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- Question 2, Exercise 2.5
- } \\ 0 & 0 & 1 \end{array} \right] \end{align*} There are $3$ non-zero rows.\\ The rank of the matrix i... nd{array} \right] \quad R3 - 12R2 \\ \end{align*} There are $2$ non-zero rows.\\ The rank of the matrix i... rray} \right] \quad \frac{3}{124}R3 \end{align*} There are $3$ non-zero rows.\\ The rank of the matrix i... \right] \quad R1 - 3R2, \, R3 - 3R2 \end{align*} There are $2$ non-zero rows.\\ The rank of the matrix i
- Question 4, Exercise 2.1
- 9 & 5 & 0 \end{array}\right] $$ Since $D^t=-D$, therefore $D$ is skew-symmetric. =====Question 4(v)===... \ 9 & 0 & 0 \end{array}\right] $$ Since $E^t=E$, therefore $E$ is symmetric. =====Question 4(vi)===== F
- Question 1, Exercise 2.3
- (-2i) (1) \\ &= 2i - 6 - 2i \\ &= -6\end{align*} Therefore, the determinant of the matrix \(\left[\begin... \))} \\ &= 22 + 8i - 6 \\ &= 16 + 8i \end{align*} Therefore, the determinant of the matrix \(\left[\begin
- Question 1, Exercise 2.6
- \end{align*} With the different values of $x_3$, there are infinite solutions. Hence solution is; \beg... {5}x_3 = 0 \\ &x_1 = -\frac{7}{5}x_3 \end{align*} Therefore, the solution set is: \begin{align*} &\left[
- Question 6, Exercise 2.6
- 1} \\ \frac{70}{11} \end{bmatrix} \end{align*} Therefore, the solution to the system of equations is: ... 7}{3} \\ \frac{11}{12} \end{bmatrix} \end{align*} Therefore, the solution to the system of equations is:
- Question 4, Exercise 2.2
- -\dfrac{11}{2} \end{array}\right]. \end{align*} Therefore, the matrix \( A \) is: \begin{align*} A &=
- Question 10, Exercise 2.2
- B\\ &=B\end{align*} Now, $$A^2 + B^2 = A + B$$ Therefore, given the conditions $AB = B$ and $BA = A$,
- Question 13, Exercise 2.2
- & 0 & 1 \\ 0 & -1 & -1 \end{pmatrix} \end{align*} Therefore, the matrices $X$ and $Y$ are: \begin{align*}
- Question 7, Exercise 2.3
- -1 & \frac{1}{4} \end{array}\right] \end{align*} Therefore, $(AB)^{-1} = B^{-1}A^{-1}$ is verified for t
- Question 3, Exercise 2.6
- frac{92}{19}\\ x &= \frac{46}{19} \end{align*} Therefore, the solution to the system is: $$x = \frac
- Question 4, Exercise 2.6
- } R_2 \text{ by 11 and } R_3 - R_2). \end{align*} There is no value of $x$. Then $x_3 = 0$. From the s
- Question 7 and 8, Exercise 2.6
- 62} & \dfrac{-11}{62} \end{bmatrix} \end{align*} Therefore, the inverse of matrix $A $ is: $$A^{-1} = \