IB exam questions: topic 7

1. It is a $3\times 3$ system with decimal coefficients, and Paper 2 allows the use of a graphic display calculator (GDC). We enter the coefficient matrix and the column of constant terms (or we use the simultaneous linear equations solver on the GDC directly).
2. The GDC gives the unique solution, which in exact form is $$x=-\frac{932}{389},y=\frac{628}{389},z=-\frac{612}{389}.$$
3. Rounding to three significant figures: $x\approx -2.40$, $y\approx 1.61$, $z\approx -1.57$.
4. Check in the first equation: $0.1(-2.396)-1.7(1.614)+0.9(-1.573)\approx -0.240-2.744-1.416=-4.40$.

Answer: $x=-2.40$, $y=1.61$, $z=-1.57$

1. We eliminate $x$: $E_2-E_1$ gives $-y+2z=-1$, i.e. $y=1+2z$. And $E_3-2E_1$ gives $y-2z=1$, which is the same equation. The system is equivalent to two independent equations in three unknowns: it has infinitely many solutions.
2. We take $z=t$ as parameter: $y=1+2t$.
3. From the first equation: $x=3+y+z=3+1+2t+t=4+3t$.
4. Check in the third equation: $2(4+3t)-(1+2t)-4t=8+6t-1-2t-4t=7$.

Answer: $(x,y,z)=(4+3t,1+2t,t)$, with $t\in \mathbb{R}$

a) 1. We row reduce. To eliminate $x$ we use the second equation (it has coefficient $1$ on $x$): $$E_1-2E_2:(-1-2a)z=0,E_3-5E_2:2y-5az=2a-25.$$ 2. The equivalent system is $$\{\begin{array}{rl}x+2y+az& =5\\ 2y-5az& =2a-25\\ (-1-2a)z& =0\end{array}$$ 3. If $-1-2a\ne 0$, then $z=0$ from the last equation; then $y$ and $x$ are obtained uniquely. There is a unique solution. 4. If $-1-2a=0$, the last equation is $0=0$ and the system is left with two independent equations for three unknowns: there is no unique solution. 5. Therefore, there is no unique solution when $a=-\frac{1}{2}$.

b) 1. With $a=2$, the last equation of the reduced system is $-5z=0$, so $z=0$. 2. From the second: $2y-0=2\cdot 2-25=-21$, so $y=-\frac{21}{2}$. 3. From the first: $x=5-2y-2z=5+21-0=26$. 4. Check in the original $E_1$: $2\cdot 26+4\left(-\frac{21}{2}\right)-0=52-42=10$. Correct.

Answer: a) $a=-\frac{1}{2}$; b) $x=26$, $y=-\frac{21}{2}$, $z=0$

1. The system is homogeneous: it always has the solution $(0,0,0)$. We look for the values of $k$ for which there are other solutions as well.
2. From the second equation, $x=-4z$.
3. We substitute into the first: $-8z-2y+kz=0$, hence $y=\frac{(k-8)z}{2}$.
4. We substitute $x$ and $y$ into the third: $$-4kz+\frac{(k-8)z}{2}+z=0\Rightarrow \frac{-8k+k-8+2}{2}z=0\Rightarrow \frac{-7k-6}{2}z=0.$$
5. If $-7k-6\ne 0$, then $z=0$ and therefore $x=0$ and $y=0$: only the zero solution.
6. If $-7k-6=0$, i.e. $k=-\frac{6}{7}$, the equation holds for any $z$. Taking $z=t$ we obtain $x=-4t$, $y=\frac{(k-8)t}{2}=-\frac{31}{7}t$: infinitely many solutions, among them some different from $(0,0,0)$.

Answer: $k=-\frac{6}{7}$

1. We call the equations $E_1$, $E_2$, $E_3$. We eliminate $x$ from $E_1$ and from $E_2$ with the help of $E_3$ (which has coefficient $1$ on $x$): $$E_1-4E_3:7y+(2-4a)z=1-14=-13,$$ $$E_2-2E_3:7y-2az=-6-7=-13.$$
2. We subtract the two equations obtained: $$(2-4a)z-(-2a)z=0\Rightarrow (2-2a)z=0.$$
3. If $a\ne 1$, then $z=0$; from $7y=-13$ we obtain $y=-\frac{13}{7}$, and from $E_3$ we solve for $x$: there is a unique solution.
4. If $a=1$, the previous equation is $0=0$: we are left with a system of two independent equations and three unknowns, which does not have a unique solution.

