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- Question 2, Exercise 2.3 @math-11-kpk:sol:unit02
- ingular and $A^{-1}$ exists. Now \begin{align} & \left[\begin{matrix} 4 & -2 & 5 \\ 2 & 1 & 0 \\ -1 & 2 & 3 \end{matrix}\left| \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0... d{matrix}\right. \right] \\ \underset{\sim}{R} & \left[\begin{matrix} 1 & 4 & 14 \\ 0 & 5 & 6 \\ -1 & 2 & 3 \end{matrix}\left| \begin{matrix} 1 & 0 & 3 \\ 0 & 1 & 2 \\ 0 & 0
- Question 8 & 9, Review Exercise 10 @math-11-kpk:sol:unit10
- n. =====Question 8===== Prove the identity $\sin \left( \dfrac{\pi }{4}-\theta \right)\sin \left( \dfrac{\pi }{4}+\theta \right)=\dfrac{1}{2}\cos 2\theta $.... ==== We know that $2\sin \alpha \sin \beta =\cos \left( \alpha -\beta \right)-\cos \left( \alpha +\beta \right)$ \begin{align}L.H.S.&=\sin \left( \dfrac{\p
- Question 3, Exercise 2.1 @math-11-kpk:sol:unit02
- egin{bmatrix}x\\y\\z\end{bmatrix}$. Verify that $\left( AB \right)C=A\left( BC \right)$. ====Solution==== Given: $A=\begin{bmatrix}x & y & z\end{bmatrix}$, $B... bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}\\ &=\left[ x( ax+hy+gz )+y( hx+by+fz )+z( gx+fy+cz ) \right] \\ &=\left[ a{{x}^{2}}+hxy+gxz+hxy+b{{y}^{2}}+fyz+gxz+fyz+c{
- Question 2, Exercise 10.1 @math-11-kpk:sol:unit10
- end{align} \begin{align} \Rightarrow \quad \sin \left( \frac{\pi }{3}-\frac{\pi }{4} \right) & =\sin \f... }{4}-\cos \frac{\pi }{3}\sin \frac{\pi }{4} \\ &=\left( \frac{\sqrt{3}}{2} \right)\left( \frac{1}{\sqrt{2}} \right)-\left( \frac{1}{2} \right)\left( \frac{1}{\sqrt{2}} \right) \\ & =\frac{\sq
- Question 5, Exercise 10.3 @math-11-kpk:sol:unit10
- == We know that\\ $2\cos \alpha \cos \beta =\cos \left( \alpha +\beta \right)+\cos \left( \alpha -\beta \right)$\\ \begin{align}L.H.S.&=\cos {{20}^{\circ }}... c }}\\ &=\cos {{20}^{\circ }}\cos {{40}^{\circ }}\left( \dfrac{1}{2} \right)\cos {{80}^{\circ }}\\ &=\df... 0}^{\circ }}\cos {{20}^{\circ }}\\ &=\dfrac{1}{4}\left( 2\,\cos {{80}^{\circ }}\cos {{40}^{\circ }} \rig
- Question 5, Exercise 10.3 @math-11-kpk:sol:unit10
- == We know that\\ $2\cos \alpha \cos \beta =\cos \left( \alpha +\beta \right)+\cos \left( \alpha -\beta \right)$\\ \begin{align}L.H.S.&=\cos {{20}^{\circ }}... c }}\\ &=\cos {{20}^{\circ }}\cos {{40}^{\circ }}\left( \dfrac{1}{2} \right)\cos {{80}^{\circ }}\\ &=\df... 0}^{\circ }}\cos {{20}^{\circ }}\\ &=\dfrac{1}{4}\left( 2\,\cos {{80}^{\circ }}\cos {{40}^{\circ }} \rig
- Question 1, Exercise 2.1 @math-11-kpk:sol:unit02
- Question 1(i)===== Express as a single matrix $$\left[ \begin{matrix} 1 & 2 & 4 \\ \end{matrix} \right] \left[ \begin{matrix} 1 & 0 & 2 \\ 2 & 0 & 1 \\ 0 & 1 & 2 \\ \end{matrix} \right] \left[ \begin{matrix} 2 \\ 4 \\ 6 \\ \end{matrix} \right]$$ ====Solution==== \begin{align}&\left[ \begin{matrix} 1 & 2 & 4 \\ \end{matrix} \ri
