Give an example of matrices A,
Question: Give an example of matrices A, B and C such that AB = AC, where A is non-zero matrix, but B C. Solution: Let $A=\left[\begin{array}{ll}1 0 \\ 0 0\end{array}\right], B=\left[\begin{array}{ll}1 1 \\ 1 0\end{array}\right]$ and $C=\left[\begin{array}{ll}1 1 \\ 1 2\end{array}\right] \quad[\because B \neq C]$ $A B=\left[\begin{array}{ll}1 0 \\ 0 0\end{array}\right]\left[\begin{array}{ll}1 1 \\ 1 0\end{array}\right]=\left[\begin{array}{ll}1 1 \\ 0 0\end{array}\right]$ $\ldots$ (i) and $A C=\l...
Read More →Mark the tick against the correct answer in the following:
Question: Mark the tick against the correct answer in the following: The value of $\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)$ is A. $\frac{2 \pi}{3}$ B. $\frac{5 \pi}{3}$ C. $\frac{\pi}{3}$ D. none of these Solution: To Find: The value of $\sin ^{-1}\left(\sin \left(\frac{2 \pi}{3}\right)\right)$ Now, let $x=\sin ^{-1}\left(\sin \left(\frac{2 \pi}{3}\right)\right)$ $\Rightarrow \sin x=\sin \left(\frac{2 \pi}{3}\right)$ Here range of principle value of sine is $\left[-\frac{\pi}{2}, \frac{\pi}{...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{x^{2}+1}{x^{4}+7 x^{2}+1} d x$ Solution: re-writing the given equation as $\int \frac{1+\frac{1}{x^{2}}}{x^{2}+7+\frac{1}{x^{2}}} d x$ $\int \frac{1+\frac{1}{x^{2}}}{\left(x-\frac{1}{x}\right)^{2}+9} d x$ Assume $t=x-\frac{1}{x}$ $\mathrm{dt}=\left(1+\frac{1}{\mathrm{x}^{2}}\right) \mathrm{d} \mathrm{x}$ $\int \frac{d t}{(t)^{2}+9}$ Using identity $\int \frac{1}{x^{2}+1} d x=\arctan (x)$ $\frac{1}{3} \arctan \left(\frac{t}{3}\right)+c$ Subst...
Read More →If A = [3 5], B = [7 3], then find a non-zero matrix
Question: If A = [3 5], B = [7 3], then find a non-zero matrix C such that AB = AC. Solution: Given, A = [3 5]12and B = [7 3]12 For AC = BC We have order of C = 2 x n For $n=1$, let $C=\left[\begin{array}{l}x \\ y\end{array}\right]$ $\therefore A C=\left[\begin{array}{ll}3 5\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=[3 x+5 y]$ And $B C=\left[\begin{array}{ll}7 3\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=[7 x+3 y]$ For $A C=B C$, $[3 x+5 y]=[7 x+3 y]$ $...
Read More →Mark the tick against the correct answer in the following:
Question: Mark the tick against the correct answer in the following: The principal value of $\cot ^{-1}(-\sqrt{3})$ is A. $\frac{2 \pi}{6}$ B. $\frac{\pi}{6}$ C. $\frac{7 \pi}{6}$ D. $\frac{5 \pi}{6}$ Solution: To Find: The Principle value of $\cot ^{-1}(-\sqrt{3})$ Let the principle value be given by $x$ Now, let $x=\cot ^{-1}(-\sqrt{3})$ $\Rightarrow \cot x=-\sqrt{3}$ $\Rightarrow \cot x=-\cot \left(\frac{\pi}{6}\right)\left(\because \cot \left(\frac{\pi}{6}\right)=\sqrt{3}\right)$ $\Rightarro...
Read More →Mark the tick against the correct answer in the following:
Question: Mark the tick against the correct answer in the following: The principal value of $\operatorname{cosec}^{-1}(-\sqrt{2})$ is A. $\frac{-\pi}{4}$ B. $\frac{3 \pi}{4}$ C. $\frac{5 \pi}{4}$ D. none of these Solution: To Find: The Principle value of $\operatorname{cosec}^{-1}(-\sqrt{2})$ Let the principle value be given by $x$ Now, let $x=\operatorname{cosec}^{-1}(-\sqrt{2})$ $\Rightarrow \operatorname{cosec} x=-\sqrt{2}$ $\Rightarrow \operatorname{cosec} x=-\operatorname{cosec}\left(\frac{...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{x^{2}-1}{x^{4}+1} d x$ Solution: re-writing the given equation as $\int \frac{1-\frac{1}{x^{2}}}{x^{2}-\frac{1}{x^{2}}} d x$ $\int \frac{1-\frac{1}{x^{2}}}{\left(x+\frac{1}{x}\right)^{2}-2} d x$ Assume $t=x+\frac{1}{x}$ $\mathrm{dt}=\left(1-\frac{1}{\mathrm{x}^{2}}\right) \mathrm{dx}$ $\int \frac{d t}{t^{2}-2}$ Using identity $\int \frac{\mathrm{dz}}{(\mathrm{z})^{2}-1}=\frac{1}{2} \log \left|\frac{\mathrm{z}-1}{\mathrm{z}+1}\right|+\mathrm{...
