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- Question 2, Exercise 2.3
- {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}.$$ Then \begin{align}|A|&=\begin{vmatrix}4 & -2 & 5 \\ 2 & 1 & 0 \\ -1 & 2 & 3 \end{vmatrix}\\ &=4(3)+2(6)+5(4+1) \\ & =49\neq 0. \end{align} This gives, $A$ is non-singular and $A^{-1
- Question 3, Exercise 2.1
- Question 3(i)===== If $A=\begin{bmatrix}x & y & z\end{bmatrix}$, $B=\begin{bmatrix}a & h & g\\h & b & f\\g & f & c\end{bmatrix}$ and $C=\begin{bmatrix}x\\y\\z\end{bmatrix}$. Verify that $\left( AB \right)C=A\left( BC \rig... ==Solution==== Given: $A=\begin{bmatrix}x & y & z\end{bmatrix}$, $B=\begin{bmatrix}a & h & g\\h & b & f
- Question 1, Exercise 2.1
- le matrix $$\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]$$ ====Solution==== \begin{align}&\left[ \begin{matrix} 1 & 2 & 4 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 & 2
- Question 1, Exercise 2.3
- gin{bmatrix}1 & 3 & -1 \\2 & 1 & 4 \\3 & 4 & -5\end{bmatrix}$. ====Solution==== \begin{align}&\begin{... bmatrix} 1 & 3 & -1 \\ 2 & 1 & 4 \\ 3 & 4 & -5 \end{bmatrix}\\ \underset{\sim}{R}&\begin{bmatrix} 1 & 3 & -1 \\ 0 & -5 & 6 \\ 0 & -5 & -2 \end{bmatrix} \text{ by } R_2-2R_1 \text{ and } R_3-3R... trix} 1 & 3 & -1 \\ 0 & -5 & 6 \\ 0 & -5 & -2 \end{bmatrix} \text{ by } R_3-R_2 \\ \underset{\sim}{R
- Question 9, Exercise 2.1
- $A=\begin{bmatrix}2 & -1 & 3 \\1 & \quad 0 & 1 \end{bmatrix},$ $B=\begin{bmatrix}1 & 2 \\2 & 2 \\ 3 & 0 \end{bmatrix}$, show that $( AB )^t=B^tA^t$. ====Solut... matrix} 2 & -1 & 3 \\ 1 & \quad 0 & 1 \\ \end{matrix} \right],$$ $$B=\left[ \begin{matrix} 1 & 2 \\ 2 & 2 \\ 3 & 0 \\ \end{matrix} \right]$$ $$A^t=\left[ \begin{matrix}
- Question 5 & 6, Exercise 2.1
- atrix} 0 & 2b & -2 \\ 3 & 1 & 3 \\ 3a & 3 & -1 \end{bmatrix}$ is given to be symmetric. Find the valu... atrix} 0 & 2b & -2 \\ 3 & 1 & 3 \\ 3a & 3 & -1 \end{bmatrix}$ Then $$A^t=\left[ \begin{matrix} 0 &... 3 & 3a \\ 2b & 1 & 3 \\ -2 & 3 & -1 \\ \end{matrix} \right]$$ Since $A$ is given to be symmet... 3 & 3a \\ 2b & 1 & 3 \\ -2 & 3 & -1 \\ \end{matrix} \right]=\left[ \begin{matrix} 0 & 2b &
- Question 16 & 17, Exercise 2.2
- ion 16===== Let $A=\begin{bmatrix}3 & -1 \\4 & 2\end{bmatrix}$. Show that $|A^{-1}|=\dfrac{1}{|A|}$. =... \left[ \begin{matrix} 3 & -1 \\ 4 & 2 \\ \end{matrix} \right]$$ $$|A|=6+4$$ $$\Rightarrow |A|=1... \left[ \begin{matrix} 2 & 1 \\ -4 & 3 \\ \end{matrix} \right]$$ $$A^{-1}=\dfrac{1}{10}\left[ \begin{matrix} 2 & 1 \\ -4 & 3 \\ \end{matrix} \right]$$ $$=\left[ \begin{matrix} \df
- Question 6, Exercise 2.2