- Question 10 Exercise 7.2 @math-11-kpk:sol:unit07
- osficient $s=2^{n-1}$. Solution: We know that $$ \left.(1+x)^n=\left(\begin{array}{l} n \\ \vdots \end{array}\right)+\left(\begin{array}{l} m \\ 1 \end{array}\right) x+\left(\begin{array}{l} n \\ 2 \end{array}\right) x^2-\ldot
- Question 3 & 4, Exercise 1.2 @math-11-kpk:sol:unit01
- dition and multiplicative. \begin{align}{{z}_{1}}\left( {{z}_{2}}+{{z}_{3}} \right)&={{z}_{1}}{{z}_{2}}+... {{z}_{2}}+{{z}_{3}}&=\sqrt{2}-\sqrt{3}i+2+3i\\ &=\left( \sqrt{2}+2 \right)+\left( 3-\sqrt{3} \right)i\\ L.H.S.&={{z}_{1}}\left( {{z}_{2}}+{{z}_{3}} \right)\\ &=\left( \sqrt{3}+\sqrt{2
- Question 5 and 6 Exercise 7.3 @math-11-kpk:sol:unit07
- x} $$ $$ \begin{aligned} & =\frac{8^{\frac{2}{3}}\left(1+\frac{3 x}{8}\right)^{\frac{2}{3}}}{2\left(1+\frac{3 x}{2}\right) \sqrt{4}\left(1-\frac{5 x}{4}\right)^{\frac{1}{2}}} \\ & =\frac{\left(2^3\right)^{\frac{2}{3}}}{2.2}\left[\left(1+\frac
- Question 5, Exercise 10.1 @math-11-kpk:sol:unit10
- $\beta$ in the first Quadrant, then find: $\sin \left( \alpha +\beta \right)$. ====Solution==== Given:... a }\\ \Rightarrow \quad \sec \alpha &=-\sqrt{1+{{\left(\frac{3}{4} \right)}^{2}}}=-\sqrt{1+\frac{9}{16}}... \cos \alpha \\ \Rightarrow \quad \sin \alpha &=\left(\frac{3}{4}\right)\left(-\frac{4}{5} \right)\\ \Rightarrow \quad \sin\alpha &= \frac{3}{5}\end{align}
- Question 7, Exercise 10.2 @math-11-kpk:sol:unit10
- &={{\cos }^{4}}\theta -{{\sin }^{4}}\theta \\ &=\left( {{\cos }^{2}}\theta -{{\sin }^{2}}\theta \right)\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)\\ &=\left( {{\cos }^{2}}\theta -{{\sin }^{2}}\theta \right)\left( 1 \right)\\ &=\cos 2\theta \quad \text{(By usin
- Question 2 & 3, Exercise 1.1 @math-11-kpk:sol:unit01
- t i^{106}+i^{112}+i^{122}+i\cdot i^{152}\\ &=i.{{\left( {{i}^{2}} \right)}^{53}}+{{\left( {{i}^{2}} \right)}^{56}}+{{\left( {{i}^{2}} \right)}^{61}}+i.{{\left( {{i}^{2}} \right)}^{76}}\\ &=i.{{\left( -1 \right)}^{53}}+{{\left(
- Question 10 Exercise 7.1 @math-11-kpk:sol:unit07
- h the formulas below by mathematical induction, $\left(\begin{array}{1}5 \\5 \end{array}\right)+\left(\begin{array}{l}6 \\ 5\end{array}\right)+\left(\begin{array}{l}7 \\ 5\end{array}\right)+\ldots+\left(\begin{array}{c}n+4 \\ 5\end{array}\right)=\left(
- Question 8, Exercise 10.1 @math-11-kpk:sol:unit10
- tan. =====Question 8(i)===== Prove that: $\tan \left( \dfrac{\pi }{4}+\theta \right)=\dfrac{\cos \the... }$ ====Solution==== \begin{align}L.H.S.&=\tan \left( \dfrac{\pi }{4}+\theta \right)\\ &=\dfrac{\sin \left( \dfrac{\pi }{4}+\theta \right)}{\cos \left( \dfrac{\pi }{4}+\theta \right)}\\ &=\dfrac{\sin\dfrac{\