Search
You can find the results of your search below.
Fulltext results:
- Question 1, Exercise 2.5
- }1 & 3 & 5 \\ -6 & 8 & 3 \\ -4 & 6 & 5\end{array}\right]$. ** Solution. ** \begin{align*} & \quad \left... }1 & 3 & 5 \\ -6 & 8 & 3 \\ -4 & 6 & 5\end{array}\right]\\ \sim & \text{R} \left[\begin{array}{ccc} 1 & 3 & 5 \\ 0 & 26 & 33 \\ 0 & 18 & 25 \end{array}\right]\quad R_2 + 6R_1 \quad R_3 + 4R_1\\ \sim & \text{... 0 & 1 & \frac{33}{26} \\ 0 & 18 & 25 \end{array}\right] \quad \frac{1}{26}R_2 \\ \sim & \text{R} \left[\
- Question 3, Exercise 2.5
- & -1 & -1 \\ -1 & 3 & 0 \\ 1 & -1 & 4\end{array}\right]$ if it exists. Also verify your answer by showin... -1 \\ -1 & 3 & 0 \\ 1 & -1 & 4 \end{array} \right]\\ |A|&=0+1(-4)-1(1-3)\\ &=-4+3\\ &=-1\neq 0\end{... & 1 & 0 \\ 1 & -1 & 4 & 0 & 0 & 1 \end{array} \right]\\ \sim &{\text{R}} \left[ \begin{array}{ccc|ccc}... & 1 & 0 \\ 0 & -1 & -1 & 1 & 0 & 0 \end{array} \right] \quad \text{by swapping } R1 \text{ and } R3\\ \
- Question 5, Exercise 2.3
- & -1 & 1 \\ 2 & 1 & -1 \\ 1 & -2 & -1\end{array}\right]$. ** Solution. ** Given \begin{align*} A &= \... & -1 & 1 \\ 2 & 1 & -1 \\ 1 & -2 & -1 \end{array}\right]\\ |A|&= 1 [-1 - 2] + 1 [-2 + 1] + 1 [-4 - 1] \\... t|\begin{array}{cc} 1 & -1 \\ -2 & -1 \end{array}\right|\\ &= (1) [(1)(-1) - (-1)(-2)] \\ &= -1 - 2 = -3 ... ft|\begin{array}{cc} 2 & -1 \\ 1 & -1 \end{array}\right|\\ &= (-1) [(2)(-1) - (-1)(1)] \\ &= -1 [-2 + 1]
- Question 4, Exercise 2.2
- eft[\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}\right] $ and $ C = \left[\begin{array}{cc} 1 & 3 \\ 2 &
- Question 7, Exercise 2.3
- =\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} &= \dfrac{1}{4} ... left[\begin{array}{ll}6 & -1 \\ -8 & 2\end{array}\right]\\ & = \left[\begin{array}{ll} \frac{6}{4} & \fra
- Question 2, Exercise 2.5
- }5 & 9 & 3 \\ 3 & -5 & 6 \\ 2 & 10 & 6\end{array}\right]$ ** Solution. ** \begin{align*}&\quad\left[ \b... 9 & 3 \\ 3 & -5 & 6 \\ 2 & 10 & 6 \end{array} \right]\\ \sim & \text{R}\left[ \begin{array}{ccc} 1 & \... 3}{5} \\ 3 & -5 & 6 \\ 2 & 10 & 6 \end{array} \right]\quad \frac{1}{5} R1\\ \sim & \text{R}\left[ \beg... 2}{5} & \frac{15}{5} \\ 2 & 10 & 6 \end{array} \right]\quad R2 - 3 \cdot R1\\ \sim & \text{R}\left[ \be
- Question 2, Exercise 2.3
- ll}3 & 2 & 3 \\ 4 & 5 & 1 \\ 2 & 1 & 0\end{array}\right]$ using cofactor method.\\ ** Solution. ** The ... } 3 & 2 & 3 \\ 4 & 5 & 1 \\ 2 & 1 & 0 \end{array}\right]\\ & A_{11} = (-1)^{1+1} \left|\begin{array}{cc} 5 & 1 \\ 1 & 0 \end{array}\right| = (-1)^{2} (5 \cdot 0 - 1 \cdot 1) = (1) (-1) = ... left|\begin{array}{cc} 4 & 1 \\ 2 & 0 \end{array}\right| = (-1)^{3} (4 \cdot 0 - 1 \cdot 2) = (-1) (-2) =
- Question 6, Exercise 2.6
- 5 & 3 & 1 \\ 2 & 1 & 3 \\ 1 & 2 & 4 \end{array} \right|\\ &=5(4 - 6) - 3(8 - 3) + 1(4 - 1)\\ &= -10 - 15... ft| \begin{array}{cc} 1 & 3 \\ 2 & 4 \end{array} \right| = 4 - 6 = -2\\ A_{12} &= (-1)^{1+2} \left| \begin{array}{cc} 2 & 3 \\ 1 & 4 \end{array} \right| = -(8 - 3) = -5\\ A_{13} &= (-1)^{1+3} \left| \begin{array}{cc} 2 & 1 \\ 1 & 2 \end{array} \right| = 4 - 1 = 3\\ A_{21} &= (-1)^{2+1} \left| \begin