- b-c & c-a \\b-c & c-a & a-b \\c-a & a-b & b-c \end{matrix} \right|=0$ ====Solution==== Let \begin{al... c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c \\ \end{matrix} \right| \\ &=\left| \begin{matrix} a-c &... c-b \\ b-c & c-a & a-b \\ c-a & a-b & b-c \\ \end{matrix} \right| \quad R_1+R_2 \\ &=-\left| \begin... b-c \\ b-c & c-a & a-b \\ c-a & a-b & b-c \\ \end{matrix} \right|\quad -R_1 \\ &=0 \quad R_1 \cong
- Question 5, Exercise 2.2
- $\begin{vmatrix}a & b & c\\l & m & n\\x & y & z \end{vmatrix}=\begin{vmatrix}a & l & x\\b & m & y\\c & n & z \end{vmatrix}$ ====Solution==== \begin{align}L.H.S.&=\... n{vmatrix} a & b & c \\ l & m & n \\ x & y & z \end{vmatrix}\\ &=\begin{vmatrix} a & b & c \\l & m & n \\x & y & z \end{vmatrix}^t \quad \because |A|=|A^t| \\ &=\begin{v
- Question 2, Exercise 2.2
- gin{matrix} 1 & 2 & 0 \\3 & 1 & 0 \\-1 & 2 & 0 \end{matrix}\right|=0$. ====Solution==== Given $$\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$$ Taking $-4$ common from $R_2$
- Question 18, Exercise 2.2
- x} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{matrix} \right]$$ $$|A|=a_{11}a_{22}-a_{12}a_{21}... a_{22} & -a_{12} \\ -a_{21} & a_{11} \\ \end{matrix} \right]$$ $$A^{-1}=\dfrac{1}{|A|}AdjA$$... a_{22} & -a_{12} \\ -a_{21} & a_{11} \\ \end{matrix} \right]$$ $$|A^{-1}|=\dfrac{1}{a_{11}}a_{... x} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{matrix} \right]$$ $$( A^{-1} )^{-1}=\dfrac{1}{|A^
- Question 2, Exercise 2.1
- === 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: $A=\begin{bmatrix}2 & -5 & 1\\ 3 & 0 & -4\end{bmatrix}$, $B=\begin{bmatrix}1 & -2 & -3 \\ 0 & -
- Question 4, Exercise 2.1
- in{bmatrix}1 & 4 & 4 \\ 4 & 1 & 4 \\ 4 & 4 & 1 \end{bmatrix}$. Show that $\dfrac{1}{3}A^2-2A-9I=0$. =... n{bmatrix} 1 & 4 & 4 \\ 4 & 1 & 4 \\ 4 & 4 & 1 \end{bmatrix}$.\\ Now \begin{align}\frac{1}{3}A^2&=\fr... 1 & 4 & 4 \\ 4 & 1 & 4 \\ 4 & 4 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 4 & 4 \\ 4 & 1 & 4 \\ 4 & 4 & 1 \\ \end{matrix} \right]\\& =\frac{1}{3}\left[ \begin{matr
- Question 7, Exercise 2.1
- & -1 & 2 \\3 & 1 & 2 & \quad 5 \\0 & -2 & 1 & 6\end{bmatrix}$ and $ B=\begin{bmatrix} 2 & -1 & 3 & 1 \\1 & 3 & -1 & 4 \\3 & 1 & 2 & -1 \end{bmatrix}$. Then show that $( A+B )^t=A^t+B^t$. ==... & 2 \\3 & 1 & 2 & \quad 5 \\0 & -2 & 1 & 6 \\\end{matrix} \right]$ and $B=\left[ \begin{matrix}2 ... & 3 & 1 \\1 & 3 & -1 & 4 \\3 & 1 & 2 & -1 \\ \end{matrix} \right]$. Then $$A^t=\left[ \begin{matri
- Question 10, Exercise 2.1
- {bmatrix}1 & -3 & 4 \\-3 & 2 & -5 \\4 & -5 & 0 \end{bmatrix}$ and $B=\begin{bmatrix}5 & 6 & 7 \\6 & -8 & 3 \\7 & 3 & 1 \end{bmatrix}$. Verify that $A$ and $B$ are symmetric.... -3 & 4 \\ -3 & 2 & -5 \\ 4 & -5 & 0 \\ \end{matrix} \right]$$ $$B=\left[ \begin{matrix} 5 & 6 & 7 \\ 6 & -8 & 3 \\ 7 & 3 & 1 \\ \end{matrix} \right]$$ For symmetric, we have to find