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# Solved Examples on Linear Transformations using Matrices

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(1.) WASSCE:FM Two linear transformations $P$ and $Q$ in the $xy$ plane are defined as:

$P:(x, y) \rightarrow (x + 2y, x) \\[3ex] Q:(x, y) \rightarrow (x + y, -x + 2y) \\[3ex]$ (a.) Write down the matrices of $P$ and $Q$
(b.) Find $M$ such that $2P + 3Q - MQ = 5I$, where $M$ is a $2 * 2$ matrix and $I$ is a $2 * 2$ identity matrix.

$(a.) \\[3ex] Let\;\;matrices \\[3ex] P = \begin{bmatrix} a & b \\[3ex] c & d \end{bmatrix} \;\;\;and\;\;\; Q = \begin{bmatrix} e & f \\[3ex] g & h \end{bmatrix} \\[3ex]$
$\underline{Transformations\;\;to\;\;Matrices} \\[3ex] \boldsymbol{P} \\[3ex] \begin{bmatrix} a & b \\[3ex] c & d \end{bmatrix} \begin{bmatrix} x \\[3ex] y \end{bmatrix} = \begin{bmatrix} x + 2y \\[3ex] x \end{bmatrix} \\[3ex]$
$ax + by = x + 2y \\[3ex] cx + dy = x \\[3ex] \implies \\[3ex] ax + by = 1x + 2y \\[3ex] cx + dy = 1x + 0y \\[3ex] \implies \\[3ex] a = 1 \\[3ex] b = 2 \\[3ex] c = 1 \\[3ex] d = 0 \\[3ex] P = \begin{bmatrix} 1 & 2 \\[3ex] 1 & 0 \end{bmatrix} \\[3ex]$
$\boldsymbol{Q} \\[3ex] \begin{bmatrix} e & f \\[3ex] g & h \end{bmatrix} \begin{bmatrix} x \\[3ex] y \end{bmatrix} = \begin{bmatrix} x + y \\[3ex] -x + 2y \end{bmatrix} \\[3ex]$
$ex + fy = x + y \\[3ex] gx + hy = -x + 2y \\[3ex] \implies \\[3ex] e = 1 \\[3ex] f = 1 \\[3ex] g = -1 \\[3ex] h = 2 \\[3ex] Q = \begin{bmatrix} 1 & 1 \\[3ex] -1 & 2 \end{bmatrix} \\[3ex]$
$(b.) \\[3ex] 2P + 3Q - MQ = 5I \\[3ex] 2P \\[3ex] = 2\begin{bmatrix} 1 & 2 \\[3ex] 1 & 0 \end{bmatrix} \\[7ex] = \begin{bmatrix} 2(1) & 2(2) \\[3ex] 2(1) & 2(0) \end{bmatrix} \\[7ex] = \begin{bmatrix} 2 & 4 \\[3ex] 2 & 0 \end{bmatrix} \\[7ex] 3Q \\[3ex] = 3\begin{bmatrix} 1 & 1 \\[3ex] -1 & 2 \end{bmatrix} \\[7ex] = \begin{bmatrix} 3(1) & 3(1) \\[3ex] 3(-1) & 3(2) \end{bmatrix} \\[7ex] = \begin{bmatrix} 3 & 3 \\[3ex] -3 & 6 \end{bmatrix} \\[7ex] Let\;\;M = \begin{bmatrix} p & v \\[3ex] m & n \end{bmatrix} \\[7ex] MQ \\[3ex] \begin{bmatrix} p & v \\[3ex] m & n \end{bmatrix} \begin{bmatrix} 1 & 1 \\[3ex] -1 & 2 \end{bmatrix} \\[7ex] = \begin{bmatrix} p(1) + v(-1) & p(1) + v(2) \\[3ex] m(1) + n(-1) & m(1) + n(2) \end{bmatrix} \\[7ex] = \begin{bmatrix} p - v & p + 2v \\[3ex] m - n & m + 2n \end{bmatrix} \\[7ex] 2P + 3Q - MQ \\[3ex] = \begin{bmatrix} 2 & 4 \\[3ex] 2 & 0 \end{bmatrix} + \begin{bmatrix} 3 & 3 \\[3ex] -3 & 6 \end{bmatrix} - \begin{bmatrix} p - v & p + 2v \\[3ex] m - n & m + 2n \end{bmatrix} \\[7ex] = \begin{bmatrix} 2 + 3 - (p - v) & 4 + 3 - (p + 2v) \\[3ex] 2 + -3 - (m - n) & 0 + 6 - (m + 2n) \end{bmatrix} \\[7ex] = \begin{bmatrix} 5 - p + v & 7 - p - 2v \\[3ex] -1 - m + n & 6 - m - 2n \end{bmatrix} \\[7ex] 5I \\[3ex] = 5\begin{bmatrix} 1 & 0 \\[3ex] 0 & 1 \end{bmatrix} \\[7ex] = \begin{bmatrix} 5(1) & 5(0) \\[3ex] 5(0) & 5(1) \end{bmatrix} \\[7ex] = \begin{bmatrix} 5 & 0 \\[3ex] 0 & 5 \end{bmatrix} \\[7ex] 2P + 3Q - MQ = 5I \\[3ex] \begin{bmatrix} 5 - p + v & 7 - p - 2v \\[3ex] -1 - m + n & 6 - m - 2n \end{bmatrix} = \begin{bmatrix} 5 & 0 \\[3ex] 0 & 5 \end{bmatrix} \\[7ex] \implies \\[3ex] 5 - p + v = 5 \\[3ex] -p + v = 5 - 5 \\[3ex] -p + v = 0...eqn.(1) \\[3ex] 7 - p - 2v = 0 \\[3ex] -p - 2v = 0 - 7 \\[3ex] -p - 2v = -7...eqn.(2) \\[3ex] eqn.(1) - eqn.(2) \implies \\[3ex] (-p + v) - (-p - 2v) = 0 - (-7) \\[3ex] -p + v + p + 2v = 7 \\[3ex] 3v = 7 \\[3ex] v = \dfrac{7}{3} \\[5ex] 2 * eqn.(1) + eqn.(2) \implies \\[3ex] 2(-p + v) + (-p - 2v) = 2(0) + (-7) \\[3ex] -2p + 2v - p - 2v = 0 - 7 \\[3ex] -3p = -7 \\[3ex] p = \dfrac{-7}{-3} \\[3ex] p = \dfrac{7}{3} \\[5ex] Also: \\[3ex] -1 - m + n = 0 \\[3ex] -m + n = 0 + 1 \\[3ex] -m + n = 1...eqn.(3) \\[3ex] 6 - m - 2n = 5 \\[3ex] -m - 2n = 5 - 6 \\[3ex] -m - 2n = -1...eqn.(4) \\[3ex] eqn.(3) - eqn.(4) \implies \\[3ex] -m + n - (-m - 2n) = 1 - (-1) \\[3ex] -m + n + m + 2n = 1 + 1 \\[3ex] 3n = 2 \\[3ex] n = \dfrac{2}{3} \\[5ex] 2 * eqn.(3) + eqn.(4) \implies \\[3ex] 2(-m + n) + (-m - 2n) = 2(1) + (-1) \\[3ex] -2m + 2n - m - 2n = 2 - 1 \\[3ex] -3m = 1 \\[3ex] m = -\dfrac{1}{3} \\[5ex] \implies \\[3ex] M = \begin{bmatrix} p & v \\[3ex] m & n \end{bmatrix} = \begin{bmatrix} \dfrac{7}{3} & \dfrac{7}{3} \\[5ex] -\dfrac{1}{3} & \dfrac{2}{3} \end{bmatrix}$
(2.) WASSCE:FM Given that $M:(x, y) \rightarrow (7y, 3x - y)$ and $N:(x, y) \rightarrow (2x - y, 5x + 3y)$

