a) Solve by Gauss elimination or Gauss seidel method: [4] 2x + 2y + z = 6 , x y + z = 0 and 4x + 2y + 3z = 4 . b) Use the Bisection method to find solutions accurate to within 10^ 2 for x^3 7x^2 +…

a) Solution of System of Linear Equations by Gauss Elimination Method

We solve the given system of linear equations using the Gauss elimination method (Gaussian elimination). The system is:

$$\{\begin{array}{l}2x+2y+z=6\text{(1)}\\ x-y+z=0\text{(2)}\\ 4x+2y+3z=4\text{(3)}\end{array}$$

Step 1: Write the augmented matrix

The augmented matrix for the system is:

$$\left[\begin{array}{ccccc}2& 2& 1& |& 6\\ 1& -1& 1& |& 0\\ 4& 2& 3& |& 4\end{array}\right]$$

Step 2: Perform row operations to obtain upper triangular form

Row Operation 1: Swap $R_1$ and $R_2$ to make the leading coefficient of the first row 1.

$$\left[\begin{array}{ccccc}1& -1& 1& |& 0\\ 2& 2& 1& |& 6\\ 4& 2& 3& |& 4\end{array}\right]$$

Row Operation 2: Eliminate $x$ from $R_2$ and $R_3$ using $R_1$.

• $R_2\to R_2-2R_1$
• $R_3\to R_3-4R_1$

$$\left[\begin{array}{ccccc}1& -1& 1& |& 0\\ 0& 4& -1& |& 6\\ 0& 6& -1& |& 4\end{array}\right]$$

Row Operation 3: Eliminate $y$ from $R_3$ using $R_2$.

• $R_3\to R_3-\frac{6}{4}{R}_2=R_3-\frac{3}{2}{R}_2$

$$\left[\begin{array}{ccccc}1& -1& 1& |& 0\\ 0& 4& -1& |& 6\\ 0& 0& \frac{1}{2}& |& -5\end{array}\right]$$

Step 3: Back substitution

From the last row:

$$\frac{1}{2}z=-5⟹z=-10$$

Substitute $z=-10$ into the second row:

$$4y-(-10)=6⟹4y+10=6⟹4y=-4⟹y=-1$$

Substitute $y=-1$ and $z=-10$ into the first row:

$$x-(-1)+(-10)=0⟹x+1-10=0⟹x=9$$

Final Solution:

$$\boxed{x=9,y=-1,z=-10}$$

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b) Solution of $x^3-7x^2+14x-6=0$ in $(0,1)$ by Bisection Method

We apply the Bisection Method to find the root of the equation $f(x)=x^3-7x^2+14x-6=0$ in the interval $(0,1)$ with an accuracy of $10^{-2}$.

Step 1: Check continuity and sign change

Evaluate $f(x)$ at the endpoints:

$$f(0)=0-0+0-6=-6(\text{Negative})$$ $$f(1)=1-7+14-6=2(\text{Positive})$$

Since $f(0)$ and $f(1)$ have opposite signs, a root exists in $(0,1)$.

Step 2: Apply the Bisection Method

The Bisection Method formula is:

$$x_{n+1}=\frac{a_n+b_n}{2}$$

where $a_n$ and $b_n$ are the endpoints of the interval at step $n$.

Iteration 1: $$x_1=\frac{0+1}{2}=0.5$$ $$f(0.5)=(0.5{)}^3-7(0.5{)}^2+14(0.5)-6=0.125-1.75+7-6=-0.625(\text{Negative})$$

New interval: $(0.5,1)$

Iteration 2: $$x_2=\frac{0.5+1}{2}=0.75$$ $$f(0.75)=(0.75{)}^3-7(0.75{)}^2+14(0.75)-6=0.421875-3.9375+10.5-6=0.984375(\text{Positive})$$

New interval: $(0.5,0.75)$

Iteration 3: $$x_3=\frac{0.5+0.75}{2}=0.625$$ $$f(0.625)=(0.625{)}^3-7(0.625{)}^2+14(0.625)-6=0.244140625-2.734375+8.75-6=0.259765625(\text{Positive})$$

New interval: $(0.5,0.625)$

Iteration 4: $$x_4=\frac{0.5+0.625}{2}=0.5625$$ $$f(0.5625)=(0.5625{)}^3-7(0.5625{)}^2+14(0.5625)-6=0.177734375-2.24609375+7.875-6=-0.293359375(\text{Negative})$$

New interval: $(0.5625,0.625)$

Iteration 5: $$x_5=\frac{0.5625+0.625}{2}=0.59375$$ $$f(0.59375)=(0.59375{)}^3-7(0.59375{)}^2+14(0.59375)-6=0.2080078125-2.48046875+8.3125-6=0.0390390625(\text{Positive})$$

New interval: $(0.5625,0.59375)$

Iteration 6: $$x_6=\frac{0.5625+0.59375}{2}=0.578125$$ $$f(0.578125)=(0.578125{)}^3-7(0.578125{)}^2+14(0.578125)-6=0.193359375-2.3583984375+8.09375-6=-0.1713890625(\text{Negative})$$

New interval: $(0.578125,0.59375)$

Step 3: Check accuracy

The interval length at this step is:

$$0.59375-0.578125=0.015625<10^{-2}$$

Thus, the root is approximated as the midpoint of the final interval:

$$x\approx \frac{0.578125+0.59375}{2}=0.5859375$$

Rounding to two decimal places:

$$\boxed{x\approx 0.59}$$