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- Question 2, Exercise 2.3 @math-11-kpk:sol:unit02
- {bmatrix}4 & -2 & 5 \\ 2 & 1 & 0 \\ -1 & 2 & 3 \end{bmatrix}$$ ====Solution==== Let $$A=\begin{bmatrix} 4 & -2 & 5 \\ 2 & 1 & 0 \\ -1 & 2 & 3 \end{bmatrix}.$$ Then \begin{align}|A|&=\begin{vmatrix}4 & -2 & 5 \\ 2 & 1 & 0 \\ -1 & 2 & 3 \end{vmatrix}\\ &=4(3)+2(6)+5(4+1) \\ & =49\neq 0. \end{align} This gives, $A$ is non-singular and $A^{-1
- Question 3, Exercise 2.1 @math-11-kpk:sol:unit02
- Question 3(i)===== If $A=\begin{bmatrix}x & y & z\end{bmatrix}$, $B=\begin{bmatrix}a & h & g\\h & b & f\\g & f & c\end{bmatrix}$ and $C=\begin{bmatrix}x\\y\\z\end{bmatrix}$. Verify that $\left( AB \right)C=A\left( BC \rig... ==Solution==== Given: $A=\begin{bmatrix}x & y & z\end{bmatrix}$, $B=\begin{bmatrix}a & h & g\\h & b & f
- Question 4 Exercise 8.2 @math-11-nbf:sol:unit08
- {9}{25}}\\ &= \sqrt{\frac{16}{25}} = \frac{4}{5} \end{align*} (a) $\sin 2 \theta$ \begin{align*} \sin ... (\frac{4}{5} \right) \left(\frac{3}{5} \right)\\ \end{align*} \begin{align*} \implies \boxed{\sin 2\theta = \frac{24}{25}}. \end{align*} (b) $\cos 2 \theta$ \begin{align*} \cos ... \right)^2\\ &= 1-2\left(\frac{16}{25} \right)\\ \end{align*} \begin{align*} \implies \boxed{\cos 2\the
- Exercise 6.1 @matric:9th_science
- 6&=x^2+3x+2x+6,\\ &=x(x+3)+2(x+3)\\ &=(x+3)(x+2) \end{align}$ $\begin{align} x^2-4x-12&=x^2-6x+2x-12,\\ &=x(x-6)+2(x-6)\\ &=(x-6)(x+2) \end{align}$ H.C.F= $x+2$ (ii) $\begin{align} x^3-27 &=x^3-3^3,\\ &=(x-3)(x^2+3x+9)\end{align}$ $\begin{align} x^2+6x-27&=x^2+9x-3x-27,\\ &=x(x+9)-3(x+9)\\ &=(x+6)(x-3) \end{align}$ $\begin{align} 2x^2-18&=2(x^2-9),\\ &=2(
- Question 1, Exercise 2.5 @math-11-nbf:sol:unit02
- {array}{ccc}1 & 3 & 5 \\ -6 & 8 & 3 \\ -4 & 6 & 5\end{array}\right]$. ** Solution. ** \begin{align*} ... {array}{ccc}1 & 3 & 5 \\ -6 & 8 & 3 \\ -4 & 6 & 5\end{array}\right]\\ \sim & \text{R} \left[\begin{array}{ccc} 1 & 3 & 5 \\ 0 & 26 & 33 \\ 0 & 18 & 25 \end{array}\right]\quad R_2 + 6R_1 \quad R_3 + 4R_1\\ ... & 3 & 5 \\ 0 & 1 & \frac{33}{26} \\ 0 & 18 & 25 \end{array}\right] \quad \frac{1}{26}R_2 \\ \sim & \te
- Question 6, Exercise 2.6 @math-11-nbf:sol:unit02
- gin{bmatrix} 5 & 3 & 1 \\ 2 & 1 & 3 \\ 1 & 2 & 4 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 6 \\ 19 \\ 25 \end{bmatrix} \end{align*} And \begin{align*} |A|& = \left| \begin{array}{ccc} 5 & 3 & 1 \\ 2 & 1 & 3 \\ 1
- Question 1, Exercise 2.1 @math-11-kpk:sol:unit02
- le 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]$$ ====Solution==== \begin{align}&\left[ \begin{matrix} 1 & 2 & 4 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 & 2
- Question 3, Exercise 2.5 @math-11-nbf:sol:unit02
- rray}{ccc}0 & -1 & -1 \\ -1 & 3 & 0 \\ 1 & -1 & 4\end{array}\right]$ if it exists. Also verify your ans... cc} 0 & -1 & -1 \\ -1 & 3 & 0 \\ 1 & -1 & 4 \end{array} \right]\\ |A|&=0+1(-4)-1(1-3)\\ &=-4+3\\ &=-1\neq 0\end{align*} So $A$ is non singular. Now consider \beg... & 3 & 0 & 0 & 1 & 0 \\ 1 & -1 & 4 & 0 & 0 & 1 \end{array} \right]\\ \sim &{\text{R}} \left[ \begin{a