(a.) Write down the matrices $M$ and $N$ of the linear transformations
(b.) Find the image of $P(2, -3)$ under the linear transformation $N$ followed by $M$
(c.) Find the coordinates of the point $Q$ whose image is $Q'(2, 4)$ under the linear transformation $N$

$(a.) \\[3ex] Let\;\;matrices \\[3ex] M = \begin{bmatrix} a & b \\[3ex] c & d \end{bmatrix} \;\;\;and\;\;\; N = \begin{bmatrix} e & f \\[3ex] g & h \end{bmatrix} \\[3ex]$
$\underline{Transformations\;\;to\;\;Matrices} \\[3ex] \boldsymbol{M} \\[3ex] \begin{bmatrix} a & b \\[3ex] c & d \end{bmatrix} \begin{bmatrix} x \\[3ex] y \end{bmatrix} = \begin{bmatrix} 7y \\[3ex] 3x - y \end{bmatrix} \\[3ex]$
$ax + by = 7y \\[3ex] cx + dy = 3x - y \\[3ex] \implies \\[3ex] ax + by = 0x + 7y \\[3ex] cx + dy = 3x - 1y \\[3ex] \implies \\[3ex] a = 0 \\[3ex] b = 7 \\[3ex] c = 3 \\[3ex] d = -1 \\[3ex] M = \begin{bmatrix} 0 & 7 \\[3ex] 3 & -1 \end{bmatrix} \\[3ex]$
$\boldsymbol{N} \\[3ex] \begin{bmatrix} e & f \\[3ex] g & h \end{bmatrix} \begin{bmatrix} x \\[3ex] y \end{bmatrix} = \begin{bmatrix} 2x - y \\[3ex] 5x + 3y \end{bmatrix} \\[3ex]$
$ex + fy = 2x - y \\[3ex] gx + hy = 5x + 3y \\[3ex] \implies \\[3ex] e = 2 \\[3ex] f = -1 \\[3ex] g = 5 \\[3ex] h = 3 \\[3ex] N = \begin{bmatrix} 2 & -1 \\[3ex] 5 & 3 \end{bmatrix} \\[3ex]$
$(b.) \\[3ex] P(2, -3) \\[3ex] Linear\;\;Transformation:\;\;N\;\;followed\;\;by\;\;M \\[3ex] Can\;\;be\;\;solved\;\;in\;\;at\;\;least\;\;2\;\;ways \\[3ex] \underline{1st\;\;Method} \\[3ex] N:(x, y) \rightarrow (2x - y, 5x + 3y) \\[3ex] P(2, -3) \\[3ex] \rightarrow (2(2) - (-3), 5(2) + 3(-3)) \\[3ex] \rightarrow (4 + 3, 10 - 9) \\[3ex] \rightarrow (7, 1) \\[3ex] M:(x, y) \rightarrow (7y, 3x - y) (7, 1) \\[3ex] \rightarrow (7(1), 3(7) - 1) \\[3ex] \rightarrow (7, 21 - 1) \\[3ex] \rightarrow (7, 20) \\[3ex] \underline{2nd\;\;Method} \\[3ex] P(2, -3)\;\;for\;\;N\;\;followed\;\;by\;\;M = MN * P \\[3ex] P = \begin{bmatrix} 2 \\[3ex] -3 \end{bmatrix} \\[7ex] MN \\[3ex] \begin{bmatrix} 0 & 7 \\[3ex] 3 & -1 \end{bmatrix} \begin{bmatrix} 2 & -1 \\[3ex] 5 & 3 \end{bmatrix} \\[7ex] = \begin{bmatrix} 0(2) + 7(5) & 0(-1) + 7(3) \\[3ex] 3(2) + -1(5) & 3(-1) + -1(3) \end{bmatrix} \\[7ex] = \begin{bmatrix} 0 + 35 & 0 + 21 \\[3ex] 6 - 5 & -3 - 3 \end{bmatrix} \\[7ex] = \begin{bmatrix} 35 & 21 \\[3ex] 1 & -6 \end{bmatrix} \\[7ex] MN * P \\[3ex] = \begin{bmatrix} 35 & 21 \\[3ex] 1 & -6 \end{bmatrix} * \begin{bmatrix} 2 \\[3ex] -3 \end{bmatrix} \\[7ex] = \begin{bmatrix} 35(2) + 21(-3) \\[3ex] 1(2) + -6(-3) \end{bmatrix} \\[7ex] = \begin{bmatrix} 7 \\[3ex] 20 \end{bmatrix} \\[7ex] \therefore P(2, -3)\;\;for\;\;N\;\;followed\;\;by\;\;M = (7, 20) \\[3ex] (c.) \\[3ex] \underline{Under\;\;the\;\;Linear\;\;Transformation\;\;N} \\[3ex] Matrix * Point = Image \\[3ex] Let\;\;Point\;Q = (c, d) \\[3ex] Matrix\;N * Point\;Q = Image\;Q' \\[3ex] \begin{bmatrix} 2 & -1 \\[3ex] 5 & 3 \end{bmatrix} * \begin{bmatrix} c \\[3ex] d \end{bmatrix} = \begin{bmatrix} 2 \\[3ex] 4 \end{bmatrix} \\[7ex] 2c - d = 2...eqn.(1) \\[3ex] 5c + 3d = 4...eqn.(2) \\[3ex] 3 * eqn.(1) + eqn.(2) \implies \\[3ex] 3(2c - d) + (5c + 3d) = 3(2) + 4 \\[3ex] 6c - 3d + 5c + 3d = 6 + 4 \\[3ex] 11c = 10 \\[3ex] c = \dfrac{10}{11} \\[5ex] 5 * eqn.(1) - 2 * eqn.(2) \implies \\[3ex] 5(2c - d) - 2(5c + 3d) = 5(2) - 2(4) \\[3ex] 10c - 5d - 10c - 6d = 10 - 8 \\[3ex] -11d = 2 \\[3ex] d = -\dfrac{2}{11} \\[5ex] \therefore Point\;Q = \left(\dfrac{10}{11}, -\dfrac{2}{11}\right)$
(3.)

