Question

exercises 5. (5 pt)
algebra
a) calculate the determinant of the coefficient matrix (1 pt) the rank of the coefficient (0.5 pt) and solutions does the system have? (0.5 pt) of the following system, according to the results, you may
solve it, if possible through the method of elimination (2.5 pt):
\

$$\begin{cases} 2x + y + 3z = 8\\\\ 3x + 2y + 4z = 6\\\\ 3x - y - 3z = 6 \\end{cases}$$

Explanation:

Step1: Find coefficient matrix determinant

The coefficient matrix $A$ of the system is:

$$A =
LATEXBLOCK5
$$

Calculate determinant using cofactor expansion:

$$\det(A) = 2\times
LATEXBLOCK6
-1\times
LATEXBLOCK7
+ (-1)\times
LATEXBLOCK8
$$

$$= 2[(-4)(1)-(-2)(-3)] -1[(3)(1)-(-2)(3)] -1[(3)(-3)-(-4)(3)]$$
$$= 2(-4-6) -1(3+6) -1(-9+12)$$
$$= 2(-10) -1(9) -1(3) = -20-9-3 = -32$$

Step2: Calculate ranks

• Rank of coefficient matrix $A$: Since $\det(A)

eq 0$, $A$ is invertible, so $\text{rank}(A)=3$.

• Augmented matrix $\bar{A}$:

$$\bar{A} =
LATEXBLOCK9
$$

Since $\text{rank}(A)=3$ and the matrix has 3 rows, $\text{rank}(\bar{A})=3$.
Because $\text{rank}(A)=\text{rank}(\bar{A})=3$ (equal to number of variables), the system has exactly 1 unique solution.

Step3: Gaussian elimination setup

Write the augmented matrix:

$$\begin{pmatrix} 2 & 1 & -1 & | & 6 \\ 3 & -4 & -2 & | & 8 \\ 3 & -3 & 1 & | & 6 \end{pmatrix}$$

Step4: Eliminate $x$ from row2, row3

Row2 = Row2 - $\frac{3}{2}$Row1:
$$Row2: 3-\frac{3}{2}(2) = 0,\ -4-\frac{3}{2}(1)=-\frac{11}{2},\ -2-\frac{3}{2}(-1)=-\frac{1}{2},\ 8-\frac{3}{2}(6)=-1$$
Row3 = Row3 - $\frac{3}{2}$Row1:
$$Row3: 3-\frac{3}{2}(2)=0,\ -3-\frac{3}{2}(1)=-\frac{9}{2},\ 1-\frac{3}{2}(-1)=\frac{5}{2},\ 6-\frac{3}{2}(6)=-3$$
New matrix:

$$\begin{pmatrix} 2 & 1 & -1 & | & 6 \\ 0 & -\frac{11}{2} & -\frac{1}{2} & | & -1 \\ 0 & -\frac{9}{2} & \frac{5}{2} & | & -3 \end{pmatrix}$$

Multiply row2 and row3 by 2 to eliminate fractions:

$$\begin{pmatrix} 2 & 1 & -1 & | & 6 \\ 0 & -11 & -1 & | & -2 \\ 0 & -9 & 5 & | & -6 \end{pmatrix}$$

Step5: Eliminate $y$ from row3

Row3 = Row3 - $\frac{9}{11}$Row2:
$$Row3: 0-0=0,\ -9-\frac{9}{11}(-11)=0,\ 5-\frac{9}{11}(-1)=\frac{64}{11},\ -6-\frac{9}{11}(-2)=-\frac{48}{11}$$
New matrix:

$$\begin{pmatrix} 2 & 1 & -1 & | & 6 \\ 0 & -11 & -1 & | & -2 \\ 0 & 0 & \frac{64}{11} & | & -\frac{48}{11} \end{pmatrix}$$

Multiply row3 by 11:

$$\begin{pmatrix} 2 & 1 & -1 & | & 6 \\ 0 & -11 & -1 & | & -2 \\ 0 & 0 & 64 & | & -48 \end{pmatrix}$$

Step6: Back-substitute to find $z$

Solve for $z$ from row3:
$$64z = -48 \implies z = \frac{-48}{64} = -\frac{3}{4}$$

Step7: Back-substitute to find $y$

Substitute $z=-\frac{3}{4}$ into row2:
$$-11y - (-\frac{3}{4}) = -2 \implies -11y + \frac{3}{4} = -2$$
$$-11y = -2 - \frac{3}{4} = -\frac{11}{4} \implies y = \frac{1}{4}$$

Step8: Back-substitute to find $x$

Substitute $y=\frac{1}{4}$, $z=-\frac{3}{4}$ into row1:
$$2x + \frac{1}{4} - (-\frac{3}{4}) = 6 \implies 2x + 1 = 6$$
$$2x = 5 \implies x = \frac{5}{2}$$

Answer:

a) Determinant of coefficient matrix: $\boldsymbol{-32}$; Rank of coefficient matrix: $\boldsymbol{3}$; Rank of augmented matrix: $\boldsymbol{3}$; Number of solutions: $\boldsymbol{1}$
b) Unique solution: $\boldsymbol{x=\frac{5}{2},\ y=\frac{1}{4},\ z=-\frac{3}{4}}$