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- Question 1, Exercise 2.6
- em of homogeneous linear equation for non-trivial solution if exists\\ $ 2 x_{1}-3 x_{2}+4 x_{3}=0$\\ $x_{1}... _{2}+3 x_{3}=0$\\ $4 x_{1}+x_{2}-6 x_{3}=0$\\ ** Solution. ** \begin{align*} &2 x_{1}-3 x_{2}+4 x_{3}=0\cdo... 4+36=0 \end{align*} So the system has non-trivial solution. \text{By}\quad(i)-2(ii), we have \begin{align*... es of $x_3$, there are infinite solutions. Hence solution is; \begin{align*} \left[ \begin{array}{c} x_3
- Question 2, Exercise 2.6
- homogeneous linear equation may have non-trivial solution. Also solve the system for value of $\lambda$.\\ ... _{2}-x_{3}=0$\\ $3 x_{1}-2 x_{2}+4 x_{3}=0$\\ ** Solution. ** \begin{align*} &2 x_{1}-\lambda x_{2}+x_{3}=0... \\ \end{align*} Homogenous system has non-trivial solution, if \begin{align*} &\left| \begin{array}{ccc} 2 &... homogeneous linear equation may have non-trivial solution. Also solve the system for value of $\lambda$.\\
- Question 3, Exercise 2.6
- x+3 y+4 z=2$\\ $2 x+y+z=5$\\ $3 x-2 y+z=-3$\\ ** Solution. ** Given the system of equations: \begin{align*}... x &= \frac{46}{19} \end{align*} Therefore, the solution to the system is: $$x = \frac{46}{19}, \quad y ... y+z=2$\\ $2 x+2 y+6 z=1$\\ $3 x-4 y-5 z=3$\\ ** Solution. ** Given the system of equations: \begin{align*}... h is inconsistent, the system of equations has no solution. =====Question 3(iii)===== Solve the system o
- Question 5, Exercise 2.6
- }+3 x_{3}=1$\\ $3 x_{1}-7 x_{2}+4 x_{3}=10$\\ ** Solution. ** The above system may be written as $A X=B$; w... 2}\\ &= \frac{104}{52} = 2 \end{align*} Thus, the solution set is $(3, 1, 2)$. =====Question 5(ii)===== S... {2}+2 x_{3}=1$\\ $8 x_{1}+x_{2}+4 x_{3}=-1$\\ ** Solution. ** The above system maybe written as $AX = B $, ... + 12}{42}\\ &= \frac{0}{42} = 0 \end{align*} The solution set for the given system of equations using Crame
- Question 1, Exercise 2.1
- lll}1 & 3 & 0 \\ 2 & 0 & 1\end{array}\right]$ ** Solution. ** \begin{align}\text{Order of A}&= 2\times 3\e... ll}1 & 2 \\ 2 & 3 \\ 3 & 4\end{array}\right]$ ** Solution. ** \begin{align}\text{Order of B}&= 3\times 2\e... egin{array}{l}1 \\ 6 \\ 9\end{array}\right]$. ** Solution. ** \begin{align}\text{Order of C}&= 3\times 1\e... rray}{llll}2 & 1 & 6 & 8\end{array}\right]$. ** Solution. ** \begin{align}\text{Order of D}&= 1\times 4\e
- Question 2, Exercise 2.1
- ngular matrix, row matrix or column matrix.\\ ** Solution. ** Rectangular matrix =====Question 1(ii)=====... ngular matrix, row matrix or column matrix.\\ ** Solution. ** Square matrix =====Question 1(iii)===== I... ngular matrix, row matrix or column matrix.\\ ** Solution. ** Column matrix =====Question 1(iv)===== Ide... ngular matrix, row matrix or column matrix.\\ ** Solution. ** Square matrix =====Question 1(v)===== Ident
- Question 4, Exercise 2.1
- \\ \sqrt{5} & 6 \\ 1 & 9 \end{array}\right]$$ ** Solution. ** $$ A^t=\begin{bmatrix} 2 & \sqrt{5} & 1 \\ 0... ay}{cccc} 1 & 6 & 2 & 0 \end{array}\right] $$ ** Solution. ** $$B^t=\left[\begin{array}{c} 1 \\ 6 \\ 2 \\ 0... rray}{ll} 2 & 6 \\ 9 & 2 \end{array}\right]$$ ** Solution. ** $$C^t=\left[\begin{array}{ll} 2 & 9 \\ 6 & 2... & 0 & 5 \\ -9 & -5 & 0 \end{array}\right] $$ ** Solution. ** $$D^t=\left[\begin{array}{ccc} 0 & -1 & -9 \
- Question 4, Exercise 2.2
