Prove by Mathematical Induction
Question: Prove by Mathematical Induction that (A)n= (An), wheren N for any square matrix A. Solution: Let P(n): (A)n= (An) So, P(1): (A) = (A) A = A Hence, P(1) is true. Now, let P(k) = (A)k= (Ak) , where kN And, P(k + 1): (A)k+1= (A)kA = (Ak) A = (AAk) = (Ak+1) Hence, P(1) is true and whenever P(k) is true P(k + 1) is true. Therefore, P(n) is true for all nN....
Read More →Verify that A2 = I when A =
Question: Verify that A2= I when A = $\left[\begin{array}{ccc}0 1 -1 \\ 4 -3 4 \\ 3 -3 4\end{array}\right]$ Solution: Given, $A=\left[\begin{array}{ccc}0 1 -1 \\ 4 -3 4 \\ 3 -3 4\end{array}\right]$ So, $A^{2}=\left[\begin{array}{ccc}0 1 -1 \\ 4 -3 4 \\ 3 -3 4\end{array}\right] \cdot\left[\begin{array}{ccc}0 1 -1 \\ 4 -3 4 \\ 3 -3 4\end{array}\right]$ $=\left[\begin{array}{ccc}0+4-3 0-3+3 0+4-4 \\ 0-12+12 4+9-12 -4-12+16 \\ 0-12+12 3+9-12 -3-12+16\end{array}\right]=\left[\begin{array}{lll}1 0 0 \...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{1}{\left(2 x^{2}+3\right) \sqrt{x^{2}-4}} d x$ Solution: assume $x=\frac{1}{t}$ $\mathrm{dx}=-\frac{1}{\mathrm{t}^{2}} \mathrm{dt}$ $-\int \frac{t d t}{\left(3 t^{2}+2\right)\left(\sqrt{1-4 t^{2}}\right.}$ Assume $1-4 t^{2}=u^{2}$ $-4 t d t=u d u$ $-\frac{1}{4} \int \frac{\text { udu }}{\left(\frac{11-3 u^{2}}{4}\right) u}$ $-\frac{1}{3} \int \frac{d u}{\left(\frac{11}{3}-u^{2}\right)}$ Using identity $\int \frac{d z}{(z)^{2}-1}=\frac{1}{2} ...
Read More →If and x2 = –1, then show
Question: If andx2= 1, then show that (A + B)2= A2+ B2 Solution: Given, $A=\left[\begin{array}{cc}0 -x \\ x 0\end{array}\right], B=\left[\begin{array}{ll}0 1 \\ 1 0\end{array}\right]$ and $x^{2}=-1$ So, $(A+B)=\left[\begin{array}{cc}0 -x+1 \\ x+1 0\end{array}\right]$ And, And, $(A+B)^{2}=\left[\begin{array}{cc}0 -x+1 \\ x+1 0\end{array}\right]\left[\begin{array}{cc}0 -x+1 \\ x+1 0\end{array}\right]=\left[\begin{array}{cc}1-x^{2} 0 \\ 0 1-x^{2}\end{array}\right]$ And, $(A+B)^{2}=\left[\begin{arra...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{1}{\left(1+x^{2}\right) \sqrt{1-x^{2}}} d x$ Solution: assume $\mathrm{x}=\frac{1}{\mathrm{t}}$ $d x=-\frac{1}{t^{2}} d t$ $-\int \frac{t d t}{\left(t^{2}+1\right)\left(\sqrt{t^{2}-1}\right.}$ Let $\mathrm{t}^{2}-1=\mathrm{u}^{2}$' $\mathrm{tdt}=\mathrm{udu}$ $-\int \frac{u d u}{\left(u^{2}+2\right) u}$ $-\int \frac{d u}{\left(u^{2}+2\right)}$ Using identity $\int \frac{1}{x^{2}+1} d x=\arctan (x)$ $-\frac{1}{\sqrt{2}} \arctan \left(\frac{\m...
Read More →Mark the tick against the correct answer in the following:
Question: Mark the tick against the correct answer in the following: $\sin \left(\cos ^{-1} \frac{3}{5}\right)=?$ A. $\frac{3}{4}$ B. $\frac{4}{5}$ C. $\frac{3}{5}$ D. none of these Solution: To Find: The value of $\sin \left(\cos ^{-1} \frac{3}{5}\right)$ Let, $x=\cos ^{-1} \frac{3}{5}$ $\Rightarrow \cos x=\frac{3}{5}$ Now, $\sin \left(\cos ^{-1} \frac{3}{5}\right)$ becomes $\sin (x)$ Since we know that $\sin x=\sqrt{1-\cos ^{2} x}$ $=\sqrt{1-\left(\frac{3}{5}\right)^{2}}$ $\sin \left(\cos ^{-1...
