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- Question 2, Exercise 2.3 @math-11-kpk:sol:unit02
- {bmatrix}4 & -2 & 5 \\ 2 & 1 & 0 \\ -1 & 2 & 3 \end{bmatrix}$$ ====Solution==== Let $$A=\begin{bmatrix} 4 & -2 & 5 \\ 2 & 1 & 0 \\ -1 & 2 & 3 \end{bmatrix}.$$ Then \begin{align}|A|&=\begin{vmatrix}4 & -2 & 5 \\ 2 & 1 & 0 \\ -1 & 2 & 3 \end{vmatrix}\\ &=4(3)+2(6)+5(4+1) \\ & =49\neq 0. \end{align} This gives, $A$ is non-singular and $A^{-1
- Question 3, Exercise 2.1 @math-11-kpk:sol:unit02
- Question 3(i)===== If $A=\begin{bmatrix}x & y & z\end{bmatrix}$, $B=\begin{bmatrix}a & h & g\\h & b & f\\g & f & c\end{bmatrix}$ and $C=\begin{bmatrix}x\\y\\z\end{bmatrix}$. Verify that $\left( AB \right)C=A\left( BC \rig... ==Solution==== Given: $A=\begin{bmatrix}x & y & z\end{bmatrix}$, $B=\begin{bmatrix}a & h & g\\h & b & f
- Question 1, Exercise 2.1 @math-11-kpk:sol:unit02
- le matrix $$\left[ \begin{matrix} 1 & 2 & 4 \\ \end{matrix} \right] \left[ \begin{matrix} 1 & 0 & 2 \\ 2 & 0 & 1 \\ 0 & 1 & 2 \\ \end{matrix} \right] \left[ \begin{matrix} 2 \\ 4 \\ 6 \\ \end{matrix} \right]$$ ====Solution==== \begin{align}&\left[ \begin{matrix} 1 & 2 & 4 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 & 2
- Question 10 Exercise 7.2 @math-11-kpk:sol:unit07
- \left.(1+x)^n=\left(\begin{array}{l} n \\ \vdots \end{array}\right)+\left(\begin{array}{l} m \\ 1 \end{array}\right) x+\left(\begin{array}{l} n \\ 2 \end{array}\right) x^2-\ldots+i_n^*\right) x^n \cdot $$ ... e have $(1 \div 1)^n=\left(\begin{array}{l}n \\ 0\end{array}\right)+\left(\begin{array}{l}n \\ 1\end{ar
- Question 13 Exercise 6.2 @math-11-kpk:sol:unit06
- re} &=\left(\begin{array}{c} n \\ m_1, m_2, m_3 \end{array}\right)\\&=\left(\begin{array}{c} 10 \\ 4,2,2 \end{array}\right) \\ & =\dfrac{10 !}{4 ! \cdot 2 ! \cdot 2 !}\\ &=37,800 \end{align} Begin with $\mathrm{E}$ If we have to pic... are} &=\left(\begin{array}{c} n \\ m_1, m_2, m_3 \end{array}\right)\\&=\dfrac{9 !}{3 ! \cdot 2 ! \cdot
- Question 1, Exercise 2.3 @math-11-kpk:sol:unit02
- gin{bmatrix}1 & 3 & -1 \\2 & 1 & 4 \\3 & 4 & -5\end{bmatrix}$. ====Solution==== \begin{align}&\begin{... bmatrix} 1 & 3 & -1 \\ 2 & 1 & 4 \\ 3 & 4 & -5 \end{bmatrix}\\ \underset{\sim}{R}&\begin{bmatrix} 1 & 3 & -1 \\ 0 & -5 & 6 \\ 0 & -5 & -2 \end{bmatrix} \text{ by } R_2-2R_1 \text{ and } R_3-3R... trix} 1 & 3 & -1 \\ 0 & -5 & 6 \\ 0 & -5 & -2 \end{bmatrix} \text{ by } R_3-R_2 \\ \underset{\sim}{R
- Question 9, Exercise 2.1 @math-11-kpk:sol:unit02
- $A=\begin{bmatrix}2 & -1 & 3 \\1 & \quad 0 & 1 \end{bmatrix},$ $B=\begin{bmatrix}1 & 2 \\2 & 2 \\ 3 & 0 \end{bmatrix}$, show that $( AB )^t=B^tA^t$. ====Solut... matrix} 2 & -1 & 3 \\ 1 & \quad 0 & 1 \\ \end{matrix} \right],$$ $$B=\left[ \begin{matrix} 1 & 2 \\ 2 & 2 \\ 3 & 0 \\ \end{matrix} \right]$$ $$A^t=\left[ \begin{matrix}
- Question 11 Exercise 7.1 @math-11-kpk:sol:unit07
- == \begin{align} & \left(\begin{array}{l} 2 \\ 2 \end{array}\right)+\left(\begin{array}{l} 3 \\ 2 \end{array}\right)+\left(\begin{array}{l} 4 \\ 2 \end{array}\right)+\ldots+\left(\begin{array}{l} n \\ 2 \end{array}\right)=\left(\begin{array}{c} n+1 \\ 3 \en
