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- Question 16 & 17, Exercise 2.2
- d{matrix} \right]$$ $$|A|=6+4$$ $$\Rightarrow |A|=10\ldots (1)$$ $$A^{-1}=\dfrac{1}{|A|}AdjA$$ $$AdjA=... 3 \\ \end{matrix} \right]$$ $$A^{-1}=\dfrac{1}{10}\left[ \begin{matrix} 2 & 1 \\ -4 & 3 \\ ... } \right]$$ $$=\left[ \begin{matrix} \dfrac{2}{10} & \dfrac{1}{10} \\ -\dfrac{4}{10} & \dfrac{3}{10} \\ \end{matrix} \right]$$ $$A^{-1}=\left[ \be
- Question 1, Exercise 2.1
- matrix} \right]\left[ \begin{matrix} 14 \\ 10 \\ 16 \\ \end{matrix} \right]\\ &=\left[ 14+... atrix} \right]\\ &=\left[ \begin{matrix} -19 & 10 & -3 \\ \end{matrix} \right]-\left[ \begin{matri... rix} \right]\\ &=\left[ \begin{matrix} -19-2 & 10+5 & -3-7 \\ \end{matrix} \right]\\ &=\left[ \begin{matrix} -21 & 15 & -10 \\ \end{matrix} \right] \end{align} =====Questi
- Question 14 & 15, Exercise 2.2
- \\ 0 & 5 \\ \end{matrix} \right|$$ $$A_{21}=-10$$ $$A_{22}=(-1)^{2+2}\left| \begin{matrix} 0 &... 33}=2$$ $$Adj\,\,A=\left[ \begin{matrix} 15 & -10 & -2 \\ 7 & -2 & -2 \\ -3 & 2 & 2 \\ \en... A^{-1}=\dfrac{1}{8}\left[ \begin{matrix} 15 & -10 & -2 \\ 7 & -2 & -2 \\ -3 & 2 & 2 \\ \en... =\left[ \begin{matrix} \dfrac{15}{8} & \dfrac{-10}{8} & -\dfrac{2}{8} \\ \dfrac{7}{4} & -\dfrac
- Question 2, Exercise 2.3
- }{49} \\ -\dfrac{6}{49} & \dfrac{17}{49} & \dfrac{10}{49}\\ \dfrac{5}{49} & -\dfrac{6}{49} & \dfrac{8}... {49} \\ -\dfrac{6}{49} & \dfrac{17}{49} & \dfrac{10}{49} \\ \dfrac{5}{49} & -\dfrac{6}{49} & \dfrac{... c{1}{49}\begin{bmatrix} 3 & 16 & -5 \\ -6 & 17 & 10 \\ 5 & -6 & 8 \end{bmatrix} \end{align} =====Q... \sim}{R}&\left[ \begin{matrix} 1 & 3 & 4 \\ 0 & -10 & -6 \\ 0 & 8 & 5 \end{matrix}\left| \begin{matr
- Question 4, Exercise 2.3
- & 4 & 5 \\3 & 4 & 5 & 6 \\4 & 5 & 6 & 7 \\9 & 10 & 11 & 12\end{bmatrix}$ ====Solution==== \begin{a... 4 & 5 \\ 3 & 4 & 5 & 6 \\ 4 & 5 & 6 & 7 \\ 9 & 10 & 11 & 12 \end{bmatrix}\\ \underset{\sim}{R}&\beg... 4 & 5 \\ 1 & 1 & 1 & 1 \\ 4 & 5 & 6 & 7 \\ 9 & 10 & 11 & 12 \end{bmatrix} \text{by}R_2-R_1\\ \under... 1 & 1 \\ 2 & 3 & 4 & 5 \\ 4 & 5 & 6 & 7 \\ 9 & 10 & 11 & 12 \end{bmatrix} \text{by}R_1\leftrightarr
- Question 2, Exercise 2.1
- & -1\end{bmatrix}$. Then $2A=\begin{bmatrix}4 & -10 & 2\\6 & 0 & -8\end{bmatrix}$, $3B=\begin{bmatrix... . \begin{align}&2A+3B-4C\\&=\begin{bmatrix} 4 & -10 & 2\\ 6 & 0 & -8\end{bmatrix}+\begin{bmatrix}3 & ... & -4 & -4\end{bmatrix}\\ &=\begin{bmatrix}4+3 & -10-6 & 2-9 \\6+0 & 0-3 & -8+15 \end{bmatrix} -\begin
- Question 10, Exercise 2.1
- ====== Question 10, Exercise 2.1 ====== Solutions of Question 10 of Exercise 2.1 of Unit 02: Matrices and Determinants. ... KPTB or KPTBB) Peshawar, Pakistan. =====Question 10===== Let $A=\begin{bmatrix}1 & -3 & 4 \\-3 & 2 &
- Question 8,9 & 10, Exercise 2.2
- ====== Question 8,9 & 10, Exercise 2.2 ====== Solutions of Questions 8,9 & 10 of Exercise 2.2 of Unit 02: Matrices and Determin... $=(x-p)(x-q)(x+q+p)$$ $$=R.H.S.$$ =====Question 10===== Prove that $\left| \begin{matrix}1+a & 1 & 1
- Question 11, Exercise 2.2
- 1 & 1 \\ \end{matrix} \right]$$ $$|A|=7(2+2)-1(6+10)+3(6-10)$$ $$=28-16-12$$ $$|A|=0$$ $A$ is singular. =====Question 11(ii)===== Identify singular and... ath-11-kpk:sol:unit02:ex2-2-p8 |< Question 8, 9 & 10]]</btn></text> <text align="right"><btn type="suc
- Question 3, Exercise 2.1
- B-C)&=\left[ \begin{matrix} -7 & 11 & -2 \\ -7 & 10 & 16 \\ \end{matrix} \right] ... (1) \end{align}... AB-AC&=\left[\begin{matrix} -7 & 11 & -2 \\ -7 & 10 & 16 \\ \end{matrix}\right]... (2) \end{align} Fr
- Question 12, Exercise 2.1
- ^t=\left[ \begin{matrix} 6 & 6 & -1 \\ 6 & 10 & 9 \\ -1 & 9 & 8 \\ \end{matrix} \right]$$ ... ^t=\left[ \begin{matrix} 6 & 6 & -1 \\ 6 & 10 & 9 \\ -1 & 9 & 8 \\ \end{matrix} \right]$$
- Question 8, Exercise 2.1
- matrix} \right]$$ $$A^tA=\left[ \begin{matrix} 10 & -1 & 12 \\ -1 & 5 & -8 \\ 12 & -4 & 16
- Question 9, Exercise 2.1
- xt align="right"><btn type="success">[[math-11-kpk:sol:unit02:ex2-1-p9|Question 10 >]]</btn></text>
- Question 11, Exercise 2.1
- ry">[[math-11-kpk:sol:unit02:ex2-1-p9 |< Question 10]]</btn></text> <text align="right"><btn type="suc
- Question 7, Exercise 2.2
- ="right"><btn type="success">[[math-11-kpk:sol:unit02:ex2-2-p8|Question 8,9 & 10 >]]</btn></text>