IB Maths AA HL Systems of Equations Questions

1. Identify the coefficients of the variables $x,y,z$ to form an augmented matrix or directly input into the GDC.

2. Set up the formal matrix equation: $\left[\begin{array}{ccc}2& -1& 1\\ 1& 1& 1\\ 1& -1& 2\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}3\\ 6\\ 5\end{array}\right]$.

3. Access the simultaneous equation solver on your graphic display calculator (CG50: MENU A (Equation) -> F1 (Simultaneous) -> 3 Unknowns).

4. Input the values exactly as they appear in the system, ensuring the constant terms are correctly placed on the right-hand side of the equals sign.

5. Execute the solver function to calculate the values.

6. State the final exact coordinates of the point of intersection: $\mathbf{x}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{z}\mathbf{=}\mathbf{3}$.

1. Label the equations: (1) $x+y+z=6$, (2) $2x-y+3z=9$, (3) $-x+2y-z=2$.

2. Add equation (1) and equation (3) to efficiently eliminate both $x$ and $z$ at the same time: $(x-x)+(y+2y)+(z-z)=6+2$.

3. Solve the resulting equation to immediately find $y$: $3y=8⟹y=\frac{8}{3}$.

4. Eliminate $x$ using equations (1) and (2). Take Eq(2) $-2\times$ Eq(1): $(2x-2x)+(-y-2y)+(3z-2z)=9-12$.

5. Simplify to find an equation purely in $y$ and $z$: $-3y+z=-3$.

6. Substitute $y=\frac{8}{3}$ into the new equation to find $z$: $-3\left(\frac{8}{3}\right)+z=-3⟹z=5$. Finally, use Eq(1) to find $x$: $x+\frac{8}{3}+5=6⟹x=-\frac{5}{3}$. The point is $(\mathbf{-}\frac{\mathbf{5}}{\mathbf{3}}\mathbf{,}\frac{\mathbf{8}}{\mathbf{3}}\mathbf{,}\mathbf{5})$.

1. Label the equations: (1) $x+y+2z=4$, (2) $2x+3y+5z=7$, (3) $3x+4y+kz=12$.

2. Eliminate $x$ from equation (2) by calculating Eq(2) $-2\times$ Eq(1): $(2x-2x)+(3y-2y)+(5z-4z)=7-8$, yielding $y+z=-1$.

3. Eliminate $x$ from equation (3) by calculating Eq(3) $-3\times$ Eq(1): $(3x-3x)+(4y-3y)+(kz-6z)=12-12$.

4. Simplify to yield a second equation in $y$ and $z$: $y+(k-6)z=0$.

5. Subtract the result of step 2 from the result of step 4 to eliminate $y$: $(y-y)+((k-6)z-z)=0-(-1)$. This condenses to $(k-7)z=1$.

6. Analyze the final statement. For the system to have no solutions (geometrically, planes forming a triangular prism), we require an impossible statement such as $0z=1$. This occurs when the coefficient of $z$ is zero, thus $\mathbf{k}\mathbf{=}\mathbf{7}$.