = \cos x$
We can write the elements of the matrix as

1&{\cos \left( { - \left( {\beta - \alpha } \right)} \right)}&{\cos \left( { - \left( {\gamma - \alpha } \right)} \right)} \\\ {\cos \left( {\beta - \alpha } \right)}&1&{\cos \left( { - \left( {\gamma - \beta } \right)} \right)} \\\ {\cos \left( {\gamma - \alpha } \right)}&{\cos \left( {\gamma - \beta } \right)}&1 \end{array}} \right]$$ Which is equal to $$A = \left[ {\begin{array}{*{20}{c}} 1&{\cos \left( {\beta - \alpha } \right)}&{\cos \left( {\gamma - \alpha } \right)} \\\ {\cos \left( {\beta - \alpha } \right)}&1&{\cos \left( {\gamma - \beta } \right)} \\\ {\cos \left( {\gamma - \alpha } \right)}&{\cos \left( {\gamma - \beta } \right)}&1 \end{array}} \right]$$ Also, it is known that the matrix is symmetric when $\left[ {{a_{ij}}} \right] = \left[ {{a_{ji}}} \right]$ Then, we can say that the given matrix is a symmetric matrix. **Hence, option b is correct.** **Note:** A matrix is said to be invertible if the determinant is non zero and the matrix is said to be singular if the determinant is zero. A matrix is said to be symmetric matrix when $\left[ {{a_{ij}}} \right] = \left[ {{a_{ji}}} \right]$ for every element of the matrix.