- 1 & 0 \\ 0 & 1 \end{array}\right]\end{align} ** Solution. ** Let $ B = \left[\begin{array}{cc} 2 & 1 \\ 3... & 2 \\ 2 & 4 & 1 \\ 1 & 2 & 0\end{bmatrix}.$$ ** Solution. ** =====Question 4(iii)===== If $A=\left[\begi... nd a non-zero matrix $C$ such that $A C=B C$. ** Solution. ** =====Question 4(iv)===== $\left[\begin{arr... find the values of $z, t$ and $x^{2}+y^{2}$. ** Solution. ** Given $\begin{bmatrix}x y & 4 \\ 0 & x+y\end
- Question 1, Exercise 2.5
- -6 & 8 & 3 \\ -4 & 6 & 5\end{array}\right]$. ** Solution. ** \begin{align*} & \quad \left[\begin{array}{c... l}2 & 1 \\ 3 & 2 \\ 1 & 9\end{array}\right]$. ** Solution. ** \begin{align*} & \quad \left[\begin{array}{cc... \ 4 & 7 & 8 \\ -3 & 1 & 3\end{array}\right]$. ** Solution. ** \begin{align*} & \quad \left[\begin{array}{cc... \\ 4 & 1 & 8 \\ 7 & 3 & 0\end{array}\right]$. ** Solution. ** \begin{align*} &\quad\left[\begin{array}{ccc
- Question 4, Exercise 2.6
- }+3 x_{3}=7$\\ $4 x_{1}+2 x_{2}-5 x_{3}=10$\\ ** Solution. ** \begin{align*} 2x_1 - x_2 - x_3 &= 2, \\ 3x... }\quad R_1 + \frac{1}{2}R_2\end{align*} Thus, the solution to the system of equations is: $$\boxed{x_1 = \fr... -3 x_{3}=4$\\ $10 x_{1}-4 x_{2}+18 x_{3}=7$\\ ** Solution. ** \begin{align*} 2x_1 - 3x_2 + 7x_3 &= 1, \\ ... x_1= \frac{17}{22}\end{align*} Thus, the general solution is: $$\boxed{x_1 = \frac{22}{11}, \quad x_2 = \fr
- Question 6, Exercise 2.6
- +3 y+z=6$\\ $2 x+y+3 z=19$\\ $x+2 y+4 z=25$\\ ** Solution. ** For this system of equations; we have \begin{... {11} \end{bmatrix} \end{align*} Therefore, the solution to the system of equations is: $$x = \frac{1}{11}... y-3 z=5$\\ $2 x-3 y+2 z=1$\\ $-x+2 y-5 z=-3$ ** Solution. ** For this system of equations; we have \begin... 11}{12} \end{bmatrix} \end{align*} Therefore, the solution to the system of equations is: $$ x = \frac{37}{1
- Question 1, Exercise 2.2
- es 2$ for which is $a_{i j}=\dfrac{i+3 j}{2}$ ** Solution. ** Given \( a_{ij} = \dfrac{i + 3j}{2} \). ... for which is $a_{i j}=\dfrac{i \times j}{2}$ ** Solution. ** Given \( a_{ij}=\dfrac{i \times j}{2} \). So ... \times 2$ for which is $a_{i j}=\dfrac{i}{j}$ ** Solution. ** Construct a matrix \(A=\left[a_{ij}\right]\)... 2$ for which is $a_{i j}=\dfrac{2 i-3 j}{3}$ ** Solution. ** Given \( a_{ij} = \frac{2i - 3j}{3} \), we n
- Question 1, Exercise 2.3
- \ 1 & -1 & 2 \\ 4 & 1 & 2\end{array}\right]$. ** Solution. ** Let \begin{align*} A &= \left[\begin{array}{c... s \theta & 0 \\ 0 & 0 & 1\end{array}\right]$. ** Solution. ** Let \begin{align*} A&= \left[\begin{array}{c... \\ 1 & 3 & 4 \\ 0 & 1 & 2\end{array}\right]$. ** Solution. ** \begin{align*} A &= \left[\begin{array}{ccc}... & 2 & 1 \\ -3 i & 1 & 6\end{array}\right]$. ** Solution. ** \begin{align*} A &= \left[\begin{array}{ccc}
- Question 2, Exercise 2.3
- 0\end{array}\right]$ using cofactor method.\\ ** Solution. ** The elements of \(R_1\) are \(a_{11} = 3\), ... & 4\end{array}\right]$ using cofactor method. ** Solution. ** The elements of \(R_1\) are \(a_{11} = 2\), ... 3 i\end{array}\right]$ using cofactor method. ** Solution. ** The elements of \(R_1\) are \(a_{11} = 2i\),... 3\end{array}\right]$ using cofactor method.\\ ** Solution. ** The elements of $R_1$ are $a_{11} = 1-i$, $a
- Question 3, Exercise 2.3
- ght]$ is singular and which are non-singular. ** Solution. ** \begin{align*} A &= \left[\begin{array}{ccc}... ght]$ is singular and which are non-singular. ** Solution. ** \begin{align*} A &= \left[\begin{array}{ccc}... ght]$ is singular and which are non-singular. ** Solution. ** \begin{align*} A &= \left[\begin{array}{ccc}... ht]$ is singular and which are non-singular. ** Solution. ** \begin{align*} A &= \left[\begin{array}{ccc}