Read More →Prove the following
Question: If $\mathrm{A}=\left[\begin{array}{cc}\cos \theta \sin \theta \\ -\sin \theta \cos \theta\end{array}\right]$, then show that $\mathrm{A}^{2}=\left[\begin{array}{cc}\cos 2 \theta \sin 2 \theta \\ -\sin 2 \theta \cos 2 \theta\end{array}\right]$ Solution: Given, $A=\left[\begin{array}{cc}\cos \theta \sin \theta \\ -\sin \theta \cos \theta\end{array}\right]$ Now, $A^{2}=A \cdot A$ $=\left[\begin{array}{cc}\cos \theta \sin \theta \\ -\sin \theta \cos \theta\end{array}\right]\left[\begin{arr...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{x}{\left(x^{2}+4\right) \sqrt{x^{2}+1}} d x$ Solution: assume $x^{2}+1=u^{2}$ $x d x=u d u$ $\int \frac{u d u}{\left(u^{2}+3\right) u}$ $\int \frac{d u}{\left(u^{2}+3\right)}$ Using identity $\int \frac{1}{x^{2}+1} d x=\arctan (x)$ $\frac{1}{\sqrt{3}} \arctan \left(\frac{\mathrm{u}}{\sqrt{3}}\right)+\mathrm{c}$ Substituting $u=\sqrt{1+x^{2}}$ $\frac{1}{\sqrt{3}} \arctan \left(\frac{\sqrt{1+x^{2}}}{\sqrt{3}}\right)+c$...
Read More →Mark the tick against the correct answer in the following:
Question: Mark the tick against the correct answer in the following: $\tan \frac{1}{2}\left(\cos ^{-1} \frac{\sqrt{5}}{3}\right)=?$ A. $\frac{(3-\sqrt{5})}{2}$ B. $\frac{(3+\sqrt{5})}{2}$ C. $\frac{(5-\sqrt{3})}{2}$ D. $\frac{(5+\sqrt{3})}{2}$ Solution: To Find: The value of $\tan \frac{1}{2}\left(\cos ^{-1} \frac{\sqrt{5}}{3}\right)$ Let, $x=\cos ^{-1} \frac{\sqrt{5}}{3}$ $\Rightarrow \cos x=\frac{\sqrt{5}}{3}$ Now, $\tan \frac{1}{2}\left(\cos ^{-1} \frac{\sqrt{5}}{3}\right)$ becomes $\tan \fra...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{1}{\left(x^{2}-1\right) \sqrt{x^{2}+1}} d x$ Solution: assume $\mathrm{X}=\frac{1}{\mathrm{t}}$ $\mathrm{dx}=-\frac{1}{\mathrm{t}^{2}} \mathrm{dt}$ Let $1+t^{2}=u^{2}$ $\mathrm{tdt}=\mathrm{udu}$ $\int \frac{u d u}{\left(u^{2}-2\right) u}$ $\int \frac{d u}{\left(u^{2}-2\right)}$ Using identity $\int \frac{d z}{(z)^{2}-1}=\frac{1}{2} \log \left|\frac{z-1}{z+1}\right|+c$ $\frac{1}{2 \sqrt{2}} \log \left|\frac{u-\sqrt{2}}{u+\sqrt{2}}\right|+c$ ...
Read More →Let and a = 4, b = –2.
Question: Let anda= 4,b= 2. $\mathrm{A}=\left[\begin{array}{cc}1 2 \\ -1 3\end{array}\right], \quad \mathrm{B}=\left[\begin{array}{ll}4 0 \\ 1 5\end{array}\right], \quad \mathrm{C}=\left[\begin{array}{cc}2 0 \\ 1 -2\end{array}\right]$ Show that: (a) A + (B + C) = (A + B) + C (b) A (BC) = (AB) C (c) (a+b)B =aB +bB (d)a(CA) =aC aA (e) (AT)T= A (f) (bA)T=bAT (g) (AB)T= BTAT (h) (A B)C = AC BC (i) (A B)T= AT BT Solution: (a) $A+(B+C)=\left[\begin{array}{cc}1 2 \\ -1 3\end{array}\right]+\left[\begin{...
Read More →Mark the tick against the correct answer in the following:
Question: Mark the tick against the correct answer in the following: $\tan \left[2 \tan ^{-1} \frac{1}{5}-\frac{\pi}{4}\right]=?$ A. $\frac{7}{17}$ B. $\frac{-7}{17}$ C. $\frac{7}{12}$ D. $\frac{-7}{12}$ Solution: To Find: The value of $\tan \left(2 \tan ^{-1} \frac{1}{5}-\frac{\pi}{4}\right)$ Consider, $\tan \left(2 \tan ^{-1} \frac{1}{5}-\frac{\pi}{4}\right)=\tan \left(\tan ^{-1}\left(\frac{2\left(\frac{1}{5}\right)}{1-\left(\frac{1}{5}\right)^{2}}\right)-\frac{\pi}{4}\right)$ $\left(\because ...
