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- Question 2, Exercise 2.3
- ingular and $A^{-1}$ exists. Now \begin{align} & \left[\begin{matrix} 4 & -2 & 5 \\ 2 & 1 & 0 \\ -1 & 2 & 3 \end{matrix}\left| \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0... d{matrix}\right. \right] \\ \underset{\sim}{R} & \left[\begin{matrix} 1 & 4 & 14 \\ 0 & 5 & 6 \\ -1 & 2 & 3 \end{matrix}\left| \begin{matrix} 1 & 0 & 3 \\ 0 & 1 & 2 \\ 0 & 0
- Question 3, Exercise 2.1
- egin{bmatrix}x\\y\\z\end{bmatrix}$. Verify that $\left( AB \right)C=A\left( BC \right)$. ====Solution==== Given: $A=\begin{bmatrix}x & y & z\end{bmatrix}$, $B... bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}\\ &=\left[ x( ax+hy+gz )+y( hx+by+fz )+z( gx+fy+cz ) \right] \\ &=\left[ a{{x}^{2}}+hxy+gxz+hxy+b{{y}^{2}}+fyz+gxz+fyz+c{
- Question 1, Exercise 2.1
- Question 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 \\ \end{matrix} \ri
- Question 9, Exercise 2.1
- how that $( AB )^t=B^tA^t$. ====Solution==== $$A=\left[ \begin{matrix} 2 & -1 & 3 \\ 1 & \quad 0 & 1 \\ \end{matrix} \right],$$ $$B=\left[ \begin{matrix} 1 & 2 \\ 2 & 2 \\ 3 & 0 \\ \end{matrix} \right]$$ $$A^t=\left[ \begin{matrix} \quad2 & 1 \\ -1 & 0 \\ ... \quad 3 & 1 \\ \end{matrix} \right]$$ $$B^t=\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 2 & 0 \
- Question 6, Exercise 2.2
- r, Pakistan. =====Questiopn 6(i)===== Prov that $\left| \begin{matrix}a-b & b-c & c-a \\b-c & c-a & a-b... ht|=0$ ====Solution==== Let \begin{align} L.H.S&=\left| \begin{matrix} a-b & b-c & c-a \\ b-c & c-a & a... \ c-a & a-b & b-c \\ \end{matrix} \right| \\ &=\left| \begin{matrix} a-c & b-a & c-b \\ b-c & c-a & a... b-c \\ \end{matrix} \right| \quad R_1+R_2 \\ &=-\left| \begin{matrix} c-a & a-b & b-c \\ b-c & c-a & a
- Question 16 & 17, Exercise 2.2
- -1}|=\dfrac{1}{|A|}$. ====Solution==== Given $$A=\left[ \begin{matrix} 3 & -1 \\ 4 & 2 \\ \end{m... ldots (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} \right]$$ $$=\left[ \begin{matrix} \dfrac{2}{10} & \dfrac{1}{10}
- Question 2, Exercise 2.2
- valuating state the reasons for the equalities. $\left|\begin{matrix} 1 & 2 & 0 \\3 & 1 & 0 \\-1 & 2 &... \end{matrix}\right|=0$. ====Solution==== Given $$\left| \begin{matrix} 1 & 2 & 0 \\ 3 & 1 & 0 \\... aluating state the reasons for the equalities. $\left| \begin{matrix}1 & 2 & 3 \\-8 & 4 & -12 \\2 & -... end{matrix} \right|=0$. ====Solution==== Given $$\left| \begin{matrix} 1 & 2 & 3 \\ -8 & 4 & -12
- Question 4, Exercise 2.2
- ===Question 4(i)===== Evaluate the determinant $\left| \begin{matrix}0 & 1 & 3 \\-1 & 2 & 1 \\2 & 1 &... matrix} \right|.$ ====Solution==== \begin{align}&\left| \begin{matrix} 0 & 1 & 3 \\ -1 & 2 & 1 \\ 2 & 1 & 1 \\ \end{matrix} \right| \\ =&0\left( 2-1 \right)-1\left( -1-2 \right)+3\left( -1-4 \right)\\ =&0+3-15=-12 \end{align} =====Question 4(ii)=
- Question 13, Exercise 2.1