Read More →If X and Y are 2 x 2 matrices,
Question: If X and Y are 2 x 2 matrices, then solve the following matrix equations for X and Y $2 \mathrm{X}+3 \mathrm{Y}=\left[\begin{array}{ll}2 3 \\ 4 0\end{array}\right], 3 \mathrm{X}+2 \mathrm{Y}=\left[\begin{array}{cc}-2 2 \\ 1 -5\end{array}\right]$ Solution: Given, $\quad 2 X+3 Y=\left[\begin{array}{ll}2 3 \\ 4 0\end{array}\right]$ $\ldots(\mathrm{i})$ and $3 X+2 Y=\left[\begin{array}{cc}-2 2 \\ 1 -5\end{array}\right]$ ...(ii) On subractiong equations (i) and (ii), we get $(3 X+2 Y)-(2 X+...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{x^{2}+1}{x^{4}-x^{2}+1} d x$ Solution: re-writing the given equation as $\int \frac{1+\frac{1}{x^{2}}}{x^{2}-1+\frac{1}{x^{2}}} d x$ $\int \frac{1+\frac{1}{x^{2}}}{\left(x-\frac{1}{x}\right)^{2}+1} d x$ Substituting $t$ as $x-\frac{1}{x}$ $\left(1+\frac{1}{x^{2}}\right) d x=d t$ $\int \frac{d t}{t^{2}+1}$ Using identity $\int \frac{1}{\mathrm{x}^{2}+1} \mathrm{dx}=\arctan (\mathrm{x})$ $\arctan t+c$ Substituting tas $\mathrm{x}-\frac{1}{\mat...
Read More →Mark the tick against the correct answer in the following:
Question: Mark the tick against the correct answer in the following: The principal value of $\sec ^{-1}\left(\frac{-2}{\sqrt{3}}\right)$ is A. $\frac{\pi}{6}$ B. $\frac{-\pi}{6}$ C. $\frac{5 \pi}{6}$ D. $\frac{7 \pi}{6}$ Solution: To Find: The Principle value of $\sec ^{-1}\left(\frac{-2}{\sqrt{3}}\right)$ Let the principle value be given by $x$ Now, let $x=\sec ^{-1}\left(\frac{-2}{\sqrt{3}}\right)$ $\Rightarrow \sec x=\frac{-2}{\sqrt{3}}$ $\Rightarrow \sec x=-\sec \left(\frac{\pi}{6}\right)\le...
Read More →Mark the tick against the correct answer in the following:
Question: Mark the tick against the correct answer in the following: The principal value of $\cot ^{-1}(-1)$ is A. $\frac{-\pi}{4}$ B. $\frac{\pi}{4}$ C. $\frac{5 \pi}{4}$ D. $\frac{3 \pi}{4}$ Solution: To Find: The Principle value of $\cot ^{-1}(-1)$ Let the principle value be given by $x$ Now, let $x=\cot ^{-1}(-1)$ $\Rightarrow \cot x=-1$ $\Rightarrow \cot x=-\cot \left(\frac{\pi}{4}\right)\left(\because \cot \left(\frac{\pi}{4}\right)=1\right)$ $\Rightarrow \cot x=\cot \left(\pi-\frac{\pi}{4...
Read More →Solve for x and y:
Question: Solve for x and y: $x\left[\begin{array}{l}2 \\ 1\end{array}\right]+y\left[\begin{array}{l}3 \\ 5\end{array}\right]+\left[\begin{array}{c}-8 \\ -11\end{array}\right]=\mathrm{O}$ Solution: Given, $x\left[\begin{array}{l}2 \\ 1\end{array}\right]+y\left[\begin{array}{l}3 \\ 5\end{array}\right]+\left[\begin{array}{c}-8 \\ -11\end{array}\right]=O$ $\left[\begin{array}{c}2 x \\ x\end{array}\right]+\left[\begin{array}{c}3 y \\ 5 y\end{array}\right]+\left[\begin{array}{c}-8 \\ -11\end{array}\r...