Read More →Mark the tick against the correct answer in the following:
Question: Mark the tick against the correct answer in the following: $\tan ^{-1} 1+\cos ^{-1}\left(\frac{-1}{2}\right)+\sin ^{-1}\left(\frac{-1}{2}\right)=?$ A. $\pi$ B. $\frac{2 \pi}{3}$ C. $\frac{3 \pi}{4}$ D. $\frac{\pi}{2}$ Solution: To Find: The value of $\tan ^{-1} 1+\cos ^{-1}\left(\frac{-1}{2}\right)+\sin ^{-1}\left(\frac{-1}{2}\right)$ Now, let $x=\tan ^{-1} 1+\cos ^{-1}\left(\frac{-1}{2}\right)+\sin ^{-1}\left(\frac{-1}{2}\right)$ $\Rightarrow x=\frac{\pi}{4}+\left[\pi-\cos ^{-1}\left(...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{1}{(x+1) \sqrt{x^{2}+x+1}} d x$ Solution: assume $x+1=\frac{1}{t}$ $\mathrm{dx}=-\frac{1}{\mathrm{t}^{2}} \mathrm{dt}$ $-\int \frac{d t}{\sqrt{1+t-t^{2}}}$ $-\int \frac{d t}{\sqrt{\frac{5}{4}-\left(t-\frac{1}{2}\right)^{2}}}$ Using identity $\int \frac{\mathrm{dx}}{\sqrt{\mathrm{a}^{2}-\mathrm{x}^{2}}}=\arcsin \left(\frac{\mathrm{x}}{\mathrm{a}}\right)+\mathrm{c}$ $-\arcsin \left(\frac{\left(t-\frac{1}{2}\right)}{\frac{\sqrt{5}}{2}}\right)+c...
Read More →Mark the tick against the correct answer in the following:
Question: Mark the tick against the correct answer in the following: $\cos ^{-1} \frac{1}{2}+2 \sin ^{-1} \frac{1}{2}=?$ A. $\frac{2 \pi}{3}$ B. $\frac{3 \pi}{2}$ C. $2 \pi$ D. none of these Solution: To Find: The value of $\cos ^{-1} \frac{1}{2}+2 \sin ^{-1} \frac{1}{2}$ Now, let $x=\cos ^{-1} \frac{1}{2}+2 \sin ^{-1} \frac{1}{2}$ $\Rightarrow \mathrm{x}=\frac{\pi}{3}+2\left(\frac{\pi}{6}\right)\left(\because \cos \left(\frac{\pi}{3}\right)=\frac{1}{2}\right.$ and $\left.\sin \left(\frac{\pi}{6...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{1}{(x-1) \sqrt{x^{2}+1}} d x$ Solution: assume $x-1=\frac{1}{t}$ $\mathrm{dx}=-\frac{1}{\mathrm{t}^{2}} \mathrm{dt}$ $-\int \frac{d t}{\sqrt{2 t^{2}+2 t+1}}$ $-\frac{1}{\sqrt{2}} \int \frac{d t}{\sqrt{\left(t+\frac{1}{2}\right)^{2}+\frac{1}{4}}}$ Using identity $\int \frac{d x}{\sqrt{x^{2}+a^{2}}}=\log \left(x+\sqrt{x^{2}+a^{2}}\right)+c$ $-\frac{1}{\sqrt{2}} \log \left(\mathrm{t}+\frac{1}{2}+\sqrt{\left.\left(\mathrm{t}+\frac{1}{2}\right)^{...
Read More →Mark the tick against the correct answer in the following:
Question: Mark the tick against the correct answer in the following: $\tan ^{-1}(\sqrt{3})-\sec ^{-1}(-2)=?$ A. $\frac{\pi}{3}$ B. $\frac{-\pi}{3}$ C. $\frac{5 \pi}{3}$ D. none of these Solution: To Find: The value of $\tan ^{-1}(\sqrt{3})-\sec ^{-1}(-2)$ Let, $x=\tan ^{-1}(\sqrt{3})-\sec ^{-1}(-2)$ $\Rightarrow \mathrm{x}=\frac{\pi}{3}-\left[\pi-\sec ^{-1}(2)\right]\left(\because \tan \left(\frac{\pi}{3}\right)=\sqrt{3}\right.$ and $\left.\sec ^{-1}(-\theta)=\pi-\sec ^{-1}(\theta)\right)$ $\Rig...
