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- Question 3, Exercise 2.1
- }$ and $C=\begin{bmatrix}x\\y\\z\end{bmatrix}$.\\ We have to prove that $$(AB)C=A(BC)$$ First, we take \begin{align} AB&=\begin{bmatrix}x & y & z\end{bmat... fyz+c{{z}^{2}} \right] \ldots (1) \end{align} Now we take \begin{align}BC&=\begin{bmatrix}a & h & g\\h... } \right] \ldots (2)\end{align} From (1) and (2), we have $$(AB)C=A(BC).$$ =====Question 3(ii)(a)===
- Question 6, Exercise 2.2
- ow from second row and second row from third row. We have, $$=\left| \begin{matrix} 0 & a-b & a^3-b... ow from second row and second row from third row, we have, $$=\left| \begin{matrix} 0 & a-b & a^2-b... w by $a$, second row by $b$ and third row by $c.$ We have, $$=\dfrac{1}{abc}\left| \begin{matrix} a
- Question 4, Exercise 2.1
- & 8 & 11 \\ \end{matrix} \right]\end{align} Now we take $$2A=\left[ \begin{matrix} 2 & 8 & 8 \\ ... & 0 & 9 \\ \end{matrix} \right]\end{align} Now, we have \begin{align}&\dfrac{1}{3}A^2-2A-9I \\ =&\le
- Question 12, Exercise 2.1
- & 6 & 4 \\ \end{matrix} \right]$$ For symmetric, we have, $$( A+A^t )^t=A+A^t$$ $$A+A^t=\left[ \begin... 4 \\ \end{matrix} \right]$$ For skew-symmetric, we have, $$( A-A^t )^t=-( A-A^t)$$ $$A-A^t=\left[ \b
- Question 13, Exercise 2.1
- a_{33} \\ \end{matrix} \right]$$ For symmetric, we have, $$( A+A^t )^t=( A+A^t )$$ $$A+A^t=\left[ \b... }} \\ \end{matrix} \right]$$ For skew-symmetric, we have, $${{\left( A-{{A}^{t}} \right)}^{t}}=-\left
- Question 8,9 & 10, Exercise 2.2
- rix} \right|$$ Subtract third row from first row. We have, $$=\left| \begin{matrix} 1 & 0 & -1 \\ ... first row and subtract third row from second row. We have, $$=\left| \begin{matrix} a & -b & 0 \\
- Question 14 & 15, Exercise 2.2
- 3 & 2 \\ 1 & 0 & 5 \\ \end{matrix} \right]$$ We have to find $A^{-1}$and we know that $$A^{-1}=\dfrac{Adj\,\,A}{|A|}$$ $$Adj\,\,A={{\left[ \begin{mat
- Question 7, Exercise 2.1
- rix} \right]...(2) \end{align} From (1) and (2), we have $(A+B)^t=A^t+B^t.$ ====Go To==== <text ali
- Question 10, Exercise 2.1
- & 3 & 1 \\ \end{matrix} \right]$$ For symmetric, we have to find out, $$A=A^t$$ $$B=B^t$$ $$( A+B
- Question 11, Exercise 2.1
- 0 \\ \end{matrix} \right]$$ For skew symmetric, we have $$( A+B )^t=-( A+B )$$ $$A+B=\left[ \begin{m
- Question 2, Exercise 2.2
- atrix} \right|=0$$ Taking $-4$ common from $R_2$, we have $$-4\left| \begin{matrix} 1 & 2 & 3 \\
- Question 3, Exercise 2.2
- \ldots (2) \end{align} Now comparing (1) and (2), we have $$|A|=|{{A}^{t}}|.$$ ====Go To==== <text a
- Question 2, Exercise 2.3
- }R_1+30R_3\text{ and } R_2-11R_3\end{align} Thus we have \begin{align} A^{-1}&=\begin{bmatrix} \dfrac