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- Question 4, Exercise 2.6
- {1}-4 x_{2}+3 x_{3}=7$\\ $4 x_{1}+2 x_{2}-5 x_{3}=10$\\ ** Solution. ** \begin{align*} 2x_1 - x_2 -... 3x_1 - 4x_2 + 3x_3 &= 7, \\ 4x_1 + 2x_2 - 5x_3 &= 10, \end{align*} The associative augment matrix: \b... & : & 2 \\ 3 & -4 & 3 & : & 7 \\ 4 & 2 & -5 & : & 10 \end{bmatrix}\\ &\sim \text{R}\begin{bmatrix} 1 &... \\ 3 & -4 & 3 & \vert & 7 \\ 4 & 2 & -5 & \vert & 10 \end{bmatrix}\quad \dfrac{1}{2}\\ &\sim \text{R}\
- Question 6, Exercise 2.6
- \right|\\ &=5(4 - 6) - 3(8 - 3) + 1(4 - 1)\\ &= -10 - 15 + 3\\ &=-22 \end{align*} This system is cons... 3 & 1 \\ 2 & 4 \end{array} \right| = -(12 - 2) = -10\\ A_{22} &= (-1)^{2+2} \left| \begin{array}{cc} 5... array}{cc} 5 & 3 \\ 1 & 2 \end{array} \right| = -(10 - 3) = -7\\ A_{31} &= (-1)^{3+1} \left| \begin{ar... nd{bmatrix}\\ A &=\begin{bmatrix} -2 & -5 & 3 \\ -10 & 19 & -7 \\ 8 & -18 & -1 \end{bmatrix}\\ \text{a
- Question 5, Exercise 2.6
- {1}-2 x_{2}+3 x_{3}=1$\\ $3 x_{1}-7 x_{2}+4 x_{3}=10$\\ ** Solution. ** The above system may be writt... end{bmatrix}, \quad B = \begin{bmatrix} 8 \\ 1 \\ 10 \end{bmatrix}\\ |A| &= \begin{vmatrix} 1 & 1 & 2 ... \frac{ \begin{vmatrix} 8 & 1 & 2 \\ 1 & -2 & 3 \\ 10 & -7 & 4 \end{vmatrix}}{52}\\ &=\frac{8(-8 + 21) - 1(4 - 30) + 2(-7 + 20)}{52}\\ &=\frac{104 + 26 + 26}{52}\\ & = \frac{156}{52} = 3 \end{ali
- Question 2, Exercise 2.5
- [\begin{array}{ccc}5 & 9 & 3 \\ 3 & -5 & 6 \\ 2 & 10 & 6\end{array}\right]$ ** Solution. ** \begin{a... egin{array}{ccc} 5 & 9 & 3 \\ 3 & -5 & 6 \\ 2 & 10 & 6 \end{array} \right]\\ \sim & \text{R}\left[ ... \frac{9}{5} & \frac{3}{5} \\ 3 & -5 & 6 \\ 2 & 10 & 6 \end{array} \right]\quad \frac{1}{5} R1\\ \s... }{5} \\ 0 & -\frac{52}{5} & \frac{15}{5} \\ 2 & 10 & 6 \end{array} \right]\quad R2 - 3 \cdot R1\\ \
- Question 4, Exercise 2.3
- {200}\\ & = \dfrac{-14 - 2i}{200}\\ &= -\dfrac{7}{100} - \dfrac{i}{100} \end{align*} Hence $\lambda = -\dfrac{7}{100} - \dfrac{i}{100}$ ====Go to ==== <text align="left"><btn type="primary">[[math-11-nbf:sol:unit02:
- Question 10, Exercise 2.2
- ====== Question 10, Exercise 2.2 ====== Solutions of Question 10 of Exercise 2.2 of Unit 02: Matrices and Determinants. ... tbook Board, Islamabad, Pakistan. =====Question 10===== If $A$ and $B$ are two matrices such that $A
- Question 2, Exercise 2.3
- ht| = (-1)^{4} (4 \cdot 1 - 5 \cdot 2) = (1) (4 - 10) = -6 \end{align*} Now, we use these cofactors to... 1)^{3} (-1 \cdot 4 - 2 \cdot 3) = (-1) (-4 - 6) = 10 \\ & A_{13} = (-1)^{1+3} \left|\begin{array}{cc} ... } + a_{12} A_{12} + a_{13} A_{13} \\ &= 2(-2) + 3(10) + (-1)(-1) \\ &= -4 + 30 + 1 \\ &= 27 \end{align
- Question 3, Exercise 2.3
- \\ &= 3(-9 - 1) - 1(-6 + 4) + 2(2 + 12) \\ &= 3(-10) - 1(-2) + 2(14) \\ &= -30 + 2 + 28 \\ &= 0 \end{... - 0 \cdot (-1)) \\ &= 3(0 - 5) - (-1)(2 + 1) + 2(10 - 0) \\ &= 3(-5) + 1(3) + 2(10) \\ &= -15 + 3 + 20 \\ &= 8\end{align*} Since $|A| = 8 \neq 0$, the ma
- Question 2, Exercise 2.6
- &3 x_{1}+ \frac{30}{11} x_{3}=0\\ & x_{1}=- \frac{10}{11} x_{3}\\ \end{align*} Hence S.S$=\left[ \begin{array}{c} - \dfrac{10}{11} x_{3} \\ \dfrac{11}{13}x_3 \\ x_3 \end{a... &= 0 \\ 8x_{2} - 6x_{3} + 2x_{2} + x_{3} &= 0 \\ 10x_{2} - 5x_{3} &= 0 \\ x_{2} &= \frac{1}{2}x_{3} \
- Question 9 and 10, Exercise 2.6
- ====== Question 9 and 10, Exercise 2.6 ====== Solutions of Question 9 and 10 of Exercise 2.6 of Unit 02: Matrices and Determin... \alpha-3 \beta$. ** Solution. ** =====Question 10===== By making use of matrix of order $2$ by $2$
- Question 9, Exercise 2.2
- text align="right"><btn type="success">[[math-11-nbf:sol:unit02:ex2-2-p10|Question 10 >]]</btn></text>
- Question 11, Exercise 2.2
- tn type="primary">[[math-11-nbf:sol:unit02:ex2-2-p10|< Question 10]]</btn></text> <text align="right"><btn type="success">[[math-11-nbf:sol:unit02:ex2-2-p
- Question 1, Review Exercise
- * (d) all of these \\ <btn type="link" collapse="a10">See Answer</btn><collapse id="a10" collapsed="true">%%(d)%%: all of these</collapse> ====Go to =
- Question 7 and 8, Exercise 2.6
- t align="right"><btn type="success">[[math-11-nbf:sol:unit02:ex2-6-p8|Question 9 & 10 >]]</btn></text>
- Question 2 and 3, Review Exercise
- {cc} 1 & 2 \\ -3 & 4 \end{array} \right| = 4 +6 = 10\\ |A|&= 1(24-9)-2(-18-18)+0\\ &=15+72\\ &=87 \end