Read More →Mark the tick against the correct answer in the following
Question: Mark the tick against the correct answer in the following $\cos ^{-1}\left(\cos \frac{2 \pi}{3}\right)+\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)=?$ A. $\frac{4 \pi}{3}$ B. $\frac{\pi}{2}$ C. $\frac{5 \pi}{3}$ D. $\pi$ Solution: To Find: The value of $\cos ^{-1}\left(\cos \left(\frac{2 \pi}{3}\right)\right)+\sin ^{-1}\left(\sin \left(\frac{2 \pi}{3}\right)\right)$ Here,consider $\cos ^{-1}\left(\cos \left(\frac{2 \pi}{3}\right)\right)$ (\because the principle value of $\cos$ lies in t...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{x}{\left(x^{2}+2 x+2\right) \sqrt{x+1}} d x$ Solution: assume $x+1=t^{2}$ $\mathrm{d} x=2 \mathrm{tdt}$ $\int \frac{2\left(t^{2}-1\right) d t}{t^{4}+1}$ Dividing by $t^{2}$ in both numerator and denominator $\int \frac{2\left(1-\frac{1}{t^{2}}\right) d t}{t^{2}+\frac{1}{t^{2}}}$ $\int \frac{2\left(1-\frac{1}{t^{2}}\right) d t}{\left(t+\frac{1}{t}\right)^{2}-2}$ Let $\left(\mathrm{t}+\frac{1}{\mathrm{t}}\right)=\mathrm{z}$ $\left(1-\frac{1}{t...
Read More →Show that if A and B are square matrices
Question: Show that if A and B are square matrices such that AB = BA, then (A + B)2= A2+ 2AB + B2. Solution: Given, A and B are square matrices such that AB = BA. So, (A + B)2= (A + B) . (A + B) = A2+ AB + BA + B2 = A2+ AB + AB + B2[Since, AB = BA] = A2+ 2AB + B2...
Read More →Let A and B be square matrices
Question: Let A and B be square matrices of the order 3 3. Is (AB)2= A2B2? Give reasons. Solution: As, A and B be square matrices of order 3 x 3. We have, (AB)2= AB . AB = A(BA)B = A(AB)B [If AB = BA] = AABB = A2B2 Thus, (AB)2= A2B2is true only if AB = BA....
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 $\sin \left(\cos ^{-1} \frac{3}{5}\right)$ is A. $\frac{2}{5}$ B. $\frac{4}{5}$ C. $\frac{-2}{5}$ D. none of these Solution: To Find: The value of $\sin \left(\cos ^{-1} \frac{3}{5}\right)$ Now, let $x=\cos ^{-1} \frac{3}{5}$ $\Rightarrow \cos x=\frac{3}{5}$ Now, $\sin x=\sqrt{1-\cos ^{2} x}$ $=\sqrt{1-\left(\frac{3}{5}\right)^{2}}$ $=\frac{4}{5}$ $\Rightarrow x=\sin ^{-1} \frac{4}{5}=\cos ^{-1} \frac{3}{5}$ Theref...
Read More →Show that A¢A and AA¢ are both symmetric matrices
Question: Show that AA and AA are both symmetric matrices for any matrix A. Solution: Let P = AA So, P = (AA) = A(A) [As (AB) = BA] Hence, AA is symmetric matrix for any matrix A. Now, let Q = AA So, Q = (AA) = (A) = AA = Q Hence, AA is symmetric matrix for any matrix A....
Read More →If then verify that:
Question: If then verify that: (i) (2A + B) = 2A + B (ii) (A B) = A B. $A=\left[\begin{array}{ll}1 2 \\ 4 1 \\ 5 6\end{array}\right], \quad B=\left[\begin{array}{ll}1 2 \\ 6 4 \\ 7 3\end{array}\right]$ Solution: Given, $A=\left[\begin{array}{ll}1 2 \\ 4 1 \\ 5 6\end{array}\right]$ and $B=\left[\begin{array}{ll}1 2 \\ 6 4 \\ 7 3\end{array}\right]$ (i) $(2 A+B)=\left[\begin{array}{cc}2 4 \\ 8 2 \\ 10 12\end{array}\right]+\left[\begin{array}{cc}1 2 \\ 6 4 \\ 7 3\end{array}\right]=\left[\begin{array...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{1}{\left(x^{2}+1\right) \sqrt{x}} d x$ Solution: let $x=t^{2}$ $\mathrm{d} \mathrm{x}=2 \mathrm{tdt}$ $\int \frac{2 d t}{t^{4}+1}$ Dividing by $t^{2}$ in both numerator and denominator $\int \frac{\left[\left(1+\frac{1}{t^{2}}\right)-\left(1-\frac{1}{t^{2}}\right)\right] d t}{t^{2}+\frac{1}{t^{2}}}$ $\int \frac{\left[\left(1+\frac{1}{\mathrm{t}^{2}}\right)\right] \mathrm{dt}}{\left(\mathrm{t}-\frac{1}{\mathrm{t}}\right)^{2}+2}-\int \frac{\le...
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