Search
You can find the results of your search below.
Fulltext results:
- Question 2, Exercise 2.3
- & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right. \right] \\ \underset{\sim}{R} & \left[\begin{matrix} 1 & 4 & 14 \\ 0 & 5 & 6 \\ -1 & 2 & 3 \end{matri... & 0 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{matrix} \right. \right]\text{ by }R_1+3R_3\text{ and }R_2+2R_3 \\ \underset{\sim}{R}&\left[\begin{matrix} 1 & 4 & 14 \
- Question 3, Exercise 2.1
- ix}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=\begin{bmat... \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{{z}^{2}} \right] \\ &=\left[ a{{x}^{2}}+2hxy+2gxz+b{{y}^{2}}+2fyz
- Question 1, Exercise 2.1
- \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]$$ ====Solution==== \begin{align}&\left[ \begin{matrix} 1 & 2 & 4 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 & 2 \\ 2 & 0 &
- Question 9, Exercise 2.1
- -1 & 3 \\ 1 & \quad 0 & 1 \\ \end{matrix} \right],$$ $$B=\left[ \begin{matrix} 1 & 2 \\ 2 & 2 \\ 3 & 0 \\ \end{matrix} \right]$$ $$A^t=\left[ \begin{matrix} \quad2 & 1 \\ -1 & 0 \\ \quad 3 & 1 \\ \end{matrix} \right]$$ $$B^t=\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 2 & 0 \\ \end{matrix} \right]$$ $$AB=\left[ \begin{matrix} 2 & -1 & 3 \\
- Question 16 & 17, Exercise 2.2
- {matrix} 3 & -1 \\ 4 & 2 \\ \end{matrix} \right]$$ $$|A|=6+4$$ $$\Rightarrow |A|=10\ldots (1)$$ $$A^{-1}=\dfrac{1}{|A|}AdjA$$ $$AdjA=\left[ \begin{matrix} 2 & 1 \\ -4 & 3 \\ \end{matrix} \right]$$ $$A^{-1}=\dfrac{1}{10}\left[ \begin{matrix} 2 & 1 \\ -4 & 3 \\ \end{matrix} \right]$$ $$=\left[ \begin{matrix} \dfrac{2}{10} & \d
- Question 6, Exercise 2.2
- b-c & c-a & a-b \\c-a & a-b & b-c \end{matrix} \right|=0$ ====Solution==== Let \begin{align} L.H.S&=\le... & c-a & a-b \\ c-a & a-b & b-c \\ \end{matrix} \right| \\ &=\left| \begin{matrix} a-c & b-a & c-b \\ ... & c-a & a-b \\ c-a & a-b & b-c \\ \end{matrix} \right| \quad R_1+R_2 \\ &=-\left| \begin{matrix} c-a & ... & c-a & a-b \\ c-a & a-b & b-c \\ \end{matrix} \right|\quad -R_1 \\ &=0 \quad R_1 \cong R_3 \\ &=R.H.S.
- Question 2, Exercise 2.2
- 1 & 2 & 0 \\3 & 1 & 0 \\-1 & 2 & 0 \end{matrix}\right|=0$. ====Solution==== Given $$\left| \begin{matri... 3 & 1 & 0 \\ -1 & 2 & 0 \\ \end{matrix} \right|=0$$ Because elements of third column are zero. ... 2 & 3 \\-8 & 4 & -12 \\2 & -1 & 3 \end{matrix} \right|=0$. ====Solution==== Given $$\left| \begin{matri... -8 & 4 & -12 \\ 2 & -1 & 3 \\ \end{matrix} \right|=0$$ Taking $-4$ common from $R_2$, we have $$-4\
- Question 4, Exercise 2.2
- & 1 & 3 \\-1 & 2 & 1 \\2 & 1 & 1 \end{matrix} \right|.$ ====Solution==== \begin{align}&\left| \begin{m... -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)===== Evaluate
- Question 18, Exercise 2.2
