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- Question 1, Exercise 2.5
- nto echelon form then into reduced echelon form $\left[\begin{array}{ccc}1 & 3 & 5 \\ -6 & 8 & 3 \\ -4 &... ight]$. ** Solution. ** \begin{align*} & \quad \left[\begin{array}{ccc}1 & 3 & 5 \\ -6 & 8 & 3 \\ -4 & 6 & 5\end{array}\right]\\ \sim & \text{R} \left[\begin{array}{ccc} 1 & 3 & 5 \\ 0 & 26 & 33 \\ 0 ... ad R_2 + 6R_1 \quad R_3 + 4R_1\\ \sim & \text{R} \left[\begin{array}{ccc} 1 & 3 & 5 \\ 0 & 1 & \frac{33}
- Question 3, Exercise 2.5
- row operations, find the inverse of the matrix $\left[\begin{array}{ccc}0 & -1 & -1 \\ -1 & 3 & 0 \\ 1 ... } A=I$.\\ ** Solution. ** Let \begin{align*} A&=\left[ \begin{array}{ccc} 0 & -1 & -1 \\ -1 & 3 & 0 ... non singular. Now consider \begin{align*} &\quad\left[ \begin{array}{ccc|ccc} 0 & -1 & -1 & 1 & 0 & 0 \... & 0 & 1 \end{array} \right]\\ \sim &{\text{R}} \left[ \begin{array}{ccc|ccc} 1 & -1 & 4 & 0 & 0 & 1 \\
- Question 5, Exercise 2.3
- llowing matrices if it exists by adjoint method $\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -1 \\ 1 &... ]$. ** Solution. ** Given \begin{align*} A &= \left[\begin{array}{ccc} 1 & -1 & 1 \\ 2 & 1 & -1 \\ 1 ... x for $A$.\\ \begin{align*} A_{11} &= (-1)^{1+1} \left|\begin{array}{cc} 1 & -1 \\ -2 & -1 \end{array}\r... )(-2)] \\ &= -1 - 2 = -3 \\ A_{12} &= (-1)^{1+2} \left|\begin{array}{cc} 2 & -1 \\ 1 & -1 \end{array}\ri
- Question 4, Exercise 2.2
- =====Question 4(i)===== Find $A$ if \begin{align}\left[\begin{array}{cc} 2 & 1 \\ 3 & 2 \end{array}\right]A\left[\begin{array}{cc} 1 & 3 \\ 2 & 4 \end{array}\right]&=\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]\end{align} ** Solution. ** Let $ B = \left[\begin{array}{cc} 2 & 1 \\ 3 & 2 \end{array}\righ
- Question 7, Exercise 2.3
- === Verify that $(A B)^{-1}=B^{-1} A^{-1}$ if $A=\left[\begin{array}{ll}2 & 1 \\ 8 & 6\end{array}\right]$ and $B=\left[\begin{array}{ll}3 & 2 \\ 0 & 2\end{array}\right]$. ** Solution. ** Given: \begin{align*} A &= \left[\begin{array}{ll}2 & 1 \\ 8 & 6\end{array}\right]... \\ |A|& = 12 - 8 = 4\\ A^{-1} &= \dfrac{1}{4} \left[\begin{array}{ll}6 & -1 \\ -8 & 2\end{array}\righ
- Question 2, Exercise 2.5
- n 2(i)===== Find the rank of each of the matrix $\left[\begin{array}{ccc}5 & 9 & 3 \\ 3 & -5 & 6 \\ 2 & ... y}\right]$ ** Solution. ** \begin{align*}&\quad\left[ \begin{array}{ccc} 5 & 9 & 3 \\ 3 & -5 & 6 \\ ... 2 & 10 & 6 \end{array} \right]\\ \sim & \text{R}\left[ \begin{array}{ccc} 1 & \frac{9}{5} & \frac{3}{5}... ay} \right]\quad \frac{1}{5} R1\\ \sim & \text{R}\left[ \begin{array}{ccc} 1 & \frac{9}{5} & \frac{3}{5}
- Question 2, Exercise 2.3
- valuate the determinant of the following matrix $\left[\begin{array}{lll}3 & 2 & 3 \\ 4 & 5 & 1 \\ 2 & 1... eir corresponding cofactors. \begin{align*} A &= \left[\begin{array}{ccc} 3 & 2 & 3 \\ 4 & 5 & 1 \\ 2 & ... 1 & 0 \end{array}\right]\\ & A_{11} = (-1)^{1+1} \left|\begin{array}{cc} 5 & 1 \\ 1 & 0 \end{array}\righ... cdot 1) = (1) (-1) = -1 \\ & A_{12} = (-1)^{1+2} \left|\begin{array}{cc} 4 & 1 \\ 2 & 0 \end{array}\righ
- Question 6, Exercise 2.6
- {bmatrix} \end{align*} And \begin{align*} |A|& = \left| \begin{array}{ccc} 5 & 3 & 1 \\ 2 & 1 & 3 \\ 1 &... h element . \begin{align*} A_{11} &= (-1)^{1+1} \left| \begin{array}{cc} 1 & 3 \\ 2 & 4 \end{array} \right| = 4 - 6 = -2\\ A_{12} &= (-1)^{1+2} \left| \begin{array}{cc} 2 & 3 \\ 1 & 4 \end{array} \right| = -(8 - 3) = -5\\ A_{13} &= (-1)^{1+3} \left| \begin{array}{cc} 2 & 1 \\ 1 & 2 \end{array} \ri