Read More →Mark the tick against the correct answer in the following:
Question: Mark the tick against the correct answer in the following: The principal value of $\tan ^{-1}(-\sqrt{3})$ is A. $\frac{2 \pi}{3}$ B. $\frac{4 \pi}{3}$ C. $\frac{-\pi}{3}$ D. none of these Solution: To Find: The Principle value of $\tan ^{-1}(-\sqrt{3})$ Let the principle value be given by $x$ Now, let $x=\tan ^{-1}(-\sqrt{3})$ $\Rightarrow \tan x=-\sqrt{3}$ $\Rightarrow \tan x=-\tan \left(\frac{\pi}{3}\right)\left(\because \tan \left(\frac{\pi}{3}\right)=-\sqrt{3}\right)$ $\Rightarrow ...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{x^{2}-3 x+1}{x^{4}+x^{2}+1} d x$ Solution: re-writing the given equation as $\int \frac{1-\frac{3}{x}+\frac{1}{x^{2}}}{x^{2}+1+\frac{1}{x^{2}}} d x$ $\int \frac{1+\frac{1}{x^{2}}}{\left(x-\frac{1}{x}\right)^{2}+3} d x-\int \frac{3 x}{x^{4}+x^{2}+1} d x$ Substituting $\mathrm{tas} \mathrm{x}-\frac{1}{\mathrm{x}}$ and $\mathrm{z}$ as $\mathrm{x}^{2}$ $\left(1+\frac{1}{\mathrm{x}^{2}}\right) \mathrm{dx}=\mathrm{dt}$ and $2 \mathrm{xdx}=\mathrm{...
Read More →Mark the tick against the correct answer in the following:
Question: Mark the tick against the correct answer in the following: The principal value of $\cos ^{-1}\left(\frac{-1}{2}\right)$ is A. $\frac{-\pi}{3}$ B. $\frac{2 \pi}{3}$ C. $\frac{4 \pi}{3}$ D. $\frac{\pi}{3}$ Solution: To Find: The Principle value of $\cos ^{-1}\left(\frac{-1}{2}\right)$ Let the principle value be given by $\mathrm{x}$ Now, let $x=\cos ^{-1}\left(\frac{-1}{2}\right)$ $\Rightarrow \cos x=\frac{-1}{2}$ $\Rightarrow \cos x=-\cos \left(\frac{\pi}{3}\right)\left(\because \cos \l...
Read More →Given Is (AB)’ = B’ A’ ?
Question: Given Is (AB) = B A ? Solution: Given, $A=\left[\begin{array}{lll}2 4 0 \\ 3 9 6\end{array}\right]_{2 \times 3}$ and $B=\left[\begin{array}{ll}1 4 \\ 2 8 \\ 1 3\end{array}\right]_{3 \times 2}$ So, their product is $A B=\left[\begin{array}{lll}2 4 0 \\ 3 9 6\end{array}\right]\left[\begin{array}{ll}1 4 \\ 2 8 \\ 1 3\end{array}\right]=\left[\begin{array}{cc}2+8+0 8+32+0 \\ 3+18+6 12+72+18\end{array}\right]=\left[\begin{array}{cc}10 40 \\ 27 102\end{array}\right]$ And, $(A B)^{\prime}=\lef...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{1}{x^{4}+x^{2}+1} d x$ Solution: re-writing the given equation as $\int \frac{\frac{1}{x^{2}}}{x^{2}+1+\frac{1}{x^{2}}} d x$ $\frac{1}{2} \int \frac{1+\frac{1}{x^{2}}+\frac{1}{x^{2}}-1}{x^{2}+1+\frac{1}{x^{2}}} d x$ $\frac{1}{2}\left[\int \frac{1+\frac{1}{x^{2}}}{x^{2}+1+\frac{1}{x^{2}}} d x+\int \frac{-1+\frac{1}{x^{2}}}{x^{2}+1+\frac{1}{x^{2}}} d x\right]$ $\frac{1}{2}\left[\int \frac{1+\frac{1}{x^{2}}}{\left(x-\frac{1}{x}\right)^{2}+3} d ...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{1}{x^{4}+x^{2}+1} d x$ Solution: re-writing the given equation as $\int \frac{\frac{1}{x^{2}}}{x^{2}+1+\frac{1}{x^{2}}} d x$ $\frac{1}{2} \int \frac{1+\frac{1}{x^{2}}+\frac{1}{x^{2}}-1}{x^{2}+1+\frac{1}{x^{2}}} d x$ $\frac{1}{2}\left[\int \frac{1+\frac{1}{x^{2}}}{x^{2}+1+\frac{1}{x^{2}}} d x+\int \frac{-1+\frac{1}{x^{2}}}{x^{2}+1+\frac{1}{x^{2}}} d x\right]$ $\frac{1}{2}\left[\int \frac{1+\frac{1}{x^{2}}}{\left(x-\frac{1}{x}\right)^{2}+3} d ...
