Exercise 5.2
Question 15
साबित करे कि दो आव्यूह (Prove that the product of two matrices)
$\left[\begin{array}{cc}\cos ^{2} \theta & \cos \theta \cdot \sin \theta \\ \cos \theta \cdot \sin \theta & \sin ^{2} \theta\end{array}\right]$ तथा (and) $\left[\begin{array}{cc}\cos ^{2} \phi & \cos \phi \cdot \sin \phi \\ \cos \phi \cdot \sin \phi & \sin ^{2} \phi\end{array}\right]$ का गुणनफल एक शून्य आव्यूह है यदि θ और ɸ का अन्तप $\dfrac{\pi}{2}$ का विषम अपवर्त्य है । (is a zero matrix when θ and ɸ differ by an odd multiple of $\dfrac{\pi}{2}$)
Sol :
$\theta-\phi=(2 n-1) \frac{\pi}{2} \quad \ldots n \in 2$
$\left[\begin{array}{cc}\cos ^{2} \theta & \cos \theta sin\theta \\ \cos \theta \sin \theta & si n^{2} \theta\end{array}\right]\left[\begin{array}{cc}\cos ^{2} \phi & \cos \phi \sin \phi \\ \cos \phi \sin \phi & \sin ^{2} \phi\end{array}\right]$
$=\left[\begin{array}{ccc}\cos ^{2} \theta \cos ^{2} \phi+\cos \theta \sin \theta & \cos ^{2} \theta \cos \phi \sin \phi+\cos \theta sin \theta \sin ^{2} \phi \\ \cos \theta \sin \theta \cos ^{2} \phi+\sin ^{2} \theta \cos \phi+sin \phi & \cos \theta \sin \theta \cos \phi \sin \phi+\sin ^{2} \theta \sin ^{2} \phi\end{array}\right]$
$\left[\begin{array}{ll}\cos \theta \cos \phi(\cos \theta \cos \phi+\sin \theta \sin \phi) & \cos \theta \sin \phi(\cos \theta \cos \phi+\sin \theta \sin \phi) \\ \sin \theta \cos \phi(\cos \theta \cos \phi+\sin \theta sin \phi) & \sin \theta \sin \phi\left(\cos \theta \cos \phi+\sin \theta sin \phi\right)\end{array}\right]$
$=\left[\begin{array}{ll}\cos \theta \cos \phi \cos (\theta-\phi) & \cos \theta \sin \phi \cos (\theta-\phi) \\ \sin \theta \cos \phi \cos (\theta-\phi) & \sin \theta \sin \phi \cos (\theta-\phi)\end{array}\right]$
$=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]=0$
Question 18
यदि (if) $A=\left[\begin{array}{rrr}-1 & 1 & -1 \\ 3 & -3 & 3 \\ 5 & -5 & 5\end{array}\right], B=\left[\begin{array}{rrr}0 & 4 & 3 \\ 1 & -3 & -3 \\ -1 & 4 & 4\end{array}\right]$
A2.B2
निकाले (compute
A2.B2)
Sol :
$A^{2}=A \cdot A=\left[\begin{array}{ccc}-1 & 1 & -1 \\ 3 & -3 & 3 \\ 5 & -5 & 5\end{array}\right]\left[\begin{array}{ccc}-1 & 1 & -1 \\ 3 & -3 & 3 \\ 5 & -5 & 5\end{array}\right]$
$=\left[\begin{array}{ccc}1+3-5 & -1-3+5 & 1+3-5 \\ -3-9+15 & 3+9-15 & -3-9+15 \\ -5-15+25 & 5+15-25 & -5-15+25\end{array}\right]$
$A^{2}=\left[\begin{array}{ccc}-1 & 1 & -1 \\ 3 & -3 & 3 \\ 5 & -5 & 5\end{array}\right]$
$B^{2}=B B=$
$A^{2} \cdot B^{2}=$
Question 19
यदि (If) $A=\begin{bmatrix}2&3&4\\1&2&3\\-1&1&2\end{bmatrix}$ , $B=\begin{bmatrix}1&3&0\\-1&2&1\\0&0&2\end{bmatrix}$ , AB तथा BA निकालें तथा दिखाएँ कि (find AB and BA and show that )AB≠BA
Sol :
Question 20
दिखाएँ कि (show that) $\left(\left[\begin{array}{ccc}1 & 0 & \omega^{2} \\ \omega & \omega^{2} & 1 \\ \omega^{2} & 1 & \omega\end{array}\right]+\left[\begin{array}{ccc}\omega & \omega^{2} & 1 \\ \omega^{2} & 1 & \omega \\ \omega & \omega^{2} & 1\end{array}\right]\right)\left[\begin{array}{c}1 \\ \omega \\ \omega^{2}\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$
Sol :
L.H.S
$=\left[\begin{array}{ccc}1+\omega & \omega+\omega^{2} & \omega^{2}+1 \\ \omega+\omega^{2} & \omega^{2}+1 & 1+\omega \\ \omega^{2}+\omega & 1+\omega^{2} & \omega+1\end{array}\right]\left[\begin{array}{c}1 \\ \omega \\ \omega^{2}\end{array}\right]$
$=\left[\begin{array}{ccc}-\omega^{2} & -1 & -\omega \\ -1 & -\omega & -\omega^{2} \\ -1 & -\omega & -\omega^{2}\end{array}\right]\left[\begin{array}{c}1 \\ \omega \\ \omega^{2}\end{array}\right]$
$=\left[\begin{array}{cc}-\omega^{2}-\omega-\omega^{3} \\ -1-\omega^{2}-\omega^{4} \\ -1-\omega^{2}-\omega^{4}\end{array}\right]=\left[\begin{array}{cc}-\omega^{2}-\omega-1 \\ -1 -\omega^{2}-\omega \\ -1-\omega^{2}-\omega\end{array}\right]$
$=\left[\begin{array}{c}-\left(\omega^{2}+\omega+1\right) \\ -\left(1+w^{2}+w\right) \\ -\left(1+w^{2}+w\right)\end{array}\right]$
$=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$
यदि (If) $A=\begin{bmatrix}1&0&-2\\3&-1&0\\-2&1&1\end{bmatrix}$ , $B=\begin{bmatrix}0&5&-4\\-2&1&3\\-1&0&2\end{bmatrix}$ तथा (and) $C=\begin{bmatrix}1&5&2\\-1&1&0\\0&-1&1\end{bmatrix}$ सत्यापित करें कि (verify that) A(B-C)=(AB-AC)
Sol :
Very helpfull
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