Question

If \(A= \begin{bmatrix}1&√3&0 \\-√3&1&0 \\ 0&0&2 \end{bmatrix}\) and \(B=\begin{bmatrix}√3&1&0 \\-1&√3&0 \\ 0&0&2 \end{bmatrix}\) then AB is equal to :

• A. \(4I\)
• B. \(-4I\)
• C. \(\begin{bmatrix} 0&0&4 \\ 0&4&0 \\ -4 &0 &0 \end{bmatrix}\)
• D. \(\begin{bmatrix} 0&4&0\\ -4&0&0 \\ 0&0 &4 \end{bmatrix}\)

Answer 1

The Correct Option is D

To find the product \(AB\), where \(A= \begin{bmatrix}1&\sqrt{3}&0 \\-\sqrt{3}&1&0 \\ 0&0&2 \end{bmatrix}\) and \(B=\begin{bmatrix}\sqrt{3}&1&0 \\-1&\sqrt{3}&0 \\ 0&0&2 \end{bmatrix}\), we will perform matrix multiplication. 
The resulting matrix \(C = AB\) is computed as follows:

For the element \(C_{11}\):

\[ C_{11} = (1)(\sqrt{3}) + (\sqrt{3})(-1) + (0)(0) = \sqrt{3} - \sqrt{3} = 0 \]

For the element \(C_{12}\):

\[ C_{12} = (1)(1) + (\sqrt{3})(\sqrt{3}) + (0)(0) = 1 + 3 = 4 \]

For the element \(C_{13}\):

\[ C_{13} = (1)(0) + (\sqrt{3})(0) + (0)(2) = 0 \]

For the element \(C_{21}\):

\[ C_{21} = (-\sqrt{3})(\sqrt{3}) + (1)(-1) + (0)(0) = -3 - 1 = -4 \]

For the element \(C_{22}\):

\[ C_{22} = (-\sqrt{3})(1) + (1)(\sqrt{3}) + (0)(0) = -\sqrt{3} + \sqrt{3} = 0 \]

For the element \(C_{23}\):

\[ C_{23} = (-\sqrt{3})(0) + (1)(0) + (0)(2) = 0 \]

For the element \(C_{31}\):

\[ C_{31} = (0)(\sqrt{3}) + (0)(-1) + (2)(0) = 0 \]

For the element \(C_{32}\):

\[ C_{32} = (0)(1) + (0)(\sqrt{3}) + (2)(0) = 0 \]

For the element \(C_{33}\):

\[ C_{33} = (0)(0) + (0)(0) + (2)(2) = 4 \]

Hence, the product matrix \(C = AB\) is:

0	4	0
-4	0	0
0	0	4