- Question 4, Exercise 2.2 @math-11-nbf:sol:unit02
- in{align}\left[\begin{array}{cc} 2 & 1 \\ 3 & 2 \end{array}\right]A\left[\begin{array}{cc} 1 & 3 \\ 2 & 4 \end{array}\right]&=\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]\end{align} ** Solution. ** Let $ B = \left[\begin{array}{cc} 2 & 1 \\ 3 & 2 \end{array
- Question 5, Exercise 2.3 @math-11-nbf:sol:unit02
- rray}{ccc}1 & -1 & 1 \\ 2 & 1 & -1 \\ 1 & -2 & -1\end{array}\right]$. ** Solution. ** Given \begin{a... ay}{ccc} 1 & -1 & 1 \\ 2 & 1 & -1 \\ 1 & -2 & -1 \end{array}\right]\\ |A|&= 1 [-1 - 2] + 1 [-2 + 1] + ... ot (-1) + 1 \cdot (-5) \\ &= -3 - 1 - 5 \\ &= -9 \end{align*} Thus, $|A| = -9 \neq 0$, so $A$ is non-si... ^{1+1} \left|\begin{array}{cc} 1 & -1 \\ -2 & -1 \end{array}\right|\\ &= (1) [(1)(-1) - (-1)(-2)] \\ &=
- Question 10 Exercise 7.2 @math-11-kpk:sol:unit07
- \left.(1+x)^n=\left(\begin{array}{l} n \\ \vdots \end{array}\right)+\left(\begin{array}{l} m \\ 1 \end{array}\right) x+\left(\begin{array}{l} n \\ 2 \end{array}\right) x^2-\ldots+i_n^*\right) x^n \cdot $$ ... e have $(1 \div 1)^n=\left(\begin{array}{l}n \\ 0\end{array}\right)+\left(\begin{array}{l}n \\ 1\end{ar
- Question 13 Exercise 6.2 @math-11-kpk:sol:unit06
- re} &=\left(\begin{array}{c} n \\ m_1, m_2, m_3 \end{array}\right)\\&=\left(\begin{array}{c} 10 \\ 4,2,2 \end{array}\right) \\ & =\dfrac{10 !}{4 ! \cdot 2 ! \cdot 2 !}\\ &=37,800 \end{align} Begin with $\mathrm{E}$ If we have to pic... are} &=\left(\begin{array}{c} n \\ m_1, m_2, m_3 \end{array}\right)\\&=\dfrac{9 !}{3 ! \cdot 2 ! \cdot
- Question 1, Exercise 8.1 @math-11-nbf:sol:unit08
- \right) \\ & = -\frac{1}{2} - 0 = -\frac{1}{2} \end{align*} \begin{align*} \cos (\alpha - \beta) & ... }\right) \\ & = -\frac{1}{2} + 0 = -\frac{1}{2} \end{align*} \begin{align*} \sin (\alpha + \beta) & ... 0-\frac{\sqrt{3}}{2} + 0 = -\frac{\sqrt{3}}{2} \end{align*} \begin{align*} \sin (\alpha - \beta) & ... = 0+\frac{\sqrt{3}}{2} + 0 = \frac{\sqrt{3}}{2} \end{align*} \begin{align*} \tan (\alpha + \beta) &
- Question 7, Exercise 2.3 @math-11-nbf:sol:unit02
- {-1}$ if $A=\left[\begin{array}{ll}2 & 1 \\ 8 & 6\end{array}\right]$ and $B=\left[\begin{array}{ll}3 & 2 \\ 0 & 2\end{array}\right]$. ** Solution. ** Given: \begin{a... lign*} A &= \left[\begin{array}{ll}2 & 1 \\ 8 & 6\end{array}\right] \\ |A|& = 12 - 8 = 4\\ A^{-1} &= ... rac{1}{4} \left[\begin{array}{ll}6 & -1 \\ -8 & 2\end{array}\right]\\ & = \left[\begin{array}{ll} \frac
- Question 2, Exercise 2.5 @math-11-nbf:sol:unit02
- {array}{ccc}5 & 9 & 3 \\ 3 & -5 & 6 \\ 2 & 10 & 6\end{array}\right]$ ** Solution. ** \begin{align*}&\... y}{ccc} 5 & 9 & 3 \\ 3 & -5 & 6 \\ 2 & 10 & 6 \end{array} \right]\\ \sim & \text{R}\left[ \begin{arr... {5} & \frac{3}{5} \\ 3 & -5 & 6 \\ 2 & 10 & 6 \end{array} \right]\quad \frac{1}{5} R1\\ \sim & \text... 0 & -\frac{52}{5} & \frac{15}{5} \\ 2 & 10 & 6 \end{array} \right]\quad R2 - 3 \cdot R1\\ \sim & \tex