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Addition and Subtraction of a Matrix

If A = begin ...

Question

If $A = [\begin{matrix} 0 & - t a n \frac{α}{2} t a n \frac{α}{2} & 0 \end{matrix}]$ and I is the identity matrix of order 2, show that I + A $= (I - A) [\begin{matrix} c o s α & - s i n α s i n α & c o s α \end{matrix}]$

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Solution

Here, $A = [\begin{matrix} 0 & - t t & 0 \end{matrix}]$ , where $t = t a n (\frac{α}{2})$
Now, $c o s α = \frac{1 - t a n^{2} (\frac{α}{2})}{1 + t a n^{2} (\frac{α}{2})} = \frac{1 - t^{2}}{1 + t^{2}} a n d s i n α = \frac{2 t a n (\frac{α}{2})}{1 + t a n^{2} (\frac{α}{2})} = \frac{2 t}{1 + t^{2}}$
$R H S = (I - A) [\begin{matrix} c o s α & - s i n α s i n α & c o s α \end{matrix}] = [\begin{matrix} (\begin{matrix} 1 & 0 0 & 1 \end{matrix}) - (\begin{matrix} 0 & - t + t & 0 \end{matrix}) \end{matrix}] ⎡ ⎢ ⎣ \begin{matrix} \frac{1 - t^{2}}{1 + t^{2}} & \frac{- 2 t}{1 + t^{2}} \frac{2 t}{1 + t^{2}} & \frac{1 - t^{2}}{1 + t^{2}} \end{matrix} ⎤ ⎥ ⎦$

$= [\begin{matrix} 1 & t - t & 1 \end{matrix}] ⎡ ⎢ ⎣ \begin{matrix} \frac{1 - t^{2}}{1 + t^{2}} & \frac{- 2 t}{1 + t^{2}} \frac{2 t}{1 +^{2}} & \frac{1 - t^{2}}{1 + t^{2}} \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎢ ⎣ \begin{matrix} \frac{1 - t^{2} + 2 t^{2}}{1 + t^{2}} & \frac{- 2 t + t (1 - t^{2})}{1 + t^{2}} \frac{- t (1 - t^{2}) + 2 t}{1 + t^{2}} & \frac{2 t^{2} + 1 - t^{2}}{1 + t^{2}} \end{matrix} ⎤ ⎥ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} \frac{1 + t^{2}}{1 + t^{2}} & \frac{- 2 t + t - t^{3}}{1 + t^{2}} \frac{- t + t^{3} + 2 t}{1 +^{2}} & \frac{2 t^{2} + 1 - t^{2}}{1 + t^{2}} \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎢ ⎣ \begin{matrix} \frac{1 + t^{2}}{1 + t^{2}} & \frac{- t (1 + t^{2})}{1 + t^{2}} \frac{t (1 + t^{2})}{1 + t^{2}} & \frac{1 + t^{2}}{1 + t^{2}} \end{matrix} ⎤ ⎥ ⎥ ⎦ = [\begin{matrix} 1 & - t t & 1 \end{matrix}] L H S = [\begin{matrix} 1 & 0 0 & 1 \end{matrix}] + [\begin{matrix} 0 & - t t & 0 \end{matrix}] = [\begin{matrix} 0 + 1 & - t + 0 t + 0 & 0 + 1 \end{matrix}] = [\begin{matrix} 1 & - t t & 1 \end{matrix}] = R H S$

Putting the value of t on both sides, we get
$[\begin{matrix} 1 & - t a n (\frac{α}{2}) t a n (\frac{α}{2}) & 1 \end{matrix}] = [\begin{matrix} 1 & - t a n (\frac{α}{2}) t a n (\frac{α}{2}) & 1 \end{matrix}]$
$∴ L H S = R H S$ . hence proved.

Alternale method Here, $A = [\begin{matrix} 0 & - t a n (\frac{α}{2}) t a n (\frac{α}{2}) & 0 \end{matrix}]$
$∴ I + A = [\begin{matrix} 1 & 0 0 & 1 \end{matrix}] + [\begin{matrix} 0 & - t a n (\frac{α}{2}) t a n (\frac{α}{2}) & 0 \end{matrix}] = [\begin{matrix} 1 & - t a n (\frac{α}{2}) t a n (\frac{α}{2}) & 0 \end{matrix}]$
and $I - A = [\begin{matrix} 1 & 0 0 & 1 \end{matrix}] - [\begin{matrix} 0 & - t a n (\frac{α}{2}) t a n (\frac{α}{2}) & 0 \end{matrix}] = [\begin{matrix} 1 & t a n (\frac{α}{2}) - t a n (\frac{α}{2}) & 1 \end{matrix}]$

$∴ (I - A) [\begin{matrix} c o s α & - s i n α s i n α & c o s α \end{matrix}] = [\begin{matrix} 1 & t a n (\frac{α}{2}) - t a n (\frac{α}{2}) & 1 \end{matrix}] [\begin{matrix} c o s α & - s i n α s i n α & c o s α \end{matrix}] = [\begin{matrix} c o s α + t a n (\frac{α}{2}) s i n α & - s i n α + t a n (\frac{α}{2}) c o s α - t a n (\frac{α}{2}) c o s α + s i n α & t a n (\frac{α}{2}) s i n α + c o s α \end{matrix}] = ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} \frac{c o s α c o s (\frac{α}{2}) + s i n (\frac{α}{2}) s i n θ}{c o s (\frac{α}{2})} & \frac{- s i n α c o s (\frac{α}{2}) + s i n (\frac{α}{2}) c o s α}{c o s (\frac{α}{2})} \frac{- s i n (\frac{α}{2}) c o s α + c o s (\frac{α}{2}) s i n α}{c o s (\frac{α}{2})} & \frac{s i n (\frac{α}{2}) s i n α + c o s (\frac{α}{2}) c o s α}{c o s (\frac{α}{2})} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦$

$= ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} \frac{c o s (α - \frac{α}{2})}{c o s (\frac{α}{2})} & \frac{- s i n (α - \frac{α}{2})}{c o s (\frac{α}{2})} \frac{s i n (α - \frac{α}{2})}{c o s (\frac{α}{2})} & \frac{c o s (α - \frac{α}{2})}{c o s (\frac{α}{2})} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} \frac{c o s (\frac{α}{2})}{c o s (\frac{α}{2})} & \frac{- s i n (\frac{α}{2})}{c o s (\frac{α}{2})} \frac{s i n (\frac{α}{2})}{c o s (\frac{α}{2})} & \frac{c o s (\frac{α}{2})}{c o s (\frac{α}{2})} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ (\begin{matrix} ∵ c o s (A - B) = c o s A c o s B + s i n A s i n B s i n (A - B) = s i n A c o s B - c o s A s i n B \end{matrix}) = [\begin{matrix} 1 & - t a n (\frac{α}{2}) t a n (\frac{α}{2}) & 1 \end{matrix}] = I + A$
Thus, $I + A = (I - A) [\begin{matrix} c o s α & - s i n α s i n α & c o s α \end{matrix}]$ Hence proved.

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