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- Question 2, Exercise 2.6
- _{2}+4 x_{3}=0\cdots(vi)\\ \end{align*} (iv)-(v), we have\\ \begin{align*} &\begin{array}{cccc} 2x_1&+... 3\\ \end{align*} Put the value of $x_2$ in (vi), we have \begin{align*} &3 x_{1}-2(\frac{11}{13})x_{3... lign*} If $\lambda=2$, put the value of $\lambda$ we have \begin{align*} x_{1} - 4x_{2} + 3x_{3} &= 0 ... = 0 \quad \text{(iii)} \end{align*} By using (i), we have \begin{align*} x_{1} &= 4x_{2} - 3x_{3} \q
- Question 2, Exercise 2.3
- 1} = 3\), \(a_{12} = 2\), and \(a_{13} = 3\). Now we find their corresponding cofactors. \begin{align*... 5 \cdot 2) = (1) (4 - 10) = -6 \end{align*} Now, we use these cofactors to find the determinant: \beg... } = 2\), \(a_{12} = 3\), and \(a_{13} = -1\). Now we find their corresponding cofactors. \begin{align*... 1 - 0 \cdot 3) = (1) (-1) = -1 \end{align*} Now, we use these cofactors to find the determinant: \beg
- Question 1, Exercise 2.6
- non-trivial solution. \text{By}\quad(i)-2(ii), we have \begin{align*} &\begin{array}{cccc} 2x_1&-3 ... \\ \end{align*} Put the value of $x_3$ in (iii), we have \begin{align*} &4 x_{1}+2x_{3}-6 x_{3}=0\\ &... {(iii)} \end{align*} For the system of equations, we have: \begin{align*} A &= \left[ \begin{array}{cc... tem has a non-trivial solution. By $(i)- 2(ii)$, we get: \begin{align*} &\begin{array}{cccc} &2x_1 &-
- Question 1, Exercise 2.2
- Method:** Given \( a_{ij}=\dfrac{i+3j}{2} \). So we have \begin{align*} A&=\begin{bmatrix}a_{11} & ... . ** Given \( a_{ij}=\dfrac{i \times j}{2} \). So we have \begin{align*} A &= \begin{bmatrix}a_{11} & ... tion. ** Given \( a_{ij} = \frac{2i - 3j}{3} \), we need to find the matrix \( A \): \[ A = \begin{b... _{12} \\ a_{21} & a_{22} \end{bmatrix} \] First, we calculate each element \( a_{ij} \): \[ \begin{a
- Question 6, Exercise 2.6
- \\ ** Solution. ** For this system of equations; we have \begin{align*} A &= \begin{bmatrix} 5 & 3 & ... This system is consistent. Now to find $A^{-1}$, we calculate the cofactors of each element . \begin{... $ ** Solution. ** For this system of equations; we have \begin{align*} A& = \begin{bmatrix} 1 & 2 & ... This system is consistent. Now to find $A^{-1}$, we calculate the cofactors of each element . \begin{
- Question 4, Exercise 2.2
- z \\ t & 6\end{bmatrix}$. Equating the elements, we get $$xy=8, z=4, t=0. x+y=6$$ As we know \begin{align*} &(x+y)^2=x^2+y^2+2xy \\ \implies & x^2+y^2 ... ies & x^2+y^2 = 6^2-2(8) =20 \end{align*} Hence, we conclude $z=4$, $t=0$ and $x^2+y^2=20$. =====Que
- Question 7, Exercise 2.2
- = \begin{bmatrix} x & 0 \\ y & 1 \end{bmatrix}.$$ We use mathematical induction to prove the given fac... c{y(x^k - 1)}{x - 1} & 1 \end{bmatrix}\end{align} We need to show that the formula holds for $ k + 1 $... 4 n \\ n & 1-2 n \end{bmatrix} $ ** Solution. ** we use mathematical induction. Put $n = 1$ \begin{a
- Question 3, Exercise 2.5
- 2} \end{array} \right]\\ \end{align*} To verify, we need to show that \( A A^{-1} = A^{-1} A = I \): ... 15-4)+5(-6)\\ &=-2+19-3-19\neq 0 \end{align*} Now we will find $A^{-1}$ \begin{align*} &\quad\left[ \b... -1 & 1 & 1 \end{array} \right] \end{align*} Now we show that $A A^{-1}=A^{-1} A=I$ \begin{align*} A
- Question 3, Exercise 2.6
- = -\frac{63}{19}\end{align*} From the second row, we have: \begin{align*} &- 2y - 3z = 3\\ \Rightarrow... 66}{19}\end{align*} Finally, from the first row, we have: \begin{align*} 2x + 3y + 4z &= 2 \\ 2x + 3(... ]\quad R_3 + 3R_2 \end{align*} From the last row, we have: \begin{align*} &-2z = 1 \\ &\Rightarrow \qu
- Question 6, Exercise 2.2
- comparing corresponding elements in the matrices, we get: \begin{align}&7 + \alpha = 2\beta \cdots (1)... cdots (2) \end{align} Using $\beta = -1$ in (1), we get: \begin{align} & 7 + \alpha = 2(-1)\\ \implie
- Question 13, Exercise 2.2
- 0 \end{pmatrix} \cdots (ii) \end{align*} From (i) we have \begin{align*} Y = 2X - \begin{pmatrix} 1 & ... d{pmatrix} \end{align*} Put value of $Y$ in (ii), we have \begin{align*} X + 3\left(2X - \begin{pmatr
- Question 6, Exercise 2.3
- find the inverse $ A^{-1} $ of the matrix $ A $, we will use the method of row reduction (Gaussian el... 0 & 0 & 1 \end{bmatrix} = I_3 \end{align*} Thus, we have shown that $ A A^{-1} = A^{-1} A = I_3 $.
- Question 7 and 8, Exercise 2.6
- This system is consistent. Now to find $A^{-1}$, we calculate the cofactors of each element . \begin{... 3\quad R_1-13R_3 \end{align*} From above equation we get, \begin{align*} x_1&=1\\ x_2&=1\\ x_3&=1 \end
- Question 10, Exercise 2.2
- fore, given the conditions $AB = B$ and $BA = A$, we find that: $$A^2 + B^2 = A + B $$ ====Go to ==
- Question 4, Exercise 2.3
- bda - 4 + 6\lambda \\ &= \lambda - 4 \end{align*} We know that the matrix $A$ is singular, then: \begi