- & a_{12} \\ a_{21} & a_{22} \\ \end{matrix} \right]$$ $$|A|=a_{11}a_{22}-a_{12}a_{21}$$ $$AdjA=\left... -a_{12} \\ -a_{21} & a_{11} \\ \end{matrix} \right]$$ $$A^{-1}=\dfrac{1}{|A|}AdjA$$ $$A^{-1}=\dfra... -a_{12} \\ -a_{21} & a_{11} \\ \end{matrix} \right]$$ $$|A^{-1}|=\dfrac{1}{a_{11}}a_{22}-a_{12}a_{21... & a_{12} \\ a_{21} & a_{22} \\ \end{matrix} \right]$$ $$( A^{-1} )^{-1}=\dfrac{1}{|A^{-1}|}AdjA^{-1}
- Question 5 & 6, Exercise 2.1
- 2b & 1 & 3 \\ -2 & 3 & -1 \\ \end{matrix} \right]$$ Since $A$ is given to be symmetirc, $A^t=A$, i... 2b & 1 & 3 \\ -2 & 3 & -1 \\ \end{matrix} \right]=\left[ \begin{matrix} 0 & 2b & -2 \\ 3 & 1 & 3 \\ 3a & 3 & -1 \\ \end{matrix} \right]$$ This gives $$ 3a=-2 \text{ and } 2b=3,$$ $$\im... atrix} 1 & 0 & 3 \\ -2 & 2 & 1 \\ \end{matrix} \right]$ and $B=\left[ \begin{matrix} 2 & 1 & \quad 1
- Question 13, Exercise 2.1
- \\ a_{31} & a_{32} & a_{33} \\ \end{matrix} \right]$$ $$A^t=\left[ \begin{matrix} a_{11} & a_{21}... \\ a_{13} & a_{23} & a_{33} \\ \end{matrix} \right]$$ For symmetric, we have, $$( A+A^t )^t=( A+A^t ... \\ a_{31} & a_{32} & a_{33} \\ \end{matrix} \right]+\left[ \begin{matrix} a_{11} & a_{21} & a_{31... \\ a_{13} & a_{23} & a_{33} \\ \end{matrix} \right]$$ $$A+A^t=\left[ \begin{matrix} a_{11}+a_{11}
- Question 10, Exercise 2.1
- -3 & 2 & -5 \\ 4 & -5 & 0 \\ \end{matrix} \right]$$ $$B=\left[ \begin{matrix} 5 & 6 & 7 \\ 6 & -8 & 3 \\ 7 & 3 & 1 \\ \end{matrix} \right]$$ For symmetric, we have to find out, $$A=A^t$$... -3 & 2 & -5 \\ 4 & -5 & 0 \\ \end{matrix} \right]$$ $$B^t=\left[ \begin{matrix} 5 & 6 & 7 \\ 6 & -8 & 3 \\ 7 & 3 & 1 \\ \end{matrix} \right]$$ $$A=\left[ \begin{matrix} 1 & -3 & 4 \\
- Question 12, Exercise 2.1
- 4 & 5 & 6 \\ -2 & 3 & 4 \\ \end{matrix} \right]$$ $$A^t=\left[ \begin{matrix} 3 & 4 & -2 \\ 2 & 5 & 3 \\ 1 & 6 & 4 \\ \end{matrix} \right]$$ For symmetric, we have, $$( A+A^t )^t=A+A^t$$ ... 4 & 5 & 6 \\ -2 & 3 & 4 \\ \end{matrix} \right]+\left[ \begin{matrix} 3 & 4 & -2 \\ 2 & 5 & 3 \\ 1 & 6 & 4 \\ \end{matrix} \right]$$ $$=\left[ \begin{matrix} 3+3 & 2+4 & 1-2 \
- Question 1, Exercise 2.2
- -1 & 2 & 0 \\ 2 & 0 & -2 \\ \end{matrix} \right]$$ $${{A}_{11}}={{\left( -1 \right)}^{1+1}}\left| \begin{matrix} 2 & 0 \\ 0 & -2 \\ \end{matrix} \right|$$ $$\implies{{A}_{11}}=-4$$ $${{A}_{21}}={{\left( -1 \right)}^{2+1}}\left| \begin{matrix} 3 & 0 \\ 1 &
- Question 14 & 15, Exercise 2.2
- -1 & 3 & 2 \\ 1 & 0 & 5 \\ \end{matrix} \right]$$ We have to find $A^{-1}$and we know that $$A^{... \\ A_{31} & A_{32} & A_{33} \\ \end{matrix} \right]}^{t}}$$ $$A_{11}=(-1)^{1+1}\left| \begin{matrix} 3 & 2 \\ 0 & 5 \\ \end{matrix} \right|$$ $$A_{11}=15$$ $$A_{12}=(-1)^{1+2}\left| \begi... {matrix} -1 & 2 \\ 1 & 5 \\ \end{matrix} \right|$$ $$A_{12}=7$$ $$A_{13}=(-1)^{1+3}\left| \begin{