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- Question 2, Exercise 2.6
- _{1}-2 x_{2}+4 x_{3}=0$\\ ** Solution. ** \begin{align*} &2 x_{1}-\lambda x_{2}+x_{3}=0 \cdots(i)\\ &2 x... i)\\ &3 x_{1}-2 x_{2}+4 x_{3}=0\cdots(iii)\\ \end{align*} Homogenous system has non-trivial solution, if \begin{align*} &\left| \begin{array}{ccc} 2 & -\lambda & 1 \\... 0+11\lambda-13=0\\ &\lambda =-\frac{7}{11}\\ \end{align*} The system becomes \begin{align*} &2 x_{1}+ \f
- Question 3, Exercise 2.6
- olution. ** Given the system of equations: \begin{align*} \begin{aligned} 2x + 3y + 4z &= 2 \\ 2x + y + z &= 5 \\ 3x - 2y + z &= -3 \end{aligned}\end{align*} The associated augmented matrix is: \begin{align*} A_{b} &=\quad \left[\begin{array}{cc
- Question 1, Exercise 2.6
- x_{1}+x_{2}-6 x_{3}=0$\\ ** Solution. ** \begin{align*} &2 x_{1}-3 x_{2}+4 x_{3}=0\cdots (i)\\ &x_{1}-2... ii)\\ &4 x_{1}+x_{2}-6 x_{3}=0\cdots (iii)\\ \end{align*} For system of equation, \begin{align*} A &= \left[ \begin{array}{ccc} 2 & -3 & 4 \\ 1 & -2 & 3 \\ 4... ight]\\ |A|&=2(9)+3(-18)+4(9)\\ &=18-54+36=0 \end{align*} So the system has non-trivial solution. \text
- Question 5, Exercise 2.3
- d{array}\right]$. ** Solution. ** Given \begin{align*} A &= \left[\begin{array}{ccc} 1 & -1 & 1 \\ 2 &... -1) + 1 \cdot (-5) \\ &= -3 - 1 - 5 \\ &= -9 \end{align*} Thus, $|A| = -9 \neq 0$, so $A$ is non-singular... \\ Let find the cofactor matrix for $A$.\\ \begin{align*} A_{11} &= (-1)^{1+1} \left|\begin{array}{cc} 1 ... |\\ &= (1) [(1)(1) - (-1)(2)]\\ &= 1 + 2 = 3 \end{align*} \begin{align*} adj(A) &= \left[\begin{array}{cc
- Question 5, Exercise 2.6
- e system may be written as $A X=B$; where, \begin{align*} &A = \begin{bmatrix} 1 & 1 & 2 \\ -1 & -2 & 3 \... - 1(-4 - 9) + 2(7 +6)\\ &= 13 + 13 + 26 = 52\end{align*} So, $A$ is non-singular. \begin{align*} x_1 &= \frac{A_1}{|A|}\\ &=\frac{ \begin{vmatrix} 8 & 1 & 2 ... {104 + 26 + 26}{52}\\ & = \frac{156}{52} = 3 \end{align*} \begin{align*} x_2& = \frac{|A_2|}{|A|} \\ &=\f
- Question 1, Exercise 2.5
- & 5\end{array}\right]$. ** Solution. ** \begin{align*} & \quad \left[\begin{array}{ccc}1 & 3 & 5 \\ -6... & 1 \end{array}\right]\quad \frac{26}{217}R_3\end{align*} This is echelon form \begin{align*} \sim & \text{R} \left[\begin{array}{ccc} 1 & 3 & 0 \\ 0 & 1 & 0 \\... 0 & 0 & 1 \end{array}\right]\quad R_1 - 3R_2 \end{align*} This is reduce echelon form. =====Question 1(i
- Question 4, Exercise 2.2
- tan. =====Question 4(i)===== Find $A$ if \begin{align}\left[\begin{array}{cc} 2 & 1 \\ 3 & 2 \end{arra... array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]\end{align} ** Solution. ** Let $ B = \left[\begin{array}{... {cc} 1 & 3 \\ 2 & 4 \end{array}\right] $ \begin{align*}BAC &= I,\\ A &= B^{-1} I C^{-1} = B^{-1} C^{-1}. \end{align*} The inverse of a 2x2 matrix $ \left[\begin{arr
- Question 2, Exercise 2.3
