Search
You can find the results of your search below.
Matching pagenames:
- Question 1, Exercise 2.1
- Question 2, Exercise 2.1
- Question 3, Exercise 2.1
- Question 4, Exercise 2.1
- Question 1, Exercise 2.2
- Question 3, Exercise 2.2
- Question 3, Exercise 2.2
- Question 4, Exercise 2.2
- Question 5, Exercise 2.2
- Question 6, Exercise 2.2
- Question 7, Exercise 2.2
- Question 8, Exercise 2.2
- Question 9, Exercise 2.2
- Question 10, Exercise 2.2
- Question 11, Exercise 2.2
- Question 12, Exercise 2.2
- Question 13, Exercise 2.2
- Question 1, Exercise 2.3
- Question 2, Exercise 2.3
- Question 3, Exercise 2.3
- Question 4, Exercise 2.3
- Question 5, Exercise 2.3
- Question 6, Exercise 2.3
- Question 7, Exercise 2.3
- Question 1, Exercise 2.5
- Question 2, Exercise 2.5
- Question 3, Exercise 2.5
- Question 1, Exercise 2.6
- Question 2, Exercise 2.6
- Question 3, Exercise 2.6
- Question 4, Exercise 2.6
- Question 5, Exercise 2.6
- Question 6, Exercise 2.6
- Question 7 and 8, Exercise 2.6
- Question 9 and 10, Exercise 2.6
- Question 1, Review Exercise
- Question 2 and 3, Review Exercise
- Question 4 and 5, Review Exercise
Fulltext results:
- Question 6, Exercise 2.6
- he cofactors of each element . \begin{align*} A_{11} &= (-1)^{1+1} \left| \begin{array}{cc} 1 & 3 \\ ... ay} \right| = 5 - 6 = -1\\ A&= \begin{bmatrix} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\... ix}\\ \implies A^{-1} &= \begin{bmatrix} \frac{1}{11} & \frac{5}{11} & \frac{-4}{11} \\ \frac{5}{22} & \frac{-19}{22} & \frac{18}{22} \\ \frac{-3}{22} & \
- Question 8, Exercise 2.2
- ). Let \begin{align*} A &= \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{bmatrix}\\ B &= \begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \\ b_{31} & b_{32} \end{bmatrix}\\ AB &= \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{bmatrix} \begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \\ b_{31} & b_{32}
- Question 2, Exercise 2.3
- ** Solution. ** The elements of \(R_1\) are \(a_{11} = 3\), \(a_{12} = 2\), and \(a_{13} = 3\). Now w... 4 & 5 & 1 \\ 2 & 1 & 0 \end{array}\right]\\ & A_{11} = (-1)^{1+1} \left|\begin{array}{cc} 5 & 1 \\ 1 ... ind the determinant: \begin{align*} \det(A) &= a_{11} A_{11} + a_{12} A_{12} + a_{13} A_{13} \\ &= 3(-1) + 2(2) + 3(-6) \\ &= -3 + 4 - 18 \\ &= -17 \end{a
- Question 9, Exercise 2.2
- ** Let: \begin{align*} A &= \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \... a_{33} \end{pmatrix} \\ B &= \begin{pmatrix} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \... gn*} \begin{align*} A + B &= \begin{pmatrix} a_{11} + b_{11} & a_{12} + b_{12} & a_{13} + b_{13} \\ a_{21} + b_{21} & a_{22} + b_{22} & a_{23} + b_{23}
- Question 4, Exercise 2.6
- frac{3}{2} & \frac{7}{2} & : & \frac{1}{2} \\ 0 & 11 & -17 & : & 2 \\ 10 & -4 & 18 & : & 7 \end{bmatri... frac{3}{2} & \frac{7}{2} & : & \frac{1}{2} \\ 0 & 11 & -17 & : & 2 0 & 11 & -17 & : & 2 \end{bmatrix}\quad \text{(} R_3 \text{ - 10}R_1)\\ &\sim \text{R}... rac{7}{2} & : & \frac{1}{2} \\ 0 & 1 & -\frac{17}{11} & : & \frac{2}{11} \\ 0 & 0 & 0 & : & 0 \end{bma
- Question 2, Exercise 2.6
- \right|=0\\ &2(12-2)+\lambda(8+3)+1(-4-9)=0\\ &20+11\lambda-13=0\\ &\lambda =-\frac{7}{11}\\ \end{align*} The system becomes \begin{align*} &2 x_{1}+ \frac{7}{11}x_{2}+x_{3}=0 \cdots(iv)\\ &2 x_{1}+3 x_{2}-x_{3}... begin{align*} &\begin{array}{cccc} 2x_1&+\frac{7}{11} x_{2}&+ x_{3}&=0\\ \mathop+\limits_{-}2x_1&\ma