Answer: $a=1$

a) 1. We call the equations $E_1$, $E_2$, $E_3$. Subtracting $E_1-E_2$ eliminates $y$: $$x+2z=-3,\text{i.e.}x=-3-2z.$$ 2. Subtracting $E_3-E_1$ also eliminates $y$: $$2x+(b-1)z=0.$$ 3. We substitute $x=-3-2z$: $$-6-4z+(b-1)z=0\Rightarrow (b-5)z=6.$$ 4. If $b\ne 5$ we can divide by $b-5$: $$z=\frac{6}{b-5}.$$

b) 1. If $b\ne 5$, the previous part gives a unique value of $z$; then $x=-3-2z$ is determined, and so is $y=1-3x-z$ (from $E_1$). The system has a unique solution. 2. If $b=5$, the equation $(b-5)z=6$ becomes $0\cdot z=6$, which is impossible: the system has no solution. 3. Therefore, the system does not have a unique solution exactly when $b=5$ (in that case it is inconsistent).

Answer: a) $z=\frac{6}{b-5}$ (for $b\ne 5$); b) $b=5$

1. A linear system has either a unique solution, no solution, or infinitely many solutions. If it has more than one, it has infinitely many, and then the determinant of the coefficient matrix is zero: $$\left|\begin{array}{ccc}2& -1& 3\\ 3& 1& 2\\ -1& 2& a\end{array}\right|=2(a-4)+(3a+2)+3(6+1)=5a+15.$$
2. $5a+15=0\Rightarrow a=-3$.
3. With $a=-3$ we solve by elimination. Adding the first two equations: $5x+5z=0\Rightarrow x=-z$.
4. From the first equation: $y=2x+3z-2=-2z+3z-2=z-2$.
5. We substitute into the third, $-x+2y-3z=b$: $$z+2(z-2)-3z=b\Rightarrow -4=b.$$
6. For $b=-4$ the system has infinitely many solutions, with general solution $(-t,t-2,t)$, $t\in \mathbb{R}$. (For $a=-3$ and $b\ne -4$ there would be no solution at all.)

Answer: $a=-3$ and $b=-4$

1. The system has a unique solution if and only if the determinant of the coefficient matrix is not zero.
2. We expand: $$\left|\begin{array}{ccc}a& 1& 1\\ 1& a& 1\\ 1& 1& a\end{array}\right|=a(a^2-1)-(a-1)+(1-a)=a^3-3a+2=(a-1{)}^2(a+2).$$ (With the GDC, in Paper 2, the same determinant is obtained.)
3. The determinant is zero for $a=1$ and $a=-2$. For any other value the system has a unique solution.
4. Check of the two exceptional cases: if $a=1$, the three equations are $x+y+z=1$ (infinitely many solutions); if $a=-2$, adding the three equations gives $0=3$ (no solution). In neither case is there a unique solution.

Answer: unique solution for $a\ne 1$ and $a\ne -2$

1. The system does not have a unique solution exactly when the determinant of the coefficient matrix is zero.
2. We expand along the first column: $$\left|\begin{array}{ccc}k& 1& 2\\ 0& -1& 4\\ 3& 4& 2\end{array}\right|=k(-2-16)+3(4+2)=-18k+18.$$
3. $-18k+18=0\Rightarrow k=1$.
4. With $k=1$ the system does not have a unique solution; in fact it is inconsistent. From the first two equations, $x+y+2z=3$ and $y=4z-5$, we get $x=8-6z$; then the third equation gives $3(8-6z)+4(4z-5)+2z=4-0z=1$, which is false.