- Question 1, Exercise 2.3
- c}2 & 3 & 1 \\ 1 & -1 & 2 \\ 4 & 1 & 2\end{array}\right]$. ** Solution. ** Let \begin{align*} A &= \left... c}2 & 3 & 1 \\ 1 & -1 & 2 \\ 4 & 1 & 2\end{array}\right]\\ |A|&=2(-2-2)-3(2-8)+1(1+4)\\ \implies |A|&=-8+... \theta & \cos \theta & 0 \\ 0 & 0 & 1\end{array}\right]$. ** Solution. ** Let \begin{align*} A&= \left... \theta & \cos \theta & 0 \\ 0 & 0 & 1\end{array}\right]\\ |A| &= cos \theta (cos \theta -0)+ sin \theta(
- Question 7, Exercise 2.2
- =\left[\begin{array}{ll}x & 0 \\ y & 1\end{array}\right]$ then prove that for all positive integers $n, A... egin{array}{cc}x^{n} & 0 \\ \dfrac{y\left(x^{n}-1\right)}{x-1} & 1\end{array}\right]$. ** Solution. ** Given: $$A = \begin{bmatrix} x & 0 \\ y & 1 \end{bmatri... left[\begin{array}{ll}3 & -4 \\ 1 & -1\end{array}\right]$ then prove that for all positive integers $n$,
- Question 3, Exercise 2.6
- 2 \\ 2 & 1 & 1 & 5 \\ 3 & -2 & 1 & -3 \end{array}\right]\\ & \sim \text{R}\left[\begin{array}{cccc} 2 & 3... -3 & 3 \\ 0 & -\frac{13}{2} & -5 & -6 \end{array}\right]\quad R_2 - R_1 \text{and}\quad R_3 - \frac{3}{2}... 0 & 0 & \frac{19}{4} & -\frac{63}{4} \end{array}\right]\quad R_3 - \frac{13}{4} R_2 \end{align*} Now \be... 2 \\ 2 & 2 & 6 & 1 \\ 3 & -4 & -5 & 3 \end{array}\right]\\ &\sim \text{R}\left[\begin{array}{cccc} 2 & 2
- Question 6, Exercise 2.3
- c}2 & 1 & -3 \\ 0 & 1 & 0 \\ 2 & 1 & 6\end{array}\right]$ then find $A^{-1}$ and hence show that $A A^{-1... & 0 & 1 & 0 \\ 2 & 1 & 6 & 0 & 0 & 1 \end{array}\right] \\ &=\left[\begin{array}{ccc|ccc} 2 & 1 & -3 & 1... & 0 & 1 & 0 \\ 0 & 0 & 9 & -1 & 0 & 1 \end{array}\right] \quad R_3 - R_1 \\ &=\left[\begin{array}{ccc|cc... & 1 & -\frac{1}{9} & 0 & \frac{1}{9} \end{array}\right] \quad \frac{1}{9} R_3\\ &=\left[\begin{array}{c
- Question 7 and 8, Exercise 2.6
- }3 & 2 & 1 \\ 4 & -1 & 2 \\ 7 & 3 & -3\end{array}\right]$; find $A^{-1}$ and hence solve the system of eq... | \begin{array}{cc} -1 & 2 \\ 3 & -3 \end{array} \right| = 3 - 6 = -3 \\ A_{12} &= (-1)^{1+2} \left| \begin{array}{cc} 4 & 2 \\ 7 & -3 \end{array} \right| = -(-12 - 14) = 26 \\ A_{13} &= (-1)^{1+3} \left| \begin{array}{cc} 4 & -1 \\ 7 & 3 \end{array} \right| = 12 + 7 = 19 \\ A_{21} &= (-1)^{2+1} \left| \be
- Question 4, Exercise 2.1
- }{ccc} 2 & 0 \\ \sqrt{5} & 6 \\ 1 & 9 \end{array}\right]$$ ** Solution. ** $$ A^t=\begin{bmatrix} 2 & \... eft[\begin{array}{cccc} 1 & 6 & 2 & 0 \end{array}\right] $$ ** Solution. ** $$B^t=\left[\begin{array}{c} 1 \\ 6 \\ 2 \\ 0 \end{array}\right] $$ =====Question 4(iii)===== Find the transpos... left[\begin{array}{ll} 2 & 6 \\ 9 & 2 \end{array}\right]$$ ** Solution. ** $$C^t=\left[\begin{array}{ll
- Question 3, Exercise 2.3
- }3 & 1 & 2 \\ 2 & 3 & 1 \\ -4 & 1 & -3\end{array}\right]$ is singular and which are non-singular. ** Sol... 3 & 1 & 2 \\ 2 & 3 & 1 \\ -4 & 1 & -3\end{array}\right]\end{align*} The determinant of a \(3 \times 3\) ... }3 & -1 & 2 \\ 2 & 0 & 1 \\ -1 & 5 & 1\end{array}\right]$ is singular and which are non-singular. ** Sol... 3 & -1 & 2 \\ 2 & 0 & 1 \\ -1 & 5 & 1\end{array}\right]\\ |A| &= 3(0 \cdot 1 - 1 \cdot 5) - (-1)(2 \cdot