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Home  >>  CBSE XII  >>  Math  >>  Matrices
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Simplify $cos\theta\begin{bmatrix} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{bmatrix}$ + $sin\theta\begin{bmatrix} sin\theta & -cos\theta \\ cos\theta & sin\theta \end{bmatrix} $

$\begin{array}{1 1} \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \\ \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \\ \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\ \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\end{array} $

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Toolbox:
  • The scalar multiplication $cA$ of a matrix $A$ and a number $c$ (also called a scalar in the parlance of abstract algebra) is given by multiplying every entry of $A$ by $c$.
  • The sum $A+B$ of two $m$-by-$n$ matrices $A$ and $B$ is calculated entrywise: $(A + B)_{i,j} = A_{i,j} + B_{i,j}$ where 1 ≤ i ≤ m and 1 ≤ j ≤ n.
  • $\sin^2x+\cos^2x = 1$
Let $A = cos\theta\begin{bmatrix} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{bmatrix} = \begin{bmatrix} cos^2\theta & \cos\theta sin\theta \\ -\cos\theta sin\theta & cos^2\theta \end{bmatrix}$
Let $B = sin\theta\begin{bmatrix} sin\theta & -cos\theta \\ cos\theta & sin\theta \end{bmatrix} = \begin{bmatrix} sin^2\theta & -\sin\theta cos\theta \\ \sin\theta cos\theta & sin^2\theta \end{bmatrix} $
$A+B = \begin{bmatrix} cos^2\theta & \cos\theta sin\theta \\ -\cos\theta sin\theta & cos^2\theta \end{bmatrix} + \begin{bmatrix} sin^2\theta & -\sin\theta cos\theta \\ \sin\theta cos\theta & sin^2\theta \end{bmatrix} $
$A+B = \begin{bmatrix} sin^2\theta+\cos^2\theta & \cos\theta\sin\theta-\sin\theta cos\theta \\ \sin\theta cos\theta-\cos \theta\sin\theta & sin^2\theta+\cos^2\theta \end{bmatrix} $
Given that $\sin^2x+\cos^2x = 1$, $A+B = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
answered Feb 27, 2013 by balaji.thirumalai
 

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