Answer: $k=1$

a) 1. We row reduce. $E_2-2E_1$ gives $-3y+3z=-3$, i.e. $y-z=1$. 2. $E_3+E_1$ gives $6y+(a-1)z=6$. 3. Subtracting $6$ times the equation $y-z=1$: $(a-1)z+6z=6-6$, i.e. $(a+5)z=0$. 4. If $a\ne -5$, then $z=0$ and the system has a unique solution. If $a=-5$, the equation becomes $0=0$ and the system does not have a unique solution. (With determinants: $\left|\begin{array}{ccc}1& 2& -1\\ 2& 1& 1\\ -1& 4& a\end{array}\right|=-3a-15=0\Rightarrow a=-5$.)

b) 1. It is enough to check that $(x,y,z)=(0,1,0)$ satisfies the three equations for any $a$: $$0+2-0=2,0+1+0=1,-0+4+a\cdot 0=4.$$ 2. The third holds whatever $a$ is, because $a$ multiplies $z=0$. Hence the system always has at least one solution. 3. More precisely: if $a\ne -5$ the unique solution is $(0,1,0)$; if $a=-5$ there are infinitely many, $(-t,1+t,t)$ with $t\in \mathbb{R}$.

Answer: a) $a=-5$; b) the solution $(0,1,0)$ holds for all $a$ (unique if $a\ne -5$, infinitely many if $a=-5$). Proved.

a) 1. We row reduce. $E_2-E_1$ gives $y+(1-k)z=3$. 2. $E_3-k{E}_1$ gives $(8-2k)y+(5-k^2)z=6$. 3. We substitute $y=3-(1-k)z$ into the new third equation: $$(8-2k)(3-(1-k)z)+(5-k^2)z=6\Rightarrow [(5-k^2)-(8-2k)(1-k)]z=6-24+6k.$$ 4. The bracket equals $5-k^2-8+10k-2k^2=-(3k^2-10k+3)=-(3k-1)(k-3)$, and the right-hand side is $-18+6k=6(k-3)$. We get $$(3k-1)(k-3)z=-6(k-3).$$ 5. If $k\ne \frac{1}{3}$ and $k\ne 3$, we solve $z=\frac{-6}{3k-1}$ and the other unknowns follow by substitution: unique solution. 6. If $k=\frac{1}{3}$ or $k=3$ the coefficient of $z$ is $0$: there is no unique solution.

b) 1. Case $k=3$: the equation becomes $0\cdot z=0$, which is always true. Taking $z=t$: $y=3-(1-3)t=3+2t$ and, from the first equation, $x=-2y-3z=-6-4t-3t=-6-7t$. Solution set: $\{(-6-7t,3+2t,t):t\in \mathbb{R}\}$. 2. Case $k=\frac{1}{3}$: the equation becomes $0\cdot z=-6\left(\frac{1}{3}-3\right)=16$, which is impossible. The system is inconsistent: the solution set is empty.

Answer: a) $k\in \mathbb{R}\setminus \left\{\frac{1}{3},3\right\}$; b) if $k=3$, $(-6-7t,3+2t,t)$ with $t\in \mathbb{R}$; if $k=\frac{1}{3}$, there are no solutions (empty set)

a) 1. We row reduce. $E_2-2E_1$ gives $y+2z=4$. 2. $E_3-E_1$ gives $-3y+(\alpha -8)z=\beta -12$. 3. Adding $3$ times the equation $y+2z=4$ to the previous one: $(\alpha -8+6)z=\beta -12+12$, i.e. $$(\alpha -2)z=\beta .$$ 4. i) No solution: $\alpha =2$ and $\beta \ne 0$ (we get $0=\beta$, which is impossible). 5. ii) Unique solution: $\alpha \ne 2$, with $\beta$ arbitrary (we solve $z=\frac{\beta }{\alpha -2}$ and then $y$ and $x$). 6. iii) Infinitely many solutions: $\alpha =2$ and $\beta =0$ (we get $0=0$ and $z$ is free). (Check with the determinant of the coefficients: $2\alpha -4=0\iff \alpha =2$.)

b) 1. With $\alpha =2$ and $\beta =0$ the system is equivalent to $2x+y+6z=0$ and $y+2z=4$. 2. From the second, $y=4-2z$. In the first: $2x=-y-6z=-4-4z$, so $x=-2-2z$. 3. Taking $z=t$ as parameter: $(x,y,z)=(-2-2t,4-2t,t)$. 4. Solving for $t$ in each coordinate gives the Cartesian form: $$\frac{x+2}{-2}=\frac{y-4}{-2}=\frac{z}{1}.$$