- that $A+A^t$ is symmetric. ====Solution==== $$A=\left[ \begin{matrix} a_{11} & a_{12} & a_{13} \\ ... a_{32} & a_{33} \\ \end{matrix} \right]$$ $$A^t=\left[ \begin{matrix} a_{11} & a_{21} & a_{31} \\ ... tric, we have, $$( A+A^t )^t=( A+A^t )$$ $$A+A^t=\left[ \begin{matrix} a_{11} & a_{12} & a_{13} \\ ... _{31} & a_{32} & a_{33} \\ \end{matrix} \right]+\left[ \begin{matrix} a_{11} & a_{21} & a_{31} \\
- Question 18, Exercise 2.2
- et $A$ is $2\times 2$ non-singular matrix.\\ $$A=\left[ \begin{matrix} a_{11} & a_{12} \\ a_{21} ... ight]$$ $$|A|=a_{11}a_{22}-a_{12}a_{21}$$ $$AdjA=\left[ \begin{matrix} a_{22} & -a_{12} \\ -a_{21... A$$ $$A^{-1}=\dfrac{1}{a_{11}}a_{22}-a_{12}a_{21}\left[ \begin{matrix} a_{22} & -a_{12} \\ -a_{21... {22}-a_{12}a_{21} )$$ $$|A^{-1}|=1$$ $$AdjA^{-1}=\left[ \begin{matrix} a_{11} & a_{12} \\ a_{21}
- Question 5 & 6, Exercise 2.1
- 1 & 3 \\ 3a & 3 & -1 \end{bmatrix}$ Then $$A^t=\left[ \begin{matrix} 0 & 3 & 3a \\ 2b & 1 & 3 ... $A$ is given to be symmetirc, $A^t=A$, implies $$\left[ \begin{matrix} 0 & 3 & 3a \\ 2b & 1 & 3 \\ -2 & 3 & -1 \\ \end{matrix} \right]=\left[ \begin{matrix} 0 & 2b & -2 \\ 3 & 1 & 3 ... -1 & 4 \end{bmatrix}$. ====Solution==== Given $A=\left[ \begin{matrix} 1 & 0 & 3 \\ -2 & 2 & 1 \\ \en
- Question 1, Exercise 2.2
- $ Also find $|A|.$ =====Solution===== Given $$A=\left[ \begin{matrix} 1 & 3 & 1 \\ -1 & 2 & 0 \... 0 & -2 \\ \end{matrix} \right]$$ $${{A}_{11}}={{\left( -1 \right)}^{1+1}}\left| \begin{matrix} 2 & 0 \\ 0 & -2 \\ \end{matrix} \right|$$ $$\implies{{A}_{11}}=-4$$ $${{A}_{21}}={{\left( -1 \right)}^{2+1}}\left| \begin{matrix} 3 & 0
- Question 10, Exercise 2.1
- fy that $A+B$ is symmetric. ====Solution==== $$A=\left[ \begin{matrix} 1 & -3 & 4 \\ -3 & 2 & -5 ... \\ 4 & -5 & 0 \\ \end{matrix} \right]$$ $$B=\left[ \begin{matrix} 5 & 6 & 7 \\ 6 & -8 & 3 \... A=A^t$$ $$B=B^t$$ $$( A+B )^t=A^t+B^t$$ $$A^t=\left[ \begin{matrix} 1 & -3 & 4 \\ -3 & 2 & -5 ... \ 4 & -5 & 0 \\ \end{matrix} \right]$$ $$B^t=\left[ \begin{matrix} 5 & 6 & 7 \\ 6 & -8 & 3 \
- Question 12, Exercise 2.1
- y that$A+A^t$ is symmetric. ====Solution==== $$A=\left[ \begin{matrix} 3 & 2 & 1 \\ 4 & 5 & 6 \\ -2 & 3 & 4 \\ \end{matrix} \right]$$ $$A^t=\left[ \begin{matrix} 3 & 4 & -2 \\ 2 & 5 & 3 \... ymmetric, we have, $$( A+A^t )^t=A+A^t$$ $$A+A^t=\left[ \begin{matrix} 3 & 2 & 1 \\ 4 & 5 & 6 \\ -2 & 3 & 4 \\ \end{matrix} \right]+\left[ \begin{matrix} 3 & 4 & -2 \\ 2 & 5 & 3 \
- Question 14 & 15, Exercise 2.2
- rix}$. Find $A^{-1}$. ====Solution==== Given $$A=\left[ \begin{matrix} 0 & 2 & 2 \\ -1 & 3 & 2 \... at $$A^{-1}=\dfrac{Adj\,\,A}{|A|}$$ $$Adj\,\,A={{\left[ \begin{matrix} A_{11} & A_{12} & A_{13} \\ ... \end{matrix} \right]}^{t}}$$ $$A_{11}=(-1)^{1+1}\left| \begin{matrix} 3 & 2 \\ 0 & 5 \\ \end{ma... rix} \right|$$ $$A_{11}=15$$ $$A_{12}=(-1)^{1+2}\left| \begin{matrix} -1 & 2 \\ 1 & 5 \\ \end{m