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- Question 3, Exercise 2.1
- Pakistan. =====Question 3===== Identify the diagonal matrix, scalar matrix, identity matrix, lower ... A: Lower triangular matrix\\ B: Scalar matrix; Diagonal matrix\\ C: Lower triangular matrix\\ D: Identity matrix; Scalar matrix; Diagonal matrix\\ E: Diagonal matrix\\ F: Upper triangular matrix\\ G: Diagonal matrix\\ H: Scalar matrix;
- Question 1, Exercise 2.1
- n{align}\text{Order of F}&= 2\times 2\end{align} GOOD ====Go to ==== <text align="left"><btn type="success">[[math-11-nbf:sol:unit02:ex2-1-p2|Question 2>
- Question 5, Exercise 2.2
- & 0 & 0 \end{bmatrix} = O\\ &=R.H.S.\end{align} GOOD ====Go to ==== <text align="left"><btn type="primary">[[math-11-nbf:sol:unit02:ex2-2-p4|< Questi
- Question 6, Exercise 2.2
- end{align} Hence $\alpha = -9$ and $\beta = -1$. GOOD ====Go to ==== <text align="left"><btn type="primary">[[math-11-nbf:sol:unit02:ex2-2-p5|< Quest
- Question 7, Exercise 2.2
- ence C-2 is satisfied, and the proof is complete. GOOD =====Question 7(ii)===== If $A=\left[\begin{a... ght] \] holds for all positive integers $n$. ====Go to ==== <text align="left"><btn type="primary">[
- Question 2, Exercise 2.1
- mn matrix.\\ ** Solution. ** Square matrix ====Go to ==== <text align="left"><btn type="primary">[
- Question 4, Exercise 2.1
- 0 & 6 & 0 \\ 1 & 3 & 1 \end{array}\right]$$ ====Go to ==== <text align="left"><btn type="primary">[
- Question 1, Exercise 2.2
- \ \frac{1}{3} & -\frac{2}{3} \end{bmatrix} \] ====Go to ==== <text align="right"><btn type="success">
- Question 3, Exercise 2.2
- \\ 4 & -4 & 1\end{bmatrix}\\ \end{align*} ====Go to ==== <text align="left"><btn type="primary">[
- Question 3, Exercise 2.2
- \\ 4 & -4 & 1\end{bmatrix}\\ \end{align*} ====Go to ==== <text align="left"><btn type="primary">[
- Question 4, Exercise 2.2
- 1\end{array}\right]=0$. ** Solution. ** ====Go to ==== <text align="left"><btn type="primary">[
- Question 8, Exercise 2.2
- \end{align*} Hence $$(A B)^{t}=B^{t} A^{t}$$ ====Go to ==== <text align="left"><btn type="primary">[
- Question 9, Exercise 2.2
- lign*} Thus, $\quad(A + B)^t = A^t + B^t$. ====Go to ==== <text align="left"><btn type="primary">[
- Question 10, Exercise 2.2
- = A$, we find that: $$A^2 + B^2 = A + B $$ ====Go to ==== <text align="left"><btn type="primary">[
- Question 11, Exercise 2.2
- ji}$, hence given matrix is skew-symmetric. ====Go to ==== <text align="left"><btn type="primary">[