Read More →Mark the tick against the correct answer in the following:
Question: Mark the tick against the correct answer in the following: The principal value of $\sin ^{-1}\left(\frac{-1}{2}\right)$ is A. $\frac{-\pi}{6}$ B. $\frac{5 \pi}{6}$ C. $\frac{7 \pi}{6}$ D. none of these Solution: To Find: The Principle value of $\sin ^{-1}\left(\frac{-1}{2}\right)$ Let the principle value be given by x Now, let $x=\sin ^{-1}\left(\frac{-1}{2}\right)$ $\Rightarrow \sin x=\frac{-1}{2}$ $\Rightarrow \sin x=-\sin \left(\frac{\pi}{6}\right)\left(\because \sin \left(\frac{\pi...
Read More →Show by an example that
Question: Show by an example that for A 0, B 0, AB = 0. Solution: Let $A=\left[\begin{array}{ll}0 1 \\ 0 2\end{array}\right] \neq O$ and $B=\left[\begin{array}{cc}-1 1 \\ 0 0\end{array}\right] \neq O$ So, the product $A B=\left[\begin{array}{ll}0 1 \\ 0 2\end{array}\right]\left[\begin{array}{cc}-1 1 \\ 0 0\end{array}\right]=\left[\begin{array}{ll}0 0 \\ 0 0\end{array}\right]=O$ Hence Proved $A=\left[\begin{array}{lll}2 4 0 \\ 3 9 6\end{array}\right]$ and $B=\left[\begin{array}{ll}1 4 \\ 2 8 \\ 1...
Read More →If possible,
Question: If possible, find BA and AB where $A=\left[\begin{array}{lll}2 1 2 \\ 1 2 4\end{array}\right], B=\left[\begin{array}{ll}4 1 \\ 2 3 \\ 1 2\end{array}\right]$ Solution: Given, $A=\left[\begin{array}{lll}2 1 2 \\ 1 2 4\end{array}\right]_{2 \times 3}$ and $B=\left[\begin{array}{ll}4 1 \\ 2 3 \\ 1 2\end{array}\right]_{3 \times 2}$ So. $A B$ and $B A$ both are defined Now, $A B=\left[\begin{array}{lll}2 1 2 \\ 1 2 4\end{array}\right]\left[\begin{array}{ll}4 1 \\ 2 3 \\ 1 2\end{array}\right]=...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{x^{2}+9}{x^{4}+81} d x$ Solution: re-writing the given equation as $\int \frac{1+\frac{9}{x^{2}}}{x^{2}+\frac{81}{x^{2}}} d x$ $\int \frac{1+\frac{9}{x^{2}}}{\left(x-\frac{9}{x}\right)^{2}+18} d x$ Let $x-\frac{9}{x}=t$ $\left(1+\frac{9}{x^{2}}\right) d x=d t$ $\int \frac{d t}{t^{2}+18}$ Using identity $\int \frac{1}{x^{2}+1} d x=\arctan (x)$ $\frac{1}{3 \sqrt{2}} \arctan \left(\frac{t}{3 \sqrt{2}}\right)+c$ Substituting $\mathrm{t}$ as $\ma...
Read More →If then verify
Question: If then verify (BA)2 B2A2 Solution: The given matrices A has order 3 x 2 and B has order 2 x 3. So, BA is defined and will have order 3 x 3. But, A2and B2are not defined as the orders dont satisfy the multiplication condition. Hence, (BA)2 B2A2...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \sqrt{\cot \theta} d \theta$ Solution: let $\cot \theta$ as $x^{2}$ $-\operatorname{cosec}^{2} \theta d \theta=2 x d x$ $\mathrm{d} \theta=-\frac{2 \mathrm{x}}{1+\cot ^{2} \theta} \mathrm{dx}$ $\mathrm{d} \theta=-\frac{2 \mathrm{x}}{1+\mathrm{x}^{4}} \mathrm{dx}$ $\int-\frac{2 x^{2}}{1+x^{4}} d x$ re-writing the given equation as $\int \frac{1+\frac{1}{x^{2}}+1-\frac{1}{x^{2}}}{\frac{1}{x^{2}}+x^{2}} d x$ $-\int \frac{1+\frac{1}{x^{2}}}{\left(x-\f...
Read More →Find A, if
Question: Find A, if Solution: Given, $\left[\begin{array}{l}4 \\ 1 \\ 3\end{array}\right] A=\left[\begin{array}{rrr}-4 8 4 \\ -1 2 1 \\ -3 6 3\end{array}\right]$ Now, let $A=\left[\begin{array}{lll}x y z\end{array}\right]$ So, $\quad\left[\begin{array}{l}4 \\ 1 \\ 3\end{array}\right][x y z]=\left[\begin{array}{lll}-4 8 4 \\ -1 2 1 \\ -3 6 3\end{array}\right]$ $\Rightarrow\left[\begin{array}{ccc}4 x 4 y 4 z \\ x y z \\ 3 x 3 y 3 z\end{array}\right]=\left[\begin{array}{ccc}-4 8 4 \\ -1 2 1 \\ -3 ...
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