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- Question 2, Exercise 2.3
- & 5 \\ 2 & 1 & 0 \\ -1 & 2 & 3 \end{bmatrix}.$$ Then \begin{align}|A|&=\begin{vmatrix}4 & -2 & 5 \\ 2... & 6 \\ 1 & 3 & 4 \\ -1 & 5 & 1 \end{bmatrix}.$$ Then \begin{align}|A|&=\left| \begin{matrix} 3 & -1 & ... 3 \\ 0 & -2 & 0 \\ -2 & -2 & 2 \end{bmatrix}.$$ Then \begin{align}|A|&=\left| \begin{matrix} 1 & 2 & -... -1 \\ 0 & -1 & 3 \\ 1 & 0 & 2 \end{bmatrix}.$$ Then \begin{align}|A|&=\left| \begin{matrix} 1 & 2 & -
- Question 3, Exercise 2.2
- ion 3===== Let $A$ be square matrix of order $3,$ then verify that $|A^t|=|A|$. ====Solution==== Let $$A... \ a_{31} & a_{32} & a_{33} \\ \end{bmatrix}$$ Then \begin{align}|A|&=a_{11} \left( a_{22} a_{33}-a_{... _{23}} & {{a}_{33}} \\ \end{matrix} \right]\\ $$ Then \begin{align} |A^t|&=a_{11}\left( a_{22}a_{33}-a_
- Question 7, Exercise 2.1
- \1 & 3 & -1 & 4 \\3 & 1 & 2 & -1 \end{bmatrix}$. Then show that $( A+B )^t=A^t+B^t$. ====Solution==== G... & 4 \\3 & 1 & 2 & -1 \\ \end{matrix} \right]$. Then $$A^t=\left[ \begin{matrix}1 & 3 & 0 \\0 & 1 & -
- Question 13, Exercise 2.1
- 13(i)===== If $A$ is a square matrix of order $3$ then show that $A+A^t$ is symmetric. ====Solution==== ... 3(ii)===== If $A$ is a square matrix of order $3$ then show that $A-{{A}^{t}}$ is skew symmetric. ====So
- Question 18, Exercise 2.2
- )===== If $A$ and $B$ are non-singular matrices, then show that $( A^{-1})^{-1}=A$. ====Solution==== Le... (ii)===== $A$ and $B$ are non-singular matrices, then show that $( AB )^{-1}=B^{-1}A^{-1}$ ====Solution
- Question 2, Exercise 2.1
- n{bmatrix}0 & 1 & -2\\0 & -1 & -1\end{bmatrix}$. Then $2A=\begin{bmatrix}4 & -10 & 2\\6 & 0 & -8\end{bm
- Question 5 & 6, Exercise 2.1
- & -2 \\ 3 & 1 & 3 \\ 3a & 3 & -1 \end{bmatrix}$ Then $$A^t=\left[ \begin{matrix} 0 & 3 & 3a \\
- Question 8, Exercise 2.1
- & 0 \\ 3 & -1 & 4 \\ \end{matrix} \right]$$ Then $$A^t=\left[ \begin{matrix} 1 & 3 \\ 2 &
- Question 1, Exercise 2.2
- & 1 \\-1 & 2 & 0 \\2 & 0 & -2 \end{bmatrix}$ , then find $A_{11},A_{21},A_{23},A_{31},A_{32},A_{33}.$
- Question 11, Exercise 2.2
- $$=-1+8-13$$ $$|A|=-6$$ $A$ is not equal to zero. Then $A$ is non-singular. =====Question 11(iii)=====
- Question 14 & 15, Exercise 2.2
- 4===== Show that inverse of square matrix exists. Then it is unique. ====Solution==== =====Questi