Matrices and Systems of Equations: practice solutions, Grade 12

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Practice solutions Grade 12 : Matrices and Systems of Equations — Zyro the alien explorer of Planète Maths

Written solutions to the chapter problems. Check each step, then correct yourself.

1 Reading a matrix ★★★

(a) $M$ has 2 rows and 3 columns, so its size is $2 \times 3$.

(b) $m_{12}$ is in row 1, column 2: $m_{12} = -2$. $m_{23}$ is in row 2, column 3: $m_{23} = 9$.

2 Adding and subtracting ★★★

Combine entries in the same position.

$P + Q = \begin{bmatrix}5&3\\-6&6\end{bmatrix}$ and $P - Q = \begin{bmatrix}1&-5\\6&4\end{bmatrix}$.

3 Scalar multiples ★★★

Multiply every entry by the scalar.

(a) $-2R = \begin{bmatrix}-2&6&-4\\0&-8&10\end{bmatrix}$.

(b) $\dfrac{1}{2}S = \begin{bmatrix}3&-2\\5&1\end{bmatrix}$.

4 Is the product defined? ★★★

The inner sizes must match.

$AB$: $(2 \times 3)(3 \times 4)$, inner numbers match, so the size is $2 \times 4$.
$BA$: $(3 \times 4)(2 \times 3)$, and $4 \neq 2$, so it is not defined.
$CD$: $(3 \times 1)(1 \times 3)$ gives a $3 \times 3$ matrix.
$DC$: $(1 \times 3)(3 \times 1)$ gives a $1 \times 1$ matrix.

5 Two-by-two determinants ★★★

Use $ad - bc$.

$5(4) - 2(3) = 20 - 6 = 14$.
$(-3)(-2) - 6(1) = 6 - 6 = 0$.
$0(4) - 7(-1) = 7$.

The matrix in (b) has determinant $0$, so it has no inverse.

6 Check an inverse ★★★

Multiply: $\begin{bmatrix}3&1\\5&2\end{bmatrix}\begin{bmatrix}2&-1\\-5&3\end{bmatrix} = \begin{bmatrix}6-5&-3+3\\10-10&-5+6\end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix}$.

The product is the identity matrix $I$, so the matrices are inverses. (Also $\det A = 6 - 5 = 1$.)

7 Write a system as a matrix equation ★★★

$\begin{bmatrix}4&3\\1&-3\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}11\\-1\end{bmatrix}$.

Substitute $x = 2$, $y = 1$: $4(2) + 3(1) = 11$ and $2 - 3(1) = -1$. Both equations hold, so $(2, 1)$ is a solution.

8 Multiplying in both orders ★★★

$AB = \begin{bmatrix}1(5)+4(2)&1(0)+4(-1)\\-2(5)+3(2)&-2(0)+3(-1)\end{bmatrix} = \begin{bmatrix}13&-4\\-4&-3\end{bmatrix}$.

$BA = \begin{bmatrix}5(1)+0(-2)&5(4)+0(3)\\2(1)+(-1)(-2)&2(4)+(-1)(3)\end{bmatrix} = \begin{bmatrix}5&20\\4&5\end{bmatrix}$.

$AB \neq BA$: matrix multiplication is not commutative.

9 Store revenue ★★★

$\begin{bmatrix}30&12&20\\25&18&10\end{bmatrix}\begin{bmatrix}0.50\\3.20\\2.40\end{bmatrix} = \begin{bmatrix}101.40\\94.10\end{bmatrix}$.

East: $30(0.5) + 12(3.2) + 20(2.4) = 15 + 38.4 + 48 = 101.4$. West: $12.5 + 57.6 + 24 = 94.1$.

The East store earned $\$101.40$ and the West store earned $\$94.10$.

10 A three-by-three determinant ★★★

Expand along the first row: $1\big((-1)(5) - 4(0)\big) - 2\big(3(5) - 4(2)\big) + 0$.

$= 1(-5) - 2(7) = -5 - 14 = -19$.

11 Finding inverses ★★★

$\det = 16 - 15 = 1$, so the inverse is $\begin{bmatrix}4&-3\\-5&4\end{bmatrix}$.
$\det = 8 - 6 = 2$, so the inverse is $\dfrac{1}{2}\begin{bmatrix}4&-6\\-1&2\end{bmatrix} = \begin{bmatrix}2&-3\\-\dfrac{1}{2}&1\end{bmatrix}$. Check: the first row of the product is $(4 - 3, -6 + 6) = (1, 0)$ and the second row is $(2 - 2, -3 + 4) = (0, 1)$.

12 Solve with an inverse ★★★

$A = \begin{bmatrix}5&3\\2&1\end{bmatrix}$, $\det A = 5 - 6 = -1$, so $A^{-1} = -1 \cdot \begin{bmatrix}1&-3\\-2&5\end{bmatrix} = \begin{bmatrix}-1&3\\2&-5\end{bmatrix}$.

