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- Question 2, Exercise 2.3
- 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
- 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
- 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 1, Exercise 2.3
- )===== 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
- 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==== $$A=\left[ \begin{matrix} 2 & -1 & 3 \\ 1 & \quad 0 & 1 \\ \end{matrix} \right],$$ $$B=\left[ \begin{matrix} 1 & 2 \\ 2 & 2 \\ 3 & 0 \\ \e
- Question 5 & 6, Exercise 2.1
- awar, Pakistan. =====Question 5===== Matrix $A= \begin{bmatrix} 0 & 2b & -2 \\ 3 & 1 & 3 \\ 3a & 3 & -... alue of $a$ and $b$. ====Solution==== Given: $A=\begin{bmatrix} 0 & 2b & -2 \\ 3 & 1 & 3 \\ 3a & 3 & -1 \end{bmatrix}$ Then $$A^t=\left[ \begin{matrix} 0 & 3 & 3a \\ 2b & 1 & 3 \\ -2... given to be symmetirc, $A^t=A$, implies $$\left[ \begin{matrix} 0 & 3 & 3a \\ 2b & 1 & 3 \\ -2
- Question 16 & 17, Exercise 2.2
- eshawar, Pakistan. =====Question 16===== Let $A=\begin{bmatrix}3 & -1 \\4 & 2\end{bmatrix}$. Show that ... frac{1}{|A|}$. ====Solution==== Given $$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
- 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
- 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 2, Exercise 2.2
- 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
- Question 18, Exercise 2.2
- is $2\times 2$ non-singular matrix.\\ $$A=\left[ \begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22}... $$|A|=a_{11}a_{22}-a_{12}a_{21}$$ $$AdjA=\left[ \begin{matrix} a_{22} & -a_{12} \\ -a_{21} & a_{1... ^{-1}=\dfrac{1}{a_{11}}a_{22}-a_{12}a_{21}\left[ \begin{matrix} a_{22} & -a_{12} \\ -a_{21} & a_{1... {12}a_{21} )$$ $$|A^{-1}|=1$$ $$AdjA^{-1}=\left[ \begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22}
- Question 2, Exercise 2.1
- Peshawar, Pakistan. =====Question 2===== Let $A=\begin{bmatrix}2 & -5 & 1\\ 3 & 0 & -4\end{bmatrix}$, $B=\begin{bmatrix}1 & -2 & -3 \\ 0 & -1 & 5\end{bmatrix}$ and $C=\begin{bmatrix}0 & 1 & -2\\0 & -1 & -1\end{bmatrix}$. Find $2A+3B-4C.$ ====Solution==== Given: $A=\begin{bmatrix}2 & -5 & 1\\ 3 & 0 & -4\end{bmatrix}$, $B
- Question 4, Exercise 2.1
- shawar, Pakistan. =====Question 4===== Let $A= \begin{bmatrix}1 & 4 & 4 \\ 4 & 1 & 4 \\ 4 & 4 & 1 \en... ac{1}{3}A^2-2A-9I=0$. ====Solution==== Given: $A=\begin{bmatrix} 1 & 4 & 4 \\ 4 & 1 & 4 \\ 4 & 4 & 1 \end{bmatrix}$.\\ Now \begin{align}\frac{1}{3}A^2&=\frac{1}{3}\left[ \begin{matrix} 1 & 4 & 4 \\ 4 & 1 & 4 \\ 4 & 4 & 1 \
- Question 7, Exercise 2.1
- Peshawar, Pakistan. =====Question 7===== If $ A=\begin{bmatrix}1 & 0 & -1 & 2 \\3 & 1 & 2 & \quad 5 \\0 & -2 & 1 & 6\end{bmatrix}$ and $ B=\begin{bmatrix} 2 & -1 & 3 & 1 \\1 & 3 & -1 & 4 \\3 & ... )^t=A^t+B^t$. ====Solution==== Given $A=\left[ \begin{matrix}1 & 0 & -1 & 2 \\3 & 1 & 2 & \quad 5 \\0... & 1 & 6 \\\end{matrix} \right]$ and $B=\left[ \begin{matrix}2 & -1 & 3 & 1 \\1 & 3 & -1 & 4 \\3 & 1
- Question 10, Exercise 2.1
- eshawar, Pakistan. =====Question 10===== Let $A=\begin{bmatrix}1 & -3 & 4 \\-3 & 2 & -5 \\4 & -5 & 0 \end{bmatrix}$ and $B=\begin{bmatrix}5 & 6 & 7 \\6 & -8 & 3 \\7 & 3 & 1 \end{... $A+B$ is symmetric. ====Solution==== $$A=\left[ \begin{matrix} 1 & -3 & 4 \\ -3 & 2 & -5 \\ 4 & -5 & 0 \\ \end{matrix} \right]$$ $$B=\left[ \begin{matrix} 5 & 6 & 7 \\ 6 & -8 & 3 \\ 7 &