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- Question 2, Exercise 2.3
- eft| \begin{matrix} -3 & 4 & -5 \\ 1 & -1 & 2 \\ \dfrac{5}{49} & -\dfrac{6}{49} & \dfrac{8}{49}\end{matrix} \right. \right]\text{by}-\dfrac{1}{49}R_3\\ \underset{\sim}{R}&\left[\begin{matrix} 1
- Question 16 & 17, Exercise 2.2
- & -1 \\4 & 2\end{bmatrix}$. Show that $|A^{-1}|=\dfrac{1}{|A|}$. ====Solution==== Given $$A=\left[ \begi... =6+4$$ $$\Rightarrow |A|=10\ldots (1)$$ $$A^{-1}=\dfrac{1}{|A|}AdjA$$ $$AdjA=\left[ \begin{matrix} 2 &... \ -4 & 3 \\ \end{matrix} \right]$$ $$A^{-1}=\dfrac{1}{10}\left[ \begin{matrix} 2 & 1 \\ -4 & ... nd{matrix} \right]$$ $$=\left[ \begin{matrix} \dfrac{2}{10} & \dfrac{1}{10} \\ -\dfrac{4}{10} & \d
- Question 19, Exercise 2.2
- -3 & 2 \\ \end{matrix} \right]$$ $$( A^t)^{-1}=\dfrac{1}{|A^t|}AdjA^t$$ $$( A^t)^{-1}=\dfrac{1}{5}\left[ \begin{matrix} 1 & 1 \\ -3 & 2 \\ \end{matrix... right]$$ $$( A^t )^{-1}=\left[ \begin{matrix} \dfrac{1}{5} & \dfrac{1}{5} \\ -\dfrac{3}{5} & \dfrac{2}{5} \\ \end{matrix} \right]$$ $$|A|=5$$ $$AdjA=\l
- Question 18, Exercise 2.2
- } & a_{11} \\ \end{matrix} \right]$$ $$A^{-1}=\dfrac{1}{|A|}AdjA$$ $$A^{-1}=\dfrac{1}{a_{11}}a_{22}-a_{12}a_{21}\left[ \begin{matrix} a_{22} & -a_{12} \\... } & a_{11} \\ \end{matrix} \right]$$ $$|A^{-1}|=\dfrac{1}{a_{11}}a_{22}-a_{12}a_{21}( a_{11}a_{22}-a_{12... 22} \\ \end{matrix} \right]$$ $$( A^{-1} )^{-1}=\dfrac{1}{|A^{-1}|}AdjA^{-1}$$ $$( A^{-1} )^{-1}=\left[
- Question 6, Exercise 2.2
- ==== Prov that $\left| \begin{matrix} bc & a^3 & \dfrac{1}{a} \\ca & b^3 & \dfrac{1}{b} \\ab & c^3 & \dfrac{1}{c}\end{matrix} \right|=0\quad a\ne 0,\,\,b\ne 0,\,\,c\ne 0$. ====Solution=... Let $$L.H.S.=\left| \begin{matrix} bc & a^3 & \dfrac{1}{a} \\ ca & b^3 & \dfrac{1}{b} \\ ab &
- Question 14 & 15, Exercise 2.2
- e have to find $A^{-1}$and we know that $$A^{-1}=\dfrac{Adj\,\,A}{|A|}$$ $$Adj\,\,A={{\left[ \begin{matri... |=0-2(-5-2)+2(-3)$$ $$=14-6$$ $$|A|=8$$ $$A^{-1}=\dfrac{1}{8}\left[ \begin{matrix} 15 & -10 & -2 \\ ... rix} \right]$$ $$A^{-1}=\left[ \begin{matrix} \dfrac{15}{8} & \dfrac{-10}{8} & -\dfrac{2}{8} \\ \dfrac{7}{4} & -\dfrac{2}{4} & -\dfrac{2}{4} \\ -\dfr
- Question 8,9 & 10, Exercise 2.2
- b & 1 \\1 & 1 & 1+c \end{matrix} \right|=abc( 1+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} )$ ====Solution==== Let $$L.H.S.=\left| \begin{matrix} 1+a & 1 & 1 \\ 1 & 1+b ... $$=a(b+bc+c)+b(c)+0$$ $$=abc+bc+ac+ab$$ $$=abc(1+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})$$ $$=R.H.S.$$
- Question 5 & 6, Exercise 2.1
- ives $$ 3a=-2 \text{ and } 2b=3,$$ $$\implies a=-\dfrac{2}{3} \text{ and } b=\dfrac{3}{2}.$$ =====Question 6(i)===== Solve the matrix equations for $X.$ Find ... } \right]$. As $$2( X-A )=B.$$ This gives $$X-A=\dfrac{B}{2}.$$ Now $$\dfrac{B}{2}=\left[ \begin{matrix}2 & 3 & \quad 1 \\0 & -2 & 1 \\ \end{matrix} \right
- Question 4, Exercise 2.1
- & 1 & 4 \\ 4 & 4 & 1 \end{bmatrix}$. Show that $\dfrac{1}{3}A^2-2A-9I=0$. ====Solution==== Given: $A=\be... x} \right]\end{align} Now, we have \begin{align}&\dfrac{1}{3}A^2-2A-9I \\ =&\left[ \begin{matrix} 11 &... 0 & 0 & 0 \\ \end{matrix} \right] \\ \implies & \dfrac{1}{3}A^2-2A-9I=0 \end{align} ====Go To==== <text
- Question 5, Exercise 2.2
- b\\a & b & c \\a^2 & b^2 & c^2 \end{vmatrix}\\ &=\dfrac{1}{abc}\begin{vmatrix} abc & bca & abc \\ a^2 & ... \end{vmatrix} \text{ by } aR_1, bR_2, cR_3 \\ &=\dfrac{abc}{abc}\begin{vmatrix} 1 & 1 & 1 \\ a^2 & b^2
- Question 1, Exercise 2.3
- & 1 & -9 \\ 0 & 0 & 1 \end{bmatrix} \text{ by }\dfrac{1}{8}R_3\\ \underset{\sim}{R}&\begin{bmatrix} 1 &