- Now we find their corresponding cofactors. \begin{align*} A &= \left[\begin{array}{ccc} 3 & 2 & 3 \\ 4 & ... (4 \cdot 1 - 5 \cdot 2) = (1) (4 - 10) = -6 \end{align*} Now, we use these cofactors to find the determinant: \begin{align*} \det(A) &= a_{11} A_{11} + a_{12} A_{12} + a_{1... ) + 2(2) + 3(-6) \\ &= -3 + 4 - 18 \\ &= -17 \end{align*} Thus, the determinant of the matrix \(\left[\be
- Question 3, Exercise 2.5
- A^{-1}=A^{-1} A=I$.\\ ** Solution. ** Let \begin{align*} A&=\left[ \begin{array}{ccc} 0 & -1 & -1 \\ -... ]\\ |A|&=0+1(-4)-1(1-3)\\ &=-4+3\\ &=-1\neq 0\end{align*} So $A$ is non singular. Now consider \begin{align*} &\quad\left[ \begin{array}{ccc|ccc} 0 & -1 & -1 & ... ac{1}{2} & \frac{1}{2} \end{array} \right]\\ \end{align*} To verify, we need to show that \( A A^{-1} =
- Question 7, Exercise 2.2
- prove the given fact. For C-1, put $n = 1$ \begin{align}A^1 =\begin{bmatrix} x^1 & 0 \\ \dfrac{y(x^1 - 1)... \begin{bmatrix} x & 0 \\ y & 1 \end{bmatrix}\end{align} C-1 is satisfied. \\ For C-2, suppose given statement is true for $n=k$. \begin{align} A^k = \begin{bmatrix} x^k & 0 \\ \frac{y(x^k - 1)}{x - 1} & 1 \end{bmatrix}\end{align} We need to show that the formula holds for $ k +
- Question 4, Exercise 2.3
- & 1\end{array}\right]$. ** Solution. ** \begin{align*} A &= \left[\begin{array}{ccc} \lambda & 1 & 3 \... &= -23\lambda - 2 + 18 \\ &= -23\lambda + 16 \end{align*} $A$ is singular, Then \begin{align*} |A|&=0\\ -23\lambda + 16 &= 0\\ -23\lambda &= -16 \\ \lambda &= \dfrac{16}{23} \end{align*} Thus, $\lambda = \dfrac{16}{23}$. =====Question
- Question 4, Exercise 2.6
- }+2 x_{2}-5 x_{3}=10$\\ ** Solution. ** \begin{align*} 2x_1 - x_2 - x_3 &= 2, \\ 3x_1 - 4x_2 + 3x_3 &= 7, \\ 4x_1 + 2x_2 - 5x_3 &= 10, \end{align*} The associative augment matrix: \begin{align*} A_b &= \begin{bmatrix} 2 & -1 & -1 & : & 2 \\ 3 & -4 ... }{21} \end{bmatrix}\quad R_1 + \frac{1}{2}R_2\end{align*} Thus, the solution to the system of equations i
- Question 6, Exercise 2.6
- . ** For this system of equations; we have \begin{align*} A &= \begin{bmatrix} 5 & 3 & 1 \\ 2 & 1 & 3 \\ ... \begin{bmatrix} 6 \\ 19 \\ 25 \end{bmatrix} \end{align*} And \begin{align*} |A|& = \left| \begin{array}{ccc} 5 & 3 & 1 \\ 2 & 1 & 3 \\ 1 & 2 & 4 \end{array} \r... (8 - 3) + 1(4 - 1)\\ &= -10 - 15 + 3\\ &=-22 \end{align*} This system is consistent. Now to find $A^{-1}$
- Question 1, Exercise 2.1
- 0 & 1\end{array}\right]$ ** Solution. ** \begin{align}\text{Order of A}&= 2\times 3\end{align} ===== Question 1(ii) ===== Find the order of the following ... 3 & 4\end{array}\right]$ ** Solution. ** \begin{align}\text{Order of B}&= 3\times 2\end{align} ===== Question 1(iii)===== Find the order of the following matr
- Question 3, Exercise 2.3
- which are non-singular. ** Solution. ** \begin{align*} A &= \left[\begin{array}{ccc} 3 & 1 & 2 \\ 2 & 3 & 1 \\ -4 & 1 & -3\end{array}\right]\end{align*} The determinant of a \(3 \times 3\) matrix is calculated as follows: \begin{align*} |A| &= 3(3 \cdot (-3) - 1 \cdot 1) - 1(2 \cdot ... ) - 1(-2) + 2(14) \\ &= -30 + 2 + 28 \\ &= 0 \end{align*} Since $|A| = 0$, the matrix $A$ is a singular m