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- Question 2, Exercise 2.3 @math-11-kpk:sol:unit02
- the matrix by using elementary row operation. $$\begin{bmatrix}4 & -2 & 5 \\ 2 & 1 & 0 \\ -1 & 2 & 3 \end{bmatrix}$$ ====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) \\
- Question 3, Exercise 2.1 @math-11-kpk:sol:unit02
- shawar, Pakistan. =====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 $\left... =A\left( BC \right)$. ====Solution==== Given: $A=\begin{bmatrix}x & y & z\end{bmatrix}$, $B=\begin{bmatri
- Question 1, Exercise 2.1 @math-11-kpk:sol:unit02
- n 1(i)===== Express as a single 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 \\ \en
- Question 13 Exercise 6.2 @math-11-kpk:sol:unit06
- word "Excellence." How many of these permutations begin with $\mathrm{E}$ ? ====Solution==== The total nu... m_2=2$ are $L$ and $m_3=2$ are $C$. Therefore, \begin{align}\text{total number of permutations are} &=\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) \\ & =\d
- Question 10 Exercise 7.2 @math-11-kpk:sol:unit07
- $. Solution: We know that $$ \left.(1+x)^n=\left(\begin{array}{l} n \\ \vdots \end{array}\right)+\left(\begin{array}{l} m \\ 1 \end{array}\right) x+\left(\begin{array}{l} n \\ 2 \end{array}\right) x^2-\ldots+i_n^*\... the above equation, we have $(1 \div 1)^n=\left(\begin{array}{l}n \\ 0\end{array}\right)+\left(\begin{ar
- Question 1, Exercise 2.3 @math-11-kpk:sol:unit02
- )===== Reduce the matrices to the echelon form: $\begin{bmatrix}1 & 3 & -1 \\2 & 1 & 4 \\3 & 4 & -5\end{bmatrix}$. ====Solution==== \begin{align}&\begin{bmatrix} 1 & 3 & -1 \\ 2 & 1 & 4 \\ 3 & 4 & -5 \end{bmatrix}\\ \underset{\sim}{R}&\begin{bmatrix} 1 & 3 & -1 \\ 0 & -5 & 6 \\ 0 & -5 & -
- Question 9, Exercise 2.1 @math-11-kpk:sol:unit02
- shawar, Pakistan. =====Question 9(i)===== If $A=\begin{bmatrix}2 & -1 & 3 \\1 & \quad 0 & 1 \end{bmatrix},$ $B=\begin{bmatrix}1 & 2 \\2 & 2 \\ 3 & 0 \end{bmatrix}$, ... $( AB )^t=B^tA^t$. ====Solution==== $$A=\left[ \begin{matrix} 2 & -1 & 3 \\ 1 & \quad 0 & 1 \\ \end{matrix} \right],$$ $$B=\left[ \begin{matrix} 1 & 2 \\ 2 & 2 \\ 3 & 0 \\ \e
- Question 11 Exercise 7.1 @math-11-kpk:sol:unit07
- PTBB) Peshawar, Pakistan. =====Question 11===== \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
- Question 10 Exercise 7.1 @math-11-kpk:sol:unit07
- formulas below by mathematical 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{a
- Question 5 & 6, Exercise 2.1 @math-11-kpk:sol:unit02
- awar, Pakistan. =====Question 5===== Matrix $A= \begin{bmatrix} 0 & 2b & -2 \\ 3 & 1 & 3 \\ 3a & 3 & -... alue of $a$ and $b$. ====Solution==== Given: $A=\begin{bmatrix} 0 & 2b & -2 \\ 3 & 1 & 3 \\ 3a & 3 & -1 \end{bmatrix}$ Then $$A^t=\left[ \begin{matrix} 0 & 3 & 3a \\ 2b & 1 & 3 \\ -2... given to be symmetirc, $A^t=A$, implies $$\left[ \begin{matrix} 0 & 3 & 3a \\ 2b & 1 & 3 \\ -2
- Question 16 & 17, Exercise 2.2 @math-11-kpk:sol:unit02
- eshawar, Pakistan. =====Question 16===== Let $A=\begin{bmatrix}3 & -1 \\4 & 2\end{bmatrix}$. Show that ... frac{1}{|A|}$. ====Solution==== Given $$A=\left[ \begin{matrix} 3 & -1 \\ 4 & 2 \\ \end{matrix} \... 1)$$ $$A^{-1}=\dfrac{1}{|A|}AdjA$$ $$AdjA=\left[ \begin{matrix} 2 & 1 \\ -4 & 3 \\ \end{matrix} \right]$$ $$A^{-1}=\dfrac{1}{10}\left[ \begin{matrix} 2 & 1 \\ -4 & 3 \\ \end{matrix} \
- Question 6, Exercise 2.2 @math-11-kpk:sol:unit02
- stan. =====Questiopn 6(i)===== Prov that $\left| \begin{matrix}a-b & b-c & c-a \\b-c & c-a & a-b \\c-a ... -c \end{matrix} \right|=0$ ====Solution==== Let \begin{align} L.H.S&=\left| \begin{matrix} a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c \\ \end{matrix} \right| \\ &=\left| \begin{matrix} a-c & b-a & c-b \\ b-c & c-a & a-b \\ c
- Question 5, Exercise 2.2 @math-11-kpk:sol:unit02
- r, Pakistan. =====Question 5(i)===== Show that $\begin{vmatrix}a & b & c\\l & m & n\\x & y & z \end{vmatrix}=\begin{vmatrix}a & l & x\\b & m & y\\c & n & z \end{vmatrix}$ ====Solution==== \begin{align}L.H.S.&=\begin{vmatrix} a & b & c \\ l & m & n \\ x & y & z \end{vmatrix}\\ &=\begin{vmatrix} a
- Question 6 Exercise 4.1 @math-11-kpk:sol:unit04
- r, \text{ where } r=0,1,2,3,\ldots.$$ For $r=0$ \begin{align}&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
- Question 2, Exercise 2.2 @math-11-kpk:sol:unit02
- ing state the reasons for the equalities. $\left|\begin{matrix} 1 & 2 & 0 \\3 & 1 & 0 \\-1 & 2 & 0 \end... trix}\right|=0$. ====Solution==== Given $$\left| \begin{matrix} 1 & 2 & 0 \\ 3 & 1 & 0 \\ -1 &... g state the reasons for the equalities. $\left| \begin{matrix}1 & 2 & 3 \\-8 & 4 & -12 \\2 & -1 & 3 \e... rix} \right|=0$. ====Solution==== Given $$\left| \begin{matrix} 1 & 2 & 3 \\ -8 & 4 & -12 \\ 2