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- Question 1, Exercise 2.5
- First reduce each of the matrix into echelon form then into reduced echelon form $\left[\begin{array}{cc... First reduce each of the matrix into echelon form then into reduced echelon form $\left[\begin{array}{ll... First reduce each of the matrix into echelon form then into reduced echelon form $\left[\begin{array}{cc... First reduce each of the matrix into echelon form then into reduced echelon form $\left[\begin{array}{cc
- Question 1, Review Exercise
- is $m \times n$ and order of $B$ is $n \times p$ then order of $A B$ is:\\ * (a) $n \times p$\\ ... ii. If $A$ is a row matrix of order $1 \times n$ then order of $A^{t} A$ is:\\ * (a) $1 \times n$\... j}=M_{i j}$</collapse> iv. If $A$ is any matrix then $A$ and $A^{t}$ are always conformable for:\\ ... a square matrix of order $3 \times 3$ and $|A|=3$ then value of $|\operatorname{adj} A|$ is:\\ *
- Question 4, Exercise 2.2
- =\left[\begin{array}{ll}2 & 14\end{array}\right]$ then find a non-zero matrix $C$ such that $A C=B C$. ... begin{array}{ll}8 & z \\ t & 6\end{array}\right]$ then find the values of $z, t$ and $x^{2}+y^{2}$. ** ... begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$ then find $\alpha$ and $\beta$ so that $A^{2}+\alpha I
- Question 4, Exercise 2.3
- &= -23\lambda + 16 \end{align*} $A$ is singular, Then \begin{align*} |A|&=0\\ -23\lambda + 16 &= 0\\ -2... {align*} We know that the matrix $A$ is singular, then: \begin{align*} |A|&=0\\ \lambda - 4 = 0\\ \lambd... 14 + 2i)\lambda - 1 \end{align*} |A| is singular, then \begin{align*} (-14 + 2i)\lambda - 1 &= 0\\ (-14
- Question 7, Exercise 2.2
- begin{array}{ll}x & 0 \\ y & 1\end{array}\right]$ then prove that for all positive integers $n, A^{n}=\l... gin{array}{ll}3 & -4 \\ 1 & -1\end{array}\right]$ then prove that for all positive integers $n$, $ A^{n}
- Question 3, Exercise 2.2
- 7 \\ 0 & 2 & -1 \\ -3 & 4 & 2\end{array}\right]$ then find a matrix $C$ such that: $A+B+C=0$ ** Soluti
- Question 3, Exercise 2.2
- 7 \\ 0 & 2 & -1 \\ -3 & 4 & 2\end{array}\right]$ then find a matrix $C$ such that: $A+B+C=0$ ** Soluti
- Question 5, Exercise 2.2
- & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]$ then prove that $X^{2}-4 X-5 I=0$. ** Solution. ** G
- Question 6, Exercise 2.2
- egin{array}{cc}2 & 1 \\ 3 & -3\end{array}\right]$ then find $\alpha$ and $\beta$ such that, $A^{2}+\alph
- Question 11, Exercise 2.2
- f $a_{ij}=-a_{ji}$. Given $a_{i j}=i^{2}-j^{2}$, then \begin{align} a_{ji} & = j^2 -i^2 \\ &= - (i^2 -j
- Question 6, Exercise 2.3
- & -3 \\ 0 & 1 & 0 \\ 2 & 1 & 6\end{array}\right]$ then find $A^{-1}$ and hence show that $A A^{-1}=A^{-1
- Question 4, Exercise 2.6
- _3 - R_2). \end{align*} There is no value of $x$. Then $x_3 = 0$. From the second row: \begin{align*}
- Question 2 and 3, Review Exercise
- =====Question 3===== Prove that if $A^{-1}=A^{t}$ then $\left|A A^{t}\right|=1$.FIXME ** Solution. **