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- Question 2, Exercise 2.3 @math-11-kpk:sol:unit02
- the matrix by using elementary row operation. $$\begin{bmatrix}4 & -2 & 5 \\ 2 & 1 & 0 \\ -1 & 2 & 3 \end{bmatrix}$$ ====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) \\
- Question 3, Exercise 2.1 @math-11-kpk:sol:unit02
- shawar, Pakistan. =====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 $\left... =A\left( BC \right)$. ====Solution==== Given: $A=\begin{bmatrix}x & y & z\end{bmatrix}$, $B=\begin{bmatri
- Question 4 Exercise 8.2 @math-11-nbf:sol:unit08
- \theta$ lies in QI, $\sin$ is positive in QI, so \begin{align*} \sin\theta & = \sqrt{1-\left(\frac{3}{5}... = \frac{4}{5} \end{align*} (a) $\sin 2 \theta$ \begin{align*} \sin 2\theta & = 2 \sin\theta \cos\theta ... \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 2\theta & = 1-2\sin^2\theta \\ &= 1
- Exercise 6.1 @matric:9th_science
- 3-2x^2)$, $54(27x^4-x)$ **Solution:**\\ (i) $\begin{align} x^2+5x+6&=x^2+3x+2x+6,\\ &=x(x+3)+2(x+3)\\ &=(x+3)(x+2) \end{align}$ $\begin{align} x^2-4x-12&=x^2-6x+2x-12,\\ &=x(x-6)+2(x-6)... &=(x-6)(x+2) \end{align}$ 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)
- Question 1, Exercise 2.5 @math-11-nbf:sol:unit02
- helon form then into reduced echelon form $\left[\begin{array}{ccc}1 & 3 & 5 \\ -6 & 8 & 3 \\ -4 & 6 & 5\end{array}\right]$. ** Solution. ** \begin{align*} & \quad \left[\begin{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 &
- Question 6, Exercise 2.6 @math-11-nbf:sol:unit02
- lution. ** For this system of equations; we have \begin{align*} A &= \begin{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*} A
- Question 1, Exercise 2.1 @math-11-kpk:sol:unit02
- n 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 \\ \en
- Question 3, Exercise 2.5 @math-11-nbf:sol:unit02
- perations, find the inverse of the matrix $\left[\begin{array}{ccc}0 & -1 & -1 \\ -1 & 3 & 0 \\ 1 & -1 & ... at $A A^{-1}=A^{-1} A=I$.\\ ** Solution. ** Let \begin{align*} A&=\left[ \begin{array}{ccc} 0 & -1 & -1 \\ -1 & 3 & 0 \\ 1 & -1 & 4 \end{array} \right]\\ ... end{align*} So $A$ is non singular. Now consider \begin{align*} &\quad\left[ \begin{array}{ccc|ccc} 0 & -
- Question 4, Exercise 2.2 @math-11-nbf:sol:unit02
- Pakistan. =====Question 4(i)===== Find $A$ if \begin{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
- Question 5, Exercise 2.3 @math-11-nbf:sol:unit02
- g matrices if it exists by adjoint method $\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -1 \\ 1 & -2 & -1\end{array}\right]$. ** Solution. ** Given \begin{align*} A &= \left[\begin{array}{ccc} 1 & -1 & 1 \\ 2 & 1 & -1 \\ 1 & -2 & -1 \end{array}\right]\\ |A|&=... gular.\\ Let find the cofactor matrix for $A$.\\ \begin{align*} A_{11} &= (-1)^{1+1} \left|\begin{array}{
- Question 13 Exercise 6.2 @math-11-kpk:sol:unit06
- word "Excellence." How many of these permutations begin with $\mathrm{E}$ ? ====Solution==== The total nu... m_2=2$ are $L$ and $m_3=2$ are $C$. Therefore, \begin{align}\text{total number of permutations are} &=\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) \\ & =\d
- Question 10 Exercise 7.2 @math-11-kpk:sol:unit07
- $. 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-\ldots+i_n^*\... the above equation, we have $(1 \div 1)^n=\left(\begin{array}{l}n \\ 0\end{array}\right)+\left(\begin{ar
- Question 1, Exercise 8.1 @math-11-nbf:sol:unit08
- iven: $\alpha=180^{\circ}$, $\beta=60^{\circ}$. \begin{align*} \cos (\alpha + \beta) & = \cos \alpha \c... -\frac{1}{2} - 0 = -\frac{1}{2} \end{align*} \begin{align*} \cos (\alpha - \beta) & = \cos \alpha \c... = -\frac{1}{2} + 0 = -\frac{1}{2} \end{align*} \begin{align*} \sin (\alpha + \beta) & = \sin \alpha \c... {3}}{2} + 0 = -\frac{\sqrt{3}}{2} \end{align*} \begin{align*} \sin (\alpha - \beta) & = \sin \alpha \c
- Question 7, Exercise 2.3 @math-11-nbf:sol:unit02
- rify that $(A B)^{-1}=B^{-1} A^{-1}$ 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{align*} A &= \left[\begin{array}{ll}2 & 1 \\ 8 & 6\end{array}\right] \\ |A|& = 12 - 8 = 4\\ A^{-1} &=
- Question 2, Exercise 2.5 @math-11-nbf:sol:unit02
- ===== Find the rank of each of the matrix $\left[\begin{array}{ccc}5 & 9 & 3 \\ 3 & -5 & 6 \\ 2 & 10 & 6\end{array}\right]$ ** Solution. ** \begin{align*}&\quad\left[ \begin{array}{ccc} 5 & 9 & 3 \\ 3 & -5 & 6 \\ 2 & 10 & 6 \end{array} \right]\\ \sim & \text{R}\left[ \begin{array}{ccc} 1 & \frac{9}{5} & \frac{3}{5} \\ 3 &