Answer: a) i) $\alpha =2$ and $\beta \ne 0$; ii) $\alpha \ne 2$ and $\beta$ arbitrary; iii) $\alpha =2$ and $\beta =0$; b) $\frac{x+2}{-2}=\frac{y-4}{-2}=\frac{z}{1}$

1. We call $p$, $q$ and $m$ the price (in dollars) of one banana, one kiwi and one melon. Each box gives one equation: $$\{\begin{array}{rl}2p+3q+4m& =6.58\\ 5p+2q+8m& =12.32\\ 5p+4q& =3\end{array}$$
2. This is a $3\times 3$ system; with the GDC (Paper 2) the solution is obtained directly. By hand: to eliminate $m$, we subtract twice the first equation from the second: $$E_2-2E_1:p-4q=12.32-13.16=-\mathrm{0.84.}$$
3. We solve for $p=4q-0.84$ and substitute into the third equation: $$5(4q-0.84)+4q=3\Rightarrow 24q=7.2\Rightarrow q=\mathrm{0.3.}$$
4. Then $p=4\cdot 0.3-0.84=0.36$.
5. From the first equation: $4m=6.58-2(0.36)-3(0.3)=6.58-0.72-0.9=4.96$, so $m=1.24$.
6. Check in the second equation: $5(0.36)+2(0.3)+8(1.24)=1.8+0.6+9.92=12.32$. Correct.

Answer: bananas at $0.36, kiwis at $0.30 and melons at $1.24

a) 1. We eliminate $x$ from the second and third equations using the first as reference. We call the three equations $E_1$, $E_2$, $E_3$ and calculate $7E_1+3E_3$: $$(-21+21)x+(7+3)y+(14-6)z=7a+3c\Rightarrow 10y+8z=7a+3c.$$ 2. We now calculate $7E_2+11E_3$: $$(-77+77)x+(14+11)y+(42-22)z=7b+11c\Rightarrow 25y+20z=7b+11c.$$ 3. The system is reduced to two equations in $y$ and $z$: $$\{\begin{array}{rl}10y+8z& =7a+3c\\ 25y+20z& =7b+11c\end{array}$$ The coefficients of the second are $\frac{5}{2}$ times those of the first ($25=\frac{5}{2}\cdot 10$ and $20=\frac{5}{2}\cdot 8$). For the system to have a solution, the constant terms must keep the same ratio: $$7b+11c=\frac{5}{2}(7a+3c)\Rightarrow 14b+22c=35a+15c\Rightarrow -35a+14b+7c=0.$$ 4. Dividing by $7$: $-5a+2b+c=0$. This is the condition for the system to have a solution (if it is not satisfied, an impossible equation of the form $0=k$ with $k\ne 0$ appears).

b) 1. With $a=2$ and $b=7$, the condition from part (a) gives $-10+14+c=0$, i.e. $c=-4$. 2. The system is now $$\{\begin{array}{rl}-3x+y+2z& =2\\ -11x+2y+6z& =7\\ 7x+y-2z& =-4\end{array}$$ and, by part (a), the third equation is a combination of the other two: we are left with an equivalent system of two equations in three unknowns. 3. From $10y+8z=7a+3c=14-12=2$ we obtain $5y+4z=1$, i.e. $y=\frac{1-4z}{5}$. 4. We substitute into $E_1$: $-3x+\frac{1-4z}{5}+2z=2$. Multiplying by $5$: $-15x+1-4z+10z=10$, hence $6z=9+15x$ and $$z=\frac{3}{2}+\frac{5}{2}x.$$ 5. Then $y=\frac{1-4z}{5}=\frac{1-6-10x}{5}=-1-2x$. 6. Taking $x=t$ as parameter, the general solution is $$(x,y,z)=\left(t,-1-2t,\frac{3}{2}+\frac{5}{2}t\right),t\in \mathbb{R}.$$ For example, with $t=1$ we obtain the solution $(1,-3,4)$. 7. The solution is not unique: the system has infinitely many solutions (one for each value of $t$), because one equation depends on the other two, so there are only two independent equations for three unknowns.

Answer: a) $-5a+2b+c=0$; b) $(x,y,z)=\left(t,-1-2t,\frac{3}{2}+\frac{5}{2}t\right)$, $t\in \mathbb{R}$ (for example, $(1,-3,4)$). It is not unique: there are infinitely many solutions.