- Question 1, Exercise 2.3
- (i)===== Evaluate the determinant of the matrix $\left[\begin{array}{ccc}2 & 3 & 1 \\ 1 & -1 & 2 \\ 4 & ... ight]$. ** Solution. ** Let \begin{align*} A &= \left[\begin{array}{ccc}2 & 3 & 1 \\ 1 & -1 & 2 \\ 4 & ... ii)===== Evaluate the determinant of the matrix $\left[\begin{array}{ccc}\cos \theta & -\sin \theta & 0 ... ight]$. ** Solution. ** Let \begin{align*} A&= \left[\begin{array}{ccc}\cos \theta & -\sin \theta & 0
- Question 7, Exercise 2.2
- mabad, Pakistan. =====Question 7(i)===== If $A=\left[\begin{array}{ll}x & 0 \\ y & 1\end{array}\right]... n prove that for all positive integers $n, A^{n}=\left[\begin{array}{cc}x^{n} & 0 \\ \dfrac{y\left(x^{n}-1\right)}{x-1} & 1\end{array}\right]$. ** Solution. ... complete. GOOD =====Question 7(ii)===== If $A=\left[\begin{array}{ll}3 & -4 \\ 1 & -1\end{array}\righ
- Question 3, Exercise 2.6
- ugmented matrix is: \begin{align*} A_{b} &=\quad \left[\begin{array}{cccc} 2 & 3 & 4 & 2 \\ 2 & 1 & 1 & ... -2 & 1 & -3 \end{array}\right]\\ & \sim \text{R}\left[\begin{array}{cccc} 2 & 3 & 4 & 2 \\ 0 & -2 & -3 ... and}\quad R_3 - \frac{3}{2}R_1\\ & \sim \text{R}\left[\begin{array}{cccc} 2 & 3 & 4 & 2 \\ 0 & -2 & -3 ... augmented matrix is: \begin{align*} A_b& =\quad \left[\begin{array}{cccc} 5 & -2 & 1 & 2 \\ 2 & 2 & 6 &
- Question 4, Exercise 2.1
- e is symmetric and which is skew-symmetric. $$ A=\left[\begin{array}{ccc} 2 & 0 \\ \sqrt{5} & 6 \\ 1 & 9... ne is symmetric and which is skew-symmetric. $$B=\left[\begin{array}{cccc} 1 & 6 & 2 & 0 \end{array}\right] $$ ** Solution. ** $$B^t=\left[\begin{array}{c} 1 \\ 6 \\ 2 \\ 0 \end{array}\rig... ne is symmetric and which is skew-symmetric. $$C=\left[\begin{array}{ll} 2 & 6 \\ 9 & 2 \end{array}\righ
- Question 6, Exercise 2.3
- slamabad, Pakistan. =====Question 6===== If $A=\left[\begin{array}{ccc}2 & 1 & -3 \\ 0 & 1 & 0 \\ 2 & ... ussian elimination). \begin{align*} A \mid I & = \left[\begin{array}{ccc|ccc} 2 & 1 & -3 & 1 & 0 & 0 \\ ... \\ 2 & 1 & 6 & 0 & 0 & 1 \end{array}\right] \\ &=\left[\begin{array}{ccc|ccc} 2 & 1 & -3 & 1 & 0 & 0 \\ ... & 0 & 1 \end{array}\right] \quad R_3 - R_1 \\ &=\left[\begin{array}{ccc|ccc} 2 & 1 & -3 & 1 & 0 & 0 \\
- Question 7 and 8, Exercise 2.6
- slamabad, Pakistan. =====Question 7===== If $A=\left[\begin{array}{ccc}3 & 2 & 1 \\ 4 & -1 & 2 \\ 7 & ... ch element . \begin{align*} A_{11} &= (-1)^{1+1} \left| \begin{array}{cc} -1 & 2 \\ 3 & -3 \end{array} \right| = 3 - 6 = -3 \\ A_{12} &= (-1)^{1+2} \left| \begin{array}{cc} 4 & 2 \\ 7 & -3 \end{array} \r... ight| = -(-12 - 14) = 26 \\ A_{13} &= (-1)^{1+3} \left| \begin{array}{cc} 4 & -1 \\ 7 & 3 \end{array} \r
- Question 3, Exercise 2.3
- uestion 3(i)===== Determine which of the matrix $\left[\begin{array}{ccc}3 & 1 & 2 \\ 2 & 3 & 1 \\ -4 & ... -singular. ** Solution. ** \begin{align*} A &= \left[\begin{array}{ccc} 3 & 1 & 2 \\ 2 & 3 & 1 \\ -4 &... estion 3(ii)===== Determine which of the matrix $\left[\begin{array}{ccc}3 & -1 & 2 \\ 2 & 0 & 1 \\ -1 &... -singular. ** Solution. ** \begin{align*} A &= \left[\begin{array}{ccc} 3 & -1 & 2 \\ 2 & 0 & 1 \\ -1