Rearrange $eqn.(1)$ to "line up"

$x - 7y = -6 ...new\:\: eqn.(1) \\[3ex] \begin{bmatrix} 1 & -7 \\[3ex] 1 & -1 \end{bmatrix} \begin{bmatrix} x \\[3ex] y \end{bmatrix} = \:\:\: \begin{bmatrix} -6 \\[3ex] 6 \end{bmatrix} \\[3ex]$
$x = \dfrac{ \begin{vmatrix} -6 & -7 \\[3ex] 6 & -1 \end{vmatrix}} {\begin{vmatrix} 1 & -7 \\[3ex] 1 & -1 \end{vmatrix}} = \dfrac{6 + 42}{-1 + 7} = \dfrac{48}{6} \\[5ex] x = 8 \\[3ex]$
$y = \dfrac{ \begin{vmatrix} 1 & -6 \\[3ex] 1 & 6 \end{vmatrix}} {\begin{vmatrix} 1 & -7 \\[3ex] 1 & -1 \end{vmatrix}} = \dfrac{6 + 6}{-1 + 7} = \dfrac{12}{6} \\[5ex] y = 2 \\[3ex]$ The two-digit number is $82$
(4.)

Rearrange $eqn.(2)$ to "line up"

$1116 = x - y \\[3ex] x - y = 1116 ...new\:\: eqn.(2) \\[3ex] \begin{bmatrix} 1 & 1 \\[3ex] 1 & -1 \end{bmatrix} \begin{bmatrix} x \\[3ex] y \end{bmatrix} = \:\:\: \begin{bmatrix} 4802 \\[3ex] 1116 \end{bmatrix} \\[3ex]$
$x = \dfrac{ \begin{vmatrix} 4802 & 1 \\[3ex] 1116 & -1 \end{vmatrix}} {\begin{vmatrix} 1 & 1 \\[3ex] 1 & -1 \end{vmatrix}} = \dfrac{-4802 - 1116}{-1 - 1} = \dfrac{-5918}{-2} \\[5ex] x = 2959 \\[3ex]$
$y = \dfrac{ \begin{vmatrix} 1 & 4802 \\[3ex] 1 & 1116 \end{vmatrix}} {\begin{vmatrix} 1 & 1 \\[3ex] 1 & -1 \end{vmatrix}} = \dfrac{1116 - 4802}{-1 - 1} = \dfrac{-3686}{-2} \\[5ex] y = 1843 \\[3ex]$ The average apartment rent in the City of Santa Monica is $\$2959.00$The average apartment rent in the City of San Francisco is$\$1843.00$
(5.)

$\begin{bmatrix} 1 & -1 \\[3ex] 1 & 1 \end{bmatrix} \begin{bmatrix} x \\[3ex] y \end{bmatrix} = \:\:\: \begin{bmatrix} 141 \\[3ex] 183 \end{bmatrix} \\[3ex]$
$x = \dfrac{ \begin{vmatrix} 141 & -1 \\[3ex] 183 & 1 \end{vmatrix}} {\begin{vmatrix} 1 & -1 \\[3ex] 1 & 1 \end{vmatrix}} = \dfrac{141 + 183}{1 + 1} = \dfrac{324}{2} \\[5ex] x = 162 \\[3ex]$
$y = \dfrac{ \begin{vmatrix} 1 & 141 \\[3ex] 1 & 183 \end{vmatrix}} {\begin{vmatrix} 1 & -1 \\[3ex] 1 & 1 \end{vmatrix}} = \dfrac{183 - 141}{1 + 1} = \dfrac{42}{2} \\[5ex] y = 21 \\[3ex]$ The speed of the plane is $162\: mph$
The speed of the wind is $21\: mph$
(6.)