- Question 10 Exercise 7.1 @math-11-kpk:sol:unit07
- ematical induction, $\left(\begin{array}{1}5 \\5 \end{array}\right)+\left(\begin{array}{l}6 \\ 5\end{array}\right)+\left(\begin{array}{l}7 \\ 5\end{array}\right)+\ldots+\left(\begin{array}{c}n+4 \\ 5\end{array}\right)=\left(\begin{array}{c}n+5 \\ 6\end{
- Question 5 & 6, Exercise 2.1 @math-11-kpk:sol:unit02
- atrix} 0 & 2b & -2 \\ 3 & 1 & 3 \\ 3a & 3 & -1 \end{bmatrix}$ is given to be symmetric. Find the valu... atrix} 0 & 2b & -2 \\ 3 & 1 & 3 \\ 3a & 3 & -1 \end{bmatrix}$ Then $$A^t=\left[ \begin{matrix} 0 &... 3 & 3a \\ 2b & 1 & 3 \\ -2 & 3 & -1 \\ \end{matrix} \right]$$ Since $A$ is given to be symmet... 3 & 3a \\ 2b & 1 & 3 \\ -2 & 3 & -1 \\ \end{matrix} \right]=\left[ \begin{matrix} 0 & 2b &
- Question 16 & 17, Exercise 2.2 @math-11-kpk:sol:unit02
- ion 16===== Let $A=\begin{bmatrix}3 & -1 \\4 & 2\end{bmatrix}$. Show that $|A^{-1}|=\dfrac{1}{|A|}$. =... \left[ \begin{matrix} 3 & -1 \\ 4 & 2 \\ \end{matrix} \right]$$ $$|A|=6+4$$ $$\Rightarrow |A|=1... \left[ \begin{matrix} 2 & 1 \\ -4 & 3 \\ \end{matrix} \right]$$ $$A^{-1}=\dfrac{1}{10}\left[ \begin{matrix} 2 & 1 \\ -4 & 3 \\ \end{matrix} \right]$$ $$=\left[ \begin{matrix} \df
- Question 6, Exercise 2.2 @math-11-kpk:sol:unit02
- b-c & c-a \\b-c & c-a & a-b \\c-a & a-b & b-c \end{matrix} \right|=0$ ====Solution==== Let \begin{al... c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c \\ \end{matrix} \right| \\ &=\left| \begin{matrix} a-c &... c-b \\ b-c & c-a & a-b \\ c-a & a-b & b-c \\ \end{matrix} \right| \quad R_1+R_2 \\ &=-\left| \begin... b-c \\ b-c & c-a & a-b \\ c-a & a-b & b-c \\ \end{matrix} \right|\quad -R_1 \\ &=0 \quad R_1 \cong
- Question 5, Exercise 2.2 @math-11-kpk:sol:unit02
- $\begin{vmatrix}a & b & c\\l & m & n\\x & y & z \end{vmatrix}=\begin{vmatrix}a & l & x\\b & m & y\\c & n & z \end{vmatrix}$ ====Solution==== \begin{align}L.H.S.&=\... n{vmatrix} a & b & c \\ l & m & n \\ x & y & z \end{vmatrix}\\ &=\begin{vmatrix} a & b & c \\l & m & n \\x & y & z \end{vmatrix}^t \quad \because |A|=|A^t| \\ &=\begin{v
- Question 6 Exercise 4.1 @math-11-kpk:sol:unit04
- }&P_{0+1}=\dfrac{5-0}{0+1} P_0\\ \implies &P_1=5.\end{align} For $r=1$ \begin{align}&P_{1+1}=\dfrac{5-1}{1+1} P_1\\ \implies &P_2=2\cdot 5=10.\end{align} For $r=2$ \begin{align}&P_{2+1}=\dfrac{5-2}{2+1} P_2\\ \implies &P_3=1\cdot 10=10\end{align} For $r=3$ \begin{align}&P_{3+1}=\dfrac{5-3}{3+1} P_3\\ \implies &P_4=\dfrac{10}{2}=5.\end{align} For $r=4$ \begin{align}&P_{4+1}=\dfrac{5-4
- Question 2, Exercise 2.2 @math-11-kpk:sol:unit02
- gin{matrix} 1 & 2 & 0 \\3 & 1 & 0 \\-1 & 2 & 0 \end{matrix}\right|=0$. ====Solution==== Given $$\left... 1 & 2 & 0 \\ 3 & 1 & 0 \\ -1 & 2 & 0 \\ \end{matrix} \right|=0$$ Because elements of third col... n{matrix}1 & 2 & 3 \\-8 & 4 & -12 \\2 & -1 & 3 \end{matrix} \right|=0$. ====Solution==== Given $$\lef... 2 & 3 \\ -8 & 4 & -12 \\ 2 & -1 & 3 \\ \end{matrix} \right|=0$$ Taking $-4$ common from $R_2$