Search
You can find the results of your search below.
Matching pagenames:
Fulltext results:
- Question 18, Exercise 2.2
- atrix.\\ $$A=\left[ \begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{matrix} \right]$$ $$|A|=a_{11}a_{22}-a_{12}a_{21}$$ $$AdjA=\left[ \begin{matrix} a_{22} & -a_{12} \\ -a_{21} & a_{11} \\ \end{matrix} \right]... 1}{|A|}AdjA$$ $$A^{-1}=\dfrac{1}{a_{11}}a_{22}-a_{12}a_{21}\left[ \begin{matrix} a_{22} & -a_{12}
- Question 13, Exercise 2.1
- tion==== $$A=\left[ \begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ ... gin{matrix} a_{11} & a_{21} & a_{31} \\ a_{12} & a_{22} & a_{32} \\ a_{13} & a_{23} & a_{33... )$$ $$A+A^t=\left[ \begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ ... gin{matrix} a_{11} & a_{21} & a_{31} \\ a_{12} & a_{22} & a_{32} \\ a_{13} & a_{23} & a_{33
- Question 3, Exercise 2.2
- lution==== Let $$A=\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ ... 11} \left( a_{22} a_{33}-a_{23} a_{32} \right)-a_{12}\left( a_{21}a_{33}-a_{23}a_{31} \right)+a_{13}\l... ght)\\ &=a_{11}a_{22}a_{33}-a_{11}a_{23}a_{32}-a_{12}a_{21}a_{33}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}-a_{13}a_{22}a_{31}\\ &=a_{11}a_{22}a_{33}+a_{12}a_
- Question 1, Exercise 2.1
- \end{matrix} \right]\left[ \begin{matrix} 2+0+12 \\ 4+0+6 \\ 0+4+12 \\ \end{matrix} \right] \\ &=\left[ \begin{matrix} 1 & 2 & 4 \\ \end{... 2 & 4 & 6 \end{bmatrix} \\ &=\begin{bmatrix}4+2 & 12+4 & 20+6 \\-2-6 & -6-12 & -10-18 \end{bmatrix} \\ &=\begin{bmatrix}6 & 16 & 26 \\-8 & -18 & -28 \end
- Question 8, Exercise 2.1
- ]$$ $$A^tA=\left[ \begin{matrix} 1+9 & 2-3 & 0+12 \\ 2-3 & 4+1 & 0-8 \\ 0+12 & 0-4 & 0+16 \\ \end{matrix} \right]$$ $$A^tA=\left[ \begin{matrix} 10 & -1 & 12 \\ -1 & 5 & -8 \\ 12 & -4 & 16 \\ \end{matrix} \right]$$ $$AA^t\ne A^tA$$ ====Go To==== <t
- Question 12, Exercise 2.1
- ====== Question 12, Exercise 2.1 ====== Solutions of Question 12 of Exercise 2.1 of Unit 02: Matrices and Determinants. ... KPTB or KPTBB) Peshawar, Pakistan. =====Question 12(i)===== Let $A=\begin{bmatrix}3 & 2 & 1 \\4 & 5 ... right]$$ $$( A+A^t )^t=( A+A^t )$$ =====Question 12(ii)===== Let $A=\begin{bmatrix}3 & 2 & 1 \\ 4 & 5
- Question 4, Exercise 2.3
- \\3 & 4 & 5 & 6 \\4 & 5 & 6 & 7 \\9 & 10 & 11 & 12\end{bmatrix}$ ====Solution==== \begin{align}&\beg... 3 & 4 & 5 & 6 \\ 4 & 5 & 6 & 7 \\ 9 & 10 & 11 & 12 \end{bmatrix}\\ \underset{\sim}{R}&\begin{bmatrix... 1 & 1 & 1 & 1 \\ 4 & 5 & 6 & 7 \\ 9 & 10 & 11 & 12 \end{bmatrix} \text{by}R_2-R_1\\ \underset{\sim}{... 2 & 3 & 4 & 5 \\ 4 & 5 & 6 & 7 \\ 9 & 10 & 11 & 12 \end{bmatrix} \text{by}R_1\leftrightarrow R_2\\ \
- Question 3, Exercise 2.1
- matrix} \right]\\ &=\left[ \begin{matrix}2-0 & 1+12 & -3+6 \\-2+0 & -1+16 & 3+8 \\\end{matrix} \ri... ft[ \begin{matrix}-1-6 & 2+9 & -8+6 \\ 1-8 & -2+12 & 8+8\\ \end{matrix} \right]\\ \implies A(B-C)&=\... matrix}\right] \\ &=\left[ \begin{matrix}2+0 & 1+12 & -3+6\\ -2+0 & -1+16 & 3+8\\ \end{matrix}\right]
- Question 12, Exercise 2.2
- ====== Question 12, Exercise 2.2 ====== Solutions of Question 12 of Exercise 2.2 of Unit 02: Matrices and Determinants. ... (KPTB or KPTBB) Peshawar, Pakistan. =====Question 12===== Find the value of $\lambda $, if $A$ is sing
- Question 13, Exercise 2.2
- & x+4 \\ \end{matrix} \right|=0$$ $$(x+2)(x^2+7x+12-12)-3(2x+8-8)+4(6-2x-6)=0$$ $$(x+2)(x^2+7x)-6x-8x=0$$ $$x^3+2x^2+7x^2+14x-14x=0$$ $$x^3+9x^2=0$$ $$x^... y">[[math-11-kpk:sol:unit02:ex2-2-p10 |< Question 12]]</btn></text> <text align="right"><btn type="suc
- Question 14 & 15, Exercise 2.2
- $$Adj\,\,A={{\left[ \begin{matrix} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ ... 5 \\ \end{matrix} \right|$$ $$A_{11}=15$$ $$A_{12}=(-1)^{1+2}\left| \begin{matrix} -1 & 2 \\ 1 & 5 \\ \end{matrix} \right|$$ $$A_{12}=7$$ $$A_{13}=(-1)^{1+3}\left| \begin{matrix}
- Question 1, Exercise 2.2
- Now \begin{align}|A|&={{a}_{11}}{{A}_{11}}+{{a}_{12}}{{A}_{12}}+{{a}_{13}}{{A}_{13}}\\ &=1\left( -4 \right)+3\left( -2 \right)+1\left( -4 \right) \\ \imp
- Question 2, Exercise 2.2
- es. $\left| \begin{matrix}1 & 2 & 3 \\-8 & 4 & -12 \\2 & -1 & 3 \end{matrix} \right|=0$. ====Soluti... eft| \begin{matrix} 1 & 2 & 3 \\ -8 & 4 & -12 \\ 2 & -1 & 3 \\ \end{matrix} \right|=0$$ Ta
- Question 4, Exercise 2.2
- ft( -1-2 \right)+3\left( -1-4 \right)\\ =&0+3-15=-12 \end{align} =====Question 4(ii)===== Evaluate th... -4 & 2 & 0 \\ \end{matrix} \right|\\ =&3\left( 0+12 \right)-4\left( 0-24 \right)-2\left( 4+16 \right)
- Question 11, Exercise 2.2
- \right]$$ $$|A|=7(2+2)-1(6+10)+3(6-10)$$ $$=28-16-12$$ $$|A|=0$$ $A$ is singular. =====Question 11(i... xt align="right"><btn type="success">[[math-11-kpk:sol:unit02:ex2-2-p10|Question 12 >]]</btn></text>