Find the matrix

Algebra Level 3

P = [ cos ⁡ π 4 − sin ⁡ π 4 sin ⁡ π 4 cos ⁡ π 4 ] X = [ 1 2 1 2 ] P= \begin{bmatrix} \cos\cfrac { \pi }{ 4 } & -\sin\cfrac { \pi }{ 4 } \\ \sin\cfrac { \pi }{ 4 } & \cos\cfrac { \pi }{ 4 } \end{bmatrix} \qquad \qquad X = \left[ \begin{matrix} \cfrac { 1 }{ \sqrt { 2 } } \\ \cfrac { 1 }{ \sqrt { 2 } } \end{matrix} \right]

For matrices P P and X X as defined above, find P 3 X P^3X .

[ 0 1 ] \left[ \begin{matrix} 0 \\ 1 \end{matrix} \right] [ − 1 2 − 1 2 ] \left[\begin{matrix}-\cfrac { 1 }{ \sqrt { 2 }}\\-\cfrac{1}{\sqrt{2}}\end{matrix} \right] [ − 1 2 1 2 ] \left[\begin{matrix}\cfrac{ -1 }{\sqrt { 2 } }\\\cfrac{1}{\sqrt{ 2 } } \end{matrix} \right] [ − 1 0 ] \left[ \begin{matrix} -1 \\ 0 \end{matrix} \right]

Suyash Mittal by Suyash Mittal , 24, India

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2 solutions

Let P = [ cos ⁡ π 4 − sin ⁡ π 4 sin ⁡ π 4 cos ⁡ π 4 ] = [ 1 2 − 1 2 1 2 1 2 ] = 1 2 [ 1 − 1 1 1 ] = 1 2 ( I + A ) P = \begin{bmatrix} \cos \frac \pi 4 & - \sin \frac \pi 4 \\ \sin \frac \pi 4 & \cos \frac \pi 4 \end{bmatrix} = \begin{bmatrix} \frac 1{\sqrt 2} & - \frac 1{\sqrt 2} \\ \frac 1{\sqrt 2} & \frac 1{\sqrt 2} \end{bmatrix} = \frac 1{\sqrt 2} \begin{bmatrix} 1 & - 1 \\ 1 & 1 \end{bmatrix} = \frac 1{\sqrt 2} (I+A) , where A = [ 0 − 1 1 0 ] A = \begin{bmatrix} 0 & - 1 \\ 1 & 0 \end{bmatrix} . Then

Y = P 3 X = 1 2 2 ( I + A ) 3 [ 1 2 1 2 ] = 1 4 ( I + 3 A + 3 A 2 + A 3 ) [ 1 1 ] Note that A 2 = [ 0 − 1 1 0 ] [ 0 − 1 1 0 ] = [ − 1 0 0 − 1 ] = − I = 1 4 ( I + 3 A − 3 I − A ) [ 1 1 ] ⟹ A 3 = − I A = − A = 1 2 ( A − I ) [ 1 1 ] = 1 2 ( [ − 1 1 ] − [ 1 1 ] ) = 1 2 [ − 2 0 ] = [ − 1 0 ] \begin{aligned} Y & = P^3X \\ & = \frac 1{2\sqrt 2}(I+A)^3 \begin{bmatrix} \frac 1{\sqrt 2} \\ \frac 1{\sqrt 2} \end{bmatrix} \\ & = \frac 14\left(I+3A+3A^2+A^3\right) \begin{bmatrix} 1 \\ 1 \end{bmatrix} & \small \color{#3D99F6} \text{Note that }A^2 = \begin{bmatrix} 0 & - 1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} 0 & - 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} = - I \\ & = \frac 14\left(I+3A-3I-A\right) \begin{bmatrix} 1 \\ 1 \end{bmatrix} & \small \color{#3D99F6} \implies A^3 = -IA = - A \\ & = \frac 12\left(A-I\right) \begin{bmatrix} 1 \\ 1 \end{bmatrix} \\ & = \frac 12\left(\begin{bmatrix} -1 \\ 1 \end{bmatrix} - \begin{bmatrix} 1 \\ 1 \end{bmatrix} \right) \\ & = \frac 12\begin{bmatrix} -2 \\ 0 \end{bmatrix} \\ & = \boxed{\begin{bmatrix} -1 \\ 0 \end{bmatrix}} \end{aligned}

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Sarth Vitekar.
Feb 25, 2017

Rotation of axes. Initial angle 45degrees, final angle being 135degrees, now the determinant can be written in terms of 135 degrees(just like the given matrix) , P^3X can be easily found out.

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