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- Question 1, Exercise 2.5
- {array}{ccc}1 & 3 & 5 \\ -6 & 8 & 3 \\ -4 & 6 & 5\end{array}\right]$. ** Solution. ** \begin{align*} ... {array}{ccc}1 & 3 & 5 \\ -6 & 8 & 3 \\ -4 & 6 & 5\end{array}\right]\\ \sim & \text{R} \left[\begin{array}{ccc} 1 & 3 & 5 \\ 0 & 26 & 33 \\ 0 & 18 & 25 \end{array}\right]\quad R_2 + 6R_1 \quad R_3 + 4R_1\\ ... & 3 & 5 \\ 0 & 1 & \frac{33}{26} \\ 0 & 18 & 25 \end{array}\right] \quad \frac{1}{26}R_2 \\ \sim & \te
- Question 6, Exercise 2.6
- gin{bmatrix} 5 & 3 & 1 \\ 2 & 1 & 3 \\ 1 & 2 & 4 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 6 \\ 19 \\ 25 \end{bmatrix} \end{align*} And \begin{align*} |A|& = \left| \begin{array}{ccc} 5 & 3 & 1 \\ 2 & 1 & 3 \\ 1
- Question 3, Exercise 2.5
- rray}{ccc}0 & -1 & -1 \\ -1 & 3 & 0 \\ 1 & -1 & 4\end{array}\right]$ 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
- 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
- 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 7, Exercise 2.3
- {-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
- {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
- Question 7, Exercise 2.2
- ===== If $A=\left[\begin{array}{ll}x & 0 \\ y & 1\end{array}\right]$ then prove that for all positive i... {n} & 0 \\ \dfrac{y\left(x^{n}-1\right)}{x-1} & 1\end{array}\right]$. ** Solution. ** Given: $$A = \begin{bmatrix} x & 0 \\ y & 1 \end{bmatrix}.$$ We use mathematical induction to prov... matrix} x^1 & 0 \\ \dfrac{y(x^1 - 1)}{x - 1} & 1 \end{bmatrix} \\ = \begin{bmatrix} x & 0 \\ \dfrac{y(x
- Question 2, Exercise 2.3
- in{array}{lll}3 & 2 & 3 \\ 4 & 5 & 1 \\ 2 & 1 & 0\end{array}\right]$ using cofactor method.\\ ** Solut... {array}{ccc} 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+2} \left|\begin{array}{cc} 4 & 1 \\ 2 & 0 \end{array}\right| = (-1)^{3} (4 \cdot 0 - 1 \cdot 2)
- Question 7 and 8, Exercise 2.6
- {array}{ccc}3 & 2 & 1 \\ 4 & -1 & 2 \\ 7 & 3 & -3\end{array}\right]$; find $A^{-1}$ and hence solve the... n{bmatrix} 3 & 2 & 1 \\ 4 & -1 & 2 \\ 7 & 3 & -3 \end{bmatrix}\\ |A| &=3(-3) - 2(-26) + 19\\ &= -9 + 52 + 19\\ &= 62 \neq 0\end{align*} This system is consistent. Now to find $A... ^{1+1} \left| \begin{array}{cc} -1 & 2 \\ 3 & -3 \end{array} \right| = 3 - 6 = -3 \\ A_{12} &= (-1)^{1+
- Question 13, Exercise 2.2
- Y=\left[\begin{array}{ccc}1 & 6 & -3 \\ 2 & 1 & 7\end{array}\right]$ and $X+3 Y=\left[\begin{array}{ccc}4 & 3 & 2 \\ 1 & -3 & 0\end{array}\right]$. ** Solution. ** Given: \begin{a... 2X - Y = \begin{pmatrix} 1 & 6 & -3 \\ 2 & 1 & 7 \end{pmatrix} \cdots (i)\\ X + 3Y = \begin{pmatrix} 4 & 3 & 2 \\ 1 & -3 & 0 \end{pmatrix} \cdots (ii) \end{align*} From (i) we hav
- Question 5, Exercise 2.6
- {bmatrix} 1 & 1 & 2 \\ -1 & -2 & 3 \\ 3 & -7 & 4 \end{bmatrix}, \quad X = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}, \quad B = \begin{bmatrix} 8 \\ 1 \\ 10 \end{bmatrix}\\ |A| &= \begin{vmatrix} 1 & 1 & 2 \\ -1 & -2 & 3 \\ 3 & -7 & 4 \end{vmatrix}\\ &= 1(-8 +21) - 1(-4 - 9) + 2(7 +6)\\ &
- Question 3, Exercise 2.6
- 4z &= 2 \\ 2x + y + z &= 5 \\ 3x - 2y + z &= -3 \end{aligned}\end{align*} The associated augmented matrix is: \begin{align*} A_{b} &=\quad \left[\begin{a... & 3 & 4 & 2 \\ 2 & 1 & 1 & 5 \\ 3 & -2 & 1 & -3 \end{array}\right]\\ & \sim \text{R}\left[\begin{array... \ 0 & -2 & -3 & 3 \\ 0 & -\frac{13}{2} & -5 & -6 \end{array}\right]\quad R_2 - R_1 \text{and}\quad R_3
- Question 6, Exercise 2.3
- n{array}{ccc}2 & 1 & -3 \\ 0 & 1 & 0 \\ 2 & 1 & 6\end{array}\right]$ then find $A^{-1}$ and hence show ... in{bmatrix} 2 & 1 & -3 \\ 0 & 1 & 0 \\ 2 & 1 & 6 \end{bmatrix} \end{align*} To find the inverse $ A^{-1} $ of the matrix $ A $, we will use the method of ... \ 0 & 1 & 0 & 0 & 1 & 0 \\ 2 & 1 & 6 & 0 & 0 & 1 \end{array}\right] \\ &=\left[\begin{array}{ccc|ccc} 2
- Question 2, Exercise 2.6
- ts(ii)\\ &3 x_{1}-2 x_{2}+4 x_{3}=0\cdots(iii)\\ \end{align*} Homogenous system has non-trivial solutio... 2 & -\lambda & 1 \\ 2 & 3 & -1 \\ 3 & -2 & 4 \end{array} \right|=0\\ &2(12-2)+\lambda(8+3)+1(-4-9)=... \ &20+11\lambda-13=0\\ &\lambda =-\frac{7}{11}\\ \end{align*} The system becomes \begin{align*} &2 x_{... dots(v)\\ &3 x_{1}-2 x_{2}+4 x_{3}=0\cdots(vi)\\ \end{align*} (iv)-(v), we have\\ \begin{align*} &\begi