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- Question 1, Exercise 2.5
- helon form then into reduced echelon form $\left[\begin{array}{ccc}1 & 3 & 5 \\ -6 & 8 & 3 \\ -4 & 6 & 5\end{array}\right]$. ** 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 & 18 &
- Question 6, Exercise 2.6
- lution. ** For this system of equations; we have \begin{align*} A &= \begin{bmatrix} 5 & 3 & 1 \\ 2 & 1 & 3 \\ 1 & 2 & 4 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 6 \\ 19 \\ 25 \end{bmatrix} \end{align*} A
- Question 3, Exercise 2.5
- perations, find the inverse of the matrix $\left[\begin{array}{ccc}0 & -1 & -1 \\ -1 & 3 & 0 \\ 1 & -1 & ... at $A A^{-1}=A^{-1} A=I$.\\ ** Solution. ** Let \begin{align*} A&=\left[ \begin{array}{ccc} 0 & -1 & -1 \\ -1 & 3 & 0 \\ 1 & -1 & 4 \end{array} \right]\\ ... end{align*} So $A$ is non singular. Now consider \begin{align*} &\quad\left[ \begin{array}{ccc|ccc} 0 & -
- Question 4, Exercise 2.2
- Pakistan. =====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
- Question 5, Exercise 2.3
- g matrices if it exists by adjoint method $\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -1 \\ 1 & -2 & -1\end{array}\right]$. ** Solution. ** Given \begin{align*} A &= \left[\begin{array}{ccc} 1 & -1 & 1 \\ 2 & 1 & -1 \\ 1 & -2 & -1 \end{array}\right]\\ |A|&=... gular.\\ Let find the cofactor matrix for $A$.\\ \begin{align*} A_{11} &= (-1)^{1+1} \left|\begin{array}{
- Question 7, Exercise 2.3
- rify 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} &=
- Question 2, Exercise 2.5
- ===== Find the rank of each of the matrix $\left[\begin{array}{ccc}5 & 9 & 3 \\ 3 & -5 & 6 \\ 2 & 10 & 6\end{array}\right]$ ** 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} \\ 3 &
- Question 7, Exercise 2.2
- Pakistan. =====Question 7(i)===== If $A=\left[\begin{array}{ll}x & 0 \\ y & 1\end{array}\right]$ then ... e that for all positive integers $n, A^{n}=\left[\begin{array}{cc}x^{n} & 0 \\ \dfrac{y\left(x^{n}-1\righ... d{array}\right]$. ** Solution. ** Given: $$A = \begin{bmatrix} x & 0 \\ y & 1 \end{bmatrix}.$$ We use m... on to prove the given fact. For C-1, put $n = 1$ \begin{align}A^1 =\begin{bmatrix} x^1 & 0 \\ \dfrac{y(x^
- Question 2, Exercise 2.3
- e the determinant of the following matrix $\left[\begin{array}{lll}3 & 2 & 3 \\ 4 & 5 & 1 \\ 2 & 1 & 0\en... 3\). Now we find their 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}\right| = (-
- Question 7 and 8, Exercise 2.6
- ad, Pakistan. =====Question 7===== If $A=\left[\begin{array}{ccc}3 & 2 & 1 \\ 4 & -1 & 2 \\ 7 & 3 & -3\... =4 ; \quad x+2 y-3 z=0$. ** Solution. ** Given \begin{align*} A &= \begin{bmatrix} 3 & 2 & 1 \\ 4 & -1 & 2 \\ 7 & 3 & -3 \end{bmatrix}\\ |A| &=3(-3) - 2(-26) ... }$, we calculate the cofactors of each element . \begin{align*} A_{11} &= (-1)^{1+1} \left| \begin{array}
- Question 13, Exercise 2.2
- the matrices $X$ and $Y$ such that $2 X-Y=\left[\begin{array}{ccc}1 & 6 & -3 \\ 2 & 1 & 7\end{array}\right]$ and $X+3 Y=\left[\begin{array}{ccc}4 & 3 & 2 \\ 1 & -3 & 0\end{array}\right]$. ** Solution. ** Given: \begin{align*} 2X - Y = \begin{pmatrix} 1 & 6 & -3 \\ 2 & 1 & 7 \end{pmatrix} \cdots (i)\\ X + 3Y = \begin{pmat
- Question 5, Exercise 2.6
- e above system may be written as $A X=B$; where, \begin{align*} &A = \begin{bmatrix} 1 & 1 & 2 \\ -1 & -2 & 3 \\ 3 & -7 & 4 \end{bmatrix}, \quad X = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}, \quad B = \begin{bmatrix} 8 \\ 1 \\ 10 \end{bmatrix}\\ |A| &= \beg
- Question 3, Exercise 2.6
- ** Solution. ** Given the system of equations: \begin{align*} \begin{aligned} 2x + 3y + 4z &= 2 \\ 2x + y + z &= 5 \\ 3x - 2y + z &= -3 \end{aligned}\end{align*} The associated augmented matrix is: \begin{align*} A_{b} &=\quad \left[\begin{array}{cccc} 2 & 3 & 4 & 2 \\ 2 & 1 & 1 & 5 \\ 3 & -2 & 1 & -3 \end{a
- Question 6, Exercise 2.3
- ad, Pakistan. =====Question 6===== If $A=\left[\begin{array}{ccc}2 & 1 & -3 \\ 0 & 1 & 0 \\ 2 & 1 & 6\e... hat $A A^{-1}=A^{-1} A=I_{3}$. ** Solution. ** \begin{align*} A &= \begin{bmatrix} 2 & 1 & -3 \\ 0 & 1 & 0 \\ 2 & 1 & 6 \end{bmatrix} \end{align*} To find t... method of row reduction (Gaussian elimination). \begin{align*} A \mid I & = \left[\begin{array}{ccc|ccc}
- Question 2, Exercise 2.6
- \ $3 x_{1}-2 x_{2}+4 x_{3}=0$\\ ** Solution. ** \begin{align*} &2 x_{1}-\lambda x_{2}+x_{3}=0 \cdots(i)\... } Homogenous system has non-trivial solution, if \begin{align*} &\left| \begin{array}{ccc} 2 & -\lambda & 1 \\ 2 & 3 & -1 \\ 3 & -2 & 4 \end{array} \right|=... -\frac{7}{11}\\ \end{align*} The system becomes \begin{align*} &2 x_{1}+ \frac{7}{11}x_{2}+x_{3}=0 \cdot