Question

If $R(t) = \left[ {\begin{array}{*{20}{c}}{\cos t}&{\sin t}\\{ - \sin t}&{\cos t}\end{array}} \right] then R(s) \cdot R(t)$ =
A. $R(s) + R(t)$
B. $R(st)$
C. $R(s + t)$
D. None of these

Answer 1

Hint: If there are exactly as many columns in matrix A as there are rows in matrix B, the two matrices are said to be compatible. Accordingly, we can state that matrices A and B are compatible if A is a matrix of order m $ \times n $ and B is a matrix of order$ n \times p$ and in our case, we have a dimension of 3 $ \times 2 $ and $ 3 \times 1$ to be multiplied and $2 \times 1$ with single element and both should be added in order to get the desired answer.

Formula Used: To multiply matrix, we use the formula
${C_{xy}} = {A_{x1}}{B_{y1}} + \ldots + {A_{xb}}{B_{by}} = \sum\limits_{k = 1}^b {{A_{xk}}} {B_{ky}}$

Complete step by step solution: We are provided in the given information that,
$R(t) = \left[ {\begin{array}{*{20}{c}}{\cos t}&{\sin t}\\{ - \sin t}&{\cos t}\end{array}} \right]$
And we are to determine the value of $R(s) \cdot R(t)$
We are given the matrix of $R(t)$
From that, we can write the matrix of $R(s)$
This can be done by replacing the element t as s because only the value of function changes
Therefore, we get
$R(s) = \left[ {\begin{array}{*{20}{c}}{\cos s}&{\sin s}\\{ - \sin s}&{\cos s}\end{array}} \right]$
Now, let us find the value of $ R(s) \cdot R(t)$
We have,
$R(s)R(t) = \left[ {\begin{array}{*{20}{c}}{\cos s}&{\sin s}\\{ - \sin s}&{\cos s}\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}{\cos t}&{\sin t}\\{ - \sin t}&{\cos t}\end{array}} \right]$
For matrix multiplication, the elements in the first matrix's column and rows are multiplied by one another, and then the results are added.
$\left[ {\begin{array}{*{20}{r}}{(\cos (s)) \cdot (\cos (t)) + (\sin (s)) \cdot ( - \sin (t))}&{(\cos (s)) \cdot (\sin (t)) + (\sin (s)) \cdot (\cos (t))}\\{( - \sin (s)) \cdot (\cos (t)) + (\cos (s)) \cdot ( - \sin (t))}&{( - \sin (s)) \cdot (\sin (t)) + (\cos (s)) \cdot (\cos (t))}\end{array}} \right]$
Now, we have to solve the each terms in the determinant above by addition/subtraction, we get
$ = \left[ {\begin{array}{*{20}{c}}{\cos (s + t)}&{\sin (s + t)}\\{ - \sin (s + t)}&{\cos (s + t)}\end{array}} \right]$
From the above resultant expression, we can come to a conclusion that $ \left[ {\begin{array}{*{20}{c}}{\cos (s + t)}&{\sin (s + t)}\\{ - \sin (s + t)}&{\cos (s + t)}\end{array}} \right] = R(s + t)$
Therefore, If $R(t) = \left[ {\begin{array}{*{20}{c}}{\cos t}&{\sin t}\\{ - \sin t}&{\cos t}\end{array}} \right] then R(s) \cdot R(t) = R(s + t)$

Option ‘C’ is correct

Note: Students should be aware that in order to do matrix multiplication, the columns of the first matrix must match the rows of the second matrix. Because the resulting matrix has the same number of columns and rows as the second matrix and its number of columns corresponds to the number of rows in the first matrix, one should exercise caution when solving the matrix multiplication.