$X = A^{-1}C = \begin{bmatrix}-1&3\\2&-5\end{bmatrix}\begin{bmatrix}11\\5\end{bmatrix} = \begin{bmatrix}-11+15\\22-25\end{bmatrix} = \begin{bmatrix}4\\-3\end{bmatrix}$.

The solution is $x = 4$, $y = -3$. Check: $20 - 9 = 11$ and $8 - 3 = 5$.

13 Gaussian elimination ★★★

Augmented matrix rows: $(1, 2, -1 \mid 3)$, $(2, 3, 1 \mid 11)$, $(3, -1, 1 \mid 10)$.

$R_2 - 2R_1 = (0, -1, 3 \mid 5)$ and $R_3 - 3R_1 = (0, -7, 4 \mid 1)$.

$R_3 - 7R_2 = (0, 0, -17 \mid -34)$, so $z = 2$. Then $-y + 6 = 5$ gives $y = 1$, and $x = 3 - 2 + 2 = 3$.

The solution is $(3, 1, 2)$.

14 Cramer’s Rule practice ★★★

$D = 2(-3) - 5(4) = -26$.

$D_x = (-4)(-3) - 5(18) = 12 - 90 = -78$, so $x = \dfrac{-78}{-26} = 3$.

$D_y = 2(18) - (-4)(4) = 36 + 16 = 52$, so $y = \dfrac{52}{-26} = -2$.

The solution is $(3, -2)$.

15 Shear then reflection ★★★

(a) The shear acts first, so the combined matrix is $FH = \begin{bmatrix}1&0\\0&-1\end{bmatrix}\begin{bmatrix}1&2\\0&1\end{bmatrix} = \begin{bmatrix}1&2\\0&-1\end{bmatrix}$.

(b) $(1, 1) \to (1 + 2, -1) = (3, -1)$; $(4, 1) \to (6, -1)$; $(1, 3) \to (1 + 6, -3) = (7, -3)$.

(c) $T$ has base $3$ and height $2$, so its area is $3$. Since $\det(FH) = -1$, the image area is $|-1| \cdot 3 = 3$. Check: the image has a horizontal base of length $3$ and height $2$, area $3$.

16 Rotation and reflection: order matters ★★★

$FR = \begin{bmatrix}1&0\\0&-1\end{bmatrix}\begin{bmatrix}0&-1\\1&0\end{bmatrix} = \begin{bmatrix}0&-1\\-1&0\end{bmatrix}$ (rotate first, then reflect).

$RF = \begin{bmatrix}0&-1\\1&0\end{bmatrix}\begin{bmatrix}1&0\\0&-1\end{bmatrix} = \begin{bmatrix}0&1\\1&0\end{bmatrix}$ (reflect first, then rotate).

$FR$ sends $(2, 5)$ to $(-5, -2)$, while $RF$ sends it to $(5, 2)$. The results differ, so the order of the transformations matters.

17 Area with determinants ★★★

(a) The area is $\dfrac{1}{2}\left|5(4) - 1(2)\right| = \dfrac{18}{2} = 9$.

(b) $P' = (3(5) + 0, 1(5) + 2(1)) = (15, 7)$ and $Q' = (6, 2 + 8) = (6, 10)$. The area is $\dfrac{1}{2}\left|15(10) - 7(6)\right| = \dfrac{108}{2} = 54$.

18 A system with a parameter ★★★

(a) $D = 2(9) - k(6) = 18 - 6k = 0$ gives $k = 3$.

(b) With $k = 3$, multiplying $2x + 3y = 5$ by $3$ gives $6x + 9y = 15$, which contradicts $6x + 9y = 12$. The lines are parallel, so there is no solution.

(c) For $k = 1$: $D = 18 - 6 = 12$, $D_x = 5(9) - 1(12) = 33$, $D_y = 2(12) - 5(6) = -6$. So $x = \dfrac{33}{12} = \dfrac{11}{4}$ and $y = \dfrac{-6}{12} = -\dfrac{1}{2}$.

19 Trail mix prices ★★★

Equations: $2a + r + 3c = 40$, $a + 2r + c = 22$, $3a + r + 2c = 38$. Swap rows 1 and 2 to get a leading $1$: $(1, 2, 1 \mid 22)$, $(2, 1, 3 \mid 40)$, $(3, 1, 2 \mid 38)$.

$R_2 - 2R_1 = (0, -3, 1 \mid -4)$ and $R_3 - 3R_1 = (0, -5, -1 \mid -28)$.

$3R_3 - 5R_2 = (0, 0, -8 \mid -64)$, so $c = 8$. Then $-3r + 8 = -4$ gives $r = 4$, and $a = 22 - 8 - 8 = 6$.

Almonds cost $\$6$ per pound, raisins $\$4$, and cashews $\$8$.