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- Exercise 2.2 (Solutions)
- ====== Exercise 2.2 (Solutions) ====== The solutions of the Exercise 2.2 of book “Model Textbook of Mathematics for Class XI” published b... {2 i-3 j}{3}$\\ [[math-11-nbf:sol:unit02:ex2-2-p1|Solution:Question 1]] **Question 2.** Construct a matrix ... +j^{2}}{i+j}$\\ [[math-11-nbf:sol:unit02:ex2-2-p2|Solution:Question 2]] **Question 3.** If $A=\left[\begin{
- Question 1, Exercise 2.6
- ====== Question 1, Exercise 2.6 ====== Solutions of Question 1 of Exercise 2.6 of Unit 02: Matrices and Dete... 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*
- Exercise 2.3 (Solutions)
- ====== Exercise 2.3 (Solutions) ====== The solutions of the Exercise 2.3 of book “Model Textbook of Mathematics for Class XI” published b... rray}\right]$\\ [[math-11-nbf:sol:unit02:ex2-3-p1|Solution:Question 1]] **Question. 2** Evaluate the deter... rray}\right]$\\ [[math-11-nbf:sol:unit02:ex2-3-p2|Solution:Question 2]] **Question. 3** Determine which of
- Question 2, Exercise 2.6
- ====== Question 2, Exercise 2.6 ====== Solutions of Question 2 of Exercise 2.6 of Unit 02: Matrices and Dete... 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 &
- Review Exercise 2 (Solutions)
- ====== Review Exercise 2 (Solutions) ====== The solutions of the Review Exercise 2 of book “Model Textbook of Mathematics for Class XI” publis... m of homogeneous linear equations has non-trivial solution if:\\ (a) $|A|>0$\\ (b) $|A|<0$\\ %%(c)%% $|A|=0$... ions with three variables system will have unique solution if:\\ (a) $\operatorname{RankA}<3$\\ (b) $\operat
- Question 3, Exercise 2.6
- ====== Question 3, Exercise 2.6 ====== Solutions of Question 3 of Exercise 2.6 of Unit 02: Matrices and Dete... x+3 y+4 z=2$\\ $2 x+y+z=5$\\ $3 x-2 y+z=-3$\\ ** Solution. ** Given the system of equations: \begin{align*}... x &= \frac{46}{19} \end{align*} Therefore, the solution to the system is: $$x = \frac{46}{19}, \quad y ... y+z=2$\\ $2 x+2 y+6 z=1$\\ $3 x-4 y-5 z=3$\\ ** Solution. ** Given the system of equations: \begin{align*}
- Question 5, Exercise 2.6
- ====== Question 5, Exercise 2.6 ====== Solutions of Question 5 of Exercise 2.6 of Unit 02: Matrices and Dete... }+3 x_{3}=1$\\ $3 x_{1}-7 x_{2}+4 x_{3}=10$\\ ** Solution. ** The above system may be written as $A X=B$; w... 2}\\ &= \frac{104}{52} = 2 \end{align*} Thus, the solution set is $(3, 1, 2)$. =====Question 5(ii)===== S... {2}+2 x_{3}=1$\\ $8 x_{1}+x_{2}+4 x_{3}=-1$\\ ** Solution. ** The above system maybe written as $AX = B $,
- Question 1, Exercise 2.1
- ====== Question 1, Exercise 2.1 ====== Solutions of Question 1 of Exercise 2.1 of Unit 02: Matrices and Dete... lll}1 & 3 & 0 \\ 2 & 0 & 1\end{array}\right]$ ** Solution. ** \begin{align}\text{Order of A}&= 2\times 3\e... ll}1 & 2 \\ 2 & 3 \\ 3 & 4\end{array}\right]$ ** Solution. ** \begin{align}\text{Order of B}&= 3\times 2\e... egin{array}{l}1 \\ 6 \\ 9\end{array}\right]$. ** Solution. ** \begin{align}\text{Order of C}&= 3\times 1\e
- Question 2, Exercise 2.1
- ====== Question 2, Exercise 2.1 ====== Solutions of Question 2 of Exercise 2.1 of Unit 02: Matrices and Dete... ngular matrix, row matrix or column matrix.\\ ** Solution. ** Rectangular matrix =====Question 1(ii)=====... ngular matrix, row matrix or column matrix.\\ ** Solution. ** Square matrix =====Question 1(iii)===== I... ngular matrix, row matrix or column matrix.\\ ** Solution. ** Column matrix =====Question 1(iv)===== Ide
- Question 4, Exercise 2.1
- ====== Question 4, Exercise 2.1 ====== Solutions of Question 4 of Exercise 2.1 of Unit 02: Matrices and Dete... \\ \sqrt{5} & 6 \\ 1 & 9 \end{array}\right]$$ ** 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. ** $$C^t=\left[\begin{array}{ll} 2 & 9 \\ 6 & 2
- Question 4, Exercise 2.2
- ====== Question 4, Exercise 2.2 ====== Solutions of Question 4 of Exercise 2.2 of Unit 02: Matrices and Dete... 1 & 0 \\ 0 & 1 \end{array}\right]\end{align} ** Solution. ** Let $ B = \left[\begin{array}{cc} 2 & 1 \\ 3... & 2 \\ 2 & 4 & 1 \\ 1 & 2 & 0\end{bmatrix}.$$ ** Solution. ** =====Question 4(iii)===== If $A=\left[\begi... nd a non-zero matrix $C$ such that $A C=B C$. ** Solution. ** =====Question 4(iv)===== $\left[\begin{arr
- Question 1, Exercise 2.5
- ====== Question 1, Exercise 2.5 ====== Solutions of Question 1 of Exercise 2.5 of Unit 02: Matrices and Dete... -6 & 8 & 3 \\ -4 & 6 & 5\end{array}\right]$. ** 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
- Question 4, Exercise 2.6
- ====== Question 4, Exercise 2.6 ====== Solutions of Question 4 of Exercise 2.6 of Unit 02: Matrices and Dete... }+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: $$\boxed{x_1 = \fr... -3 x_{3}=4$\\ $10 x_{1}-4 x_{2}+18 x_{3}=7$\\ ** Solution. ** \begin{align*} 2x_1 - 3x_2 + 7x_3 &= 1, \\
- Question 6, Exercise 2.6
- ====== Question 6, Exercise 2.6 ====== Solutions of Question 6 of Exercise 2.6 of Unit 02: Matrices and Dete... +3 y+z=6$\\ $2 x+y+3 z=19$\\ $x+2 y+4 z=25$\\ ** Solution. ** For this system of equations; we have \begin{... {11} \end{bmatrix} \end{align*} Therefore, the solution to the system of equations is: $$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
- Question 7 and 8, Exercise 2.6
- ====== Question 7 and 8, Exercise 2.6 ====== Solutions of Question 7 and 8 of Exercise 2.6 of Unit 02: Matri... y+7 z=14 ; 2 x-y+3 z=4 ; \quad x+2 y-3 z=0$. ** Solution. ** Given \begin{align*} A &= \begin{bmatrix} 3 ... align*} x_1&=1\\ x_2&=1\\ x_3&=1 \end{align*} Now solutions of above equations are; $$ \begin{bmatrix} \dfra... f $\lambda$ for which the following system has no solution, unique solution or infinitely many solutions.\\