- Question 7 and 8, Exercise 2.6
- the cofactors of each element . \begin{align*} A_{11} &= (-1)^{1+1} \left| \begin{array}{cc} -1 & 2 \\... } 3 & 2 \\ 4 & -1 \end{array} \right| = -3 - 8 = -11 \\ A &= \begin{bmatrix} -3 & 26 & 19 \\ 9 & -16 & 5 \\ 5 & -2 & -11 \end{bmatrix}\\ \text{adj}(A) &= \begin{bmatrix} -3 & 9 & 5 \\ 26 & -16 & -2 \\ 19 & 5 & -11 \end{bmatrix}\\ A^{-1} &= \frac{1}{62} \begin{bma
- Question 1, Exercise 2.2
- {i + 3j}{2} \). For \( i = 1, j = 1 \): \[ a_{11} = \dfrac{1 + 3 \cdot 1}{2} = \dfrac{1 + 3}{2} = ... So we have \begin{align*} A&=\begin{bmatrix}a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \\ &=\... So we have \begin{align*} A &= \begin{bmatrix}a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \\ &= ... ac{i}{j}\): \begin{align*} A &= \begin{bmatrix}a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \\ &=
- Question 4, Exercise 2.2
- begin{array}{cc} -5 & \dfrac{7}{2} \\ 8 & -\dfrac{11}{2} \end{array}\right]. \end{align*} Therefore,... begin{array}{cc} -5 & \dfrac{7}{2} \\ 8 & -\dfrac{11}{2} \end{array}\right]. \end{align*} =====Questi... 1 \\ 2 & 0\end{bmatrix} X=\begin{bmatrix}7 / 2 & 11 & 2 \\ 2 & 4 & 1 \\ 1 & 2 & 0\end{bmatrix}.$$ **... == <text align="left"><btn type="primary">[[math-11-nbf:sol:unit02:ex2-2-p3|< Question 3]]</btn></tex
- Question 11, Exercise 2.2
- ====== Question 11, Exercise 2.2 ====== Solutions of Question 11 of Exercise 2.2 of Unit 02: Matrices and Determinants. ... tbook Board, Islamabad, Pakistan. =====Question 11===== If $A=\left[a_{i j}\right]$ is a matrix of o... == <text align="left"><btn type="primary">[[math-11-nbf:sol:unit02:ex2-2-p10|< Question 10]]</btn></t
- Question 5, Exercise 2.3
- the cofactor matrix for $A$.\\ \begin{align*} A_{11} &= (-1)^{1+1} \left|\begin{array}{cc} 1 & -1 \\ ... q 0$, so $B$ is non-singular.\\ \begin{align*} B_{11} &= (-1)^{1+1} \left|\begin{array}{cc} -1 & -i \\... == <text align="left"><btn type="primary">[[math-11-nbf:sol:unit02:ex2-3-p4|< Question 4]]</btn></tex... t> <text align="right"><btn type="success">[[math-11-nbf:sol:unit02:ex2-3-p6|Question 6 >]]</btn></tex
- Question 10, Exercise 2.2
- == <text align="left"><btn type="primary">[[math-11-nbf:sol:unit02:ex2-2-p9|< Question 9]]</btn></tex... t> <text align="right"><btn type="success">[[math-11-nbf:sol:unit02:ex2-2-p11|Question 11 >]]</btn></text>
- Question 12, Exercise 2.2
- == <text align="left"><btn type="primary">[[math-11-nbf:sol:unit02:ex2-2-p11|< Question 11]]</btn></text> <text align="right"><btn type="success">[[math-11-nbf:sol:unit02:ex2-2-p13|Question 13 >]]</btn></t
- Question 2 and 3, Review Exercise
- } -3 & 4 \\ 2 & 1 \end{array} \right| = -3 - 8 = -11\\ A_{23} &= (-1)^{2+3} \left| \begin{array}{cc} 1... == <text align="left"><btn type="primary">[[math-11-nbf:sol:unit02:Re-ex-p1|< Question 1]]</btn></tex... t> <text align="right"><btn type="success">[[math-11-nbf:sol:unit02:Re-ex-p3|Question 4 & 5 >]]</btn><
- Question 2, Exercise 2.1
- == <text align="left"><btn type="primary">[[math-11-nbf:sol:unit02:ex2-1-p1|< Question 1]]</btn></tex... t> <text align="right"><btn type="success">[[math-11-nbf:sol:unit02:ex2-1-p3|Question 3 >]]</btn></tex