Rearrange $eqns. (1.) \:\:and\:\: (2)$ to "line up"

$x - y = -10 ...new\:\: eqn.(1) \\[3ex] -25 = 6x - 4y \\[3ex] 6x - 4y = -25 ...new\:\: eqn.(2) \\[3ex] \begin{bmatrix} 1 & -1 \\[3ex] 6 & -4 \end{bmatrix} \begin{bmatrix} x \\[3ex] y \end{bmatrix} = \:\:\: \begin{bmatrix} -10 \\[3ex] -25 \end{bmatrix} \\[3ex]$
$x = \dfrac{ \begin{vmatrix} -10 & -1 \\[3ex] -25 & -4 \end{vmatrix}} {\begin{vmatrix} 1 & -1 \\[3ex] 6 & -4 \end{vmatrix}} = \dfrac{40 - 25}{-4 + 6} = \dfrac{15}{2} \\[5ex] x = 7.5 \\[3ex]$
$y = \dfrac{ \begin{vmatrix} 1 & -10 \\[3ex] 6 & -25 \end{vmatrix}} {\begin{vmatrix} 1 & -1 \\[3ex] 6 & -4 \end{vmatrix}} = \dfrac{-25 + 60}{-4 + 6} = \dfrac{35}{2} \\[5ex] y = 17.5 \\[3ex]$ $7.5$ pounds of cashews needs to be mixed with $10$ pounds of peanuts to give $17.5$ pounds of cashews and peanuts so that the mixture will produce the same revenue as would selling the nuts separately.

(21.)

$A.\:\: -5\:\:and\:\:4 \\[3ex] B.\:\: -4\:\:and\:\:5 \\[3ex] C.\:\: -3\:\:and\:\:2 \\[3ex] D.\:\: -1\:\:and\:\:9 \\[3ex] E.\:\: -\sqrt{20}\:\:and\:\:\sqrt{20} \\[3ex]$

Compare

$\begin{bmatrix} a & b \\[3ex] c & d \end{bmatrix} \:\:and\:\: \begin{bmatrix} (x + 3) & 7 \\[3ex] 2 & (x - 2) \end{bmatrix} \\[7ex] a = x + 3 \\[3ex] b = 7 \\[3ex] c = 2 \\[3ex] d = x - 2 \\[3ex] determinant = ad - bc \\[3ex] determinant = (x + 3)(x - 2) - 7(2) \\[3ex] determinant = 0...from\:\:the\:\:Question \\[3ex] \rightarrow (x + 3)(x - 2) - 7(2) = 0 \\[3ex] x^2 - 2x + 3x - 6 - 14 = 0 \\[3ex] x^2 + x - 20 = 0 \\[3ex] (x + 5)(x - 4) = 0 \\[3ex] x + 5 = 0 \:\:OR\:\: x - 4 = 0 \\[3ex] x = -5 \:\:OR\:\: x = 4$
(22.)

Rearrange $eqns. (1.) \:\:and\:\: (2)$ to "line up"

$x - y = -10 ...new\:\: eqn.(1) \\[3ex] -25 = 6x - 4y \\[3ex] 6x - 4y = -25 ...new\:\: eqn.(2) \\[3ex] \begin{bmatrix} 1 & -1 \\[3ex] 6 & -4 \end{bmatrix} \begin{bmatrix} x \\[3ex] y \end{bmatrix} = \:\:\: \begin{bmatrix} -10 \\[3ex] -25 \end{bmatrix} \\[3ex]$
$x = \dfrac{ \begin{vmatrix} -10 & -1 \\[3ex] -25 & -4 \end{vmatrix}} {\begin{vmatrix} 1 & -1 \\[3ex] 6 & -4 \end{vmatrix}} = \dfrac{40 - 25}{-4 + 6} = \dfrac{15}{2} \\[5ex] x = 7.5 \\[3ex]$
$y = \dfrac{ \begin{vmatrix} 1 & -10 \\[3ex] 6 & -25 \end{vmatrix}} {\begin{vmatrix} 1 & -1 \\[3ex] 6 & -4 \end{vmatrix}} = \dfrac{-25 + 60}{-4 + 6} = \dfrac{35}{2} \\[5ex] y = 17.5 \\[3ex]$ $7.5$ pounds of cashews needs to be mixed with $10$ pounds of peanuts to give $17.5$ pounds of cashews and peanuts so that the mixture will produce the same revenue as would selling the nuts separately.