Interpretations of Solving Linear System
2 x 2:
Solve:
\[\begin{cases} 2x - y = 4 & (1) \\ x + 3y = 9 & (2) \end{cases}\]Solve by elimination & substitution:
\[ \begin{gathered} (1) \implies y = 2x - 4 \\[0.5em] (2) \implies x + 3(2x - 4) = 9 \\[0.5em] 7x = 9 + 12 = 21 \end{gathered} \]Hence \( x = 3, \quad y = 2(3) - 4 = 2 \)
Solution: \( (x, y) = (3, 2) \), or \( \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \end{pmatrix} \)
(I) Finding intersection between 2 st. lines (Look at each equation or row)
Figure 1
A coordinate graph with a blue vertical
y-axis and horizontal x-axis. Two black
straight lines are drawn. One line, labeled
\(2x - y = 4\) as equation (1), rises
steeply from lower left through the y-axis
near \(-4\) and crosses the x-axis near \(x
= 2\). The other line, labeled \(x + 3y =
9\) as equation (2), descends gently from
upper left crossing the y-axis near \(y =
3\) and the x-axis near \(x = 9\). The two
lines cross at a red dot marked with an
arrow labeled \((3,2)\) intersection between
(1) and (2). Marked axis values in red
include 3 and -4 on the y-axis and 2 and 9
on the x-axis.
(II) Writing vector as Linear Combinations (Look at "columns")
\[\begin{cases} 2x - y = 4 \\ x + 3y = 9 \end{cases}\]vector addition
\[\begin{pmatrix} 2x \\ x \end{pmatrix} + \begin{pmatrix} -y \\ 3y \end{pmatrix} = \begin{pmatrix} 4 \\ 9 \end{pmatrix}\]scalar multiplication
\[x\begin{pmatrix} 2 \\ 1 \end{pmatrix} + y\begin{pmatrix} -1 \\ 3 \end{pmatrix} = \begin{pmatrix} 4 \\ 9 \end{pmatrix}\]x and y are unknowns
(II) Writing vector as Linear Combinations (Look at "columns")
Figure 2
A hand-drawn vector diagram on x-y axes. The
y-axis (labeled y-axis, in blue) points up
and the x-axis (labeled x-axis, in blue)
points right. From the origin, a black
vector points to the upper-left labeled
\(\begin{pmatrix} -1 \\ 3 \end{pmatrix}\),
and another black vector points to the
lower-right labeled \(\begin{pmatrix} 2 \\ 1
\end{pmatrix}\). A longer black vector
points diagonally up-right to the point
labeled \(\begin{pmatrix} 4 \\ 9
\end{pmatrix}\). Red dashed lines complete a
parallelogram, illustrating vector addition.
Red annotations along the axes label the
scaled component vectors \(x\begin{pmatrix}
2 \\ 1 \end{pmatrix}\) and
\(y\begin{pmatrix} -1 \\ 3 \end{pmatrix}\),
showing the resultant as a linear
combination reaching \(\begin{pmatrix} 4 \\
9 \end{pmatrix}\).
Find x & y:
\[ \begin{gathered} x\begin{pmatrix} 2 \\ 1 \end{pmatrix} + y\begin{pmatrix} -1 \\ 3 \end{pmatrix} = \begin{pmatrix} 4 \\ 9 \end{pmatrix} \\[0.5em] 3\begin{pmatrix} 2 \\ 1 \end{pmatrix} + 2\begin{pmatrix} -1 \\ 3 \end{pmatrix} = \begin{pmatrix} 4 \\ 9 \end{pmatrix} \end{gathered} \](these 2 x's are not the same: the x in the diagram and the x as in x-axis.)
(III) Linear Transformation of Vectors
\[\begin{pmatrix} x \\ y \end{pmatrix} \xrightarrow{T} T\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 2x - y \\ x + 3y \end{pmatrix} = \begin{pmatrix} p \\ q \end{pmatrix}\]
Figure 3
Two separate x-y coordinate systems drawn
side by side to illustrate a linear
transformation T. On the left plane, an
arrow from the origin points to a
red-labeled point \(\begin{pmatrix} x \\ y
\end{pmatrix}\). A large curved arrow
labeled T points from the left plane to the
right plane. On the right plane, an arrow
from the origin points to a red-labeled
point, with the equation \(\begin{pmatrix} p
\\ q \end{pmatrix} = T\begin{pmatrix} x \\ y
\end{pmatrix} = \begin{pmatrix} 2x - y \\ x
+ 3y \end{pmatrix}\) written beside it. Both
axes are drawn in blue with the vertical
axis labeled y on the left plane and q on
the right plane.
(III) Linear Transformation of Vectors
\[\begin{pmatrix} x \\ y \end{pmatrix} \xrightarrow{T} T\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 2x - y \\ x + 3y \end{pmatrix} = \begin{pmatrix} p \\ q \end{pmatrix}\]Solving \( \begin{cases} 2x - y = 4 \\ x + 3y = 9 \end{cases} \) is equivalent to finding \( \begin{pmatrix} x \\ y \end{pmatrix} \) such that \( T\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 4 \\ 9 \end{pmatrix} \)
(ie. \( \begin{pmatrix} 4 \\ 9 \end{pmatrix} \) is the image of \( \begin{pmatrix} x \\ y \end{pmatrix} \) under \( T \).)
(III) Linear Transformation of Vectors
\[\begin{pmatrix} x \\ y \end{pmatrix} \xrightarrow{T} T\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 2x - y \\ x + 3y \end{pmatrix} = \begin{pmatrix} p \\ q \end{pmatrix}\]Solving \( \begin{cases} 2x - y = 4 \\ x + 3y = 9 \end{cases} \) is equivalent to finding \( \begin{pmatrix} x \\ y \end{pmatrix} \) such that \( T\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 4 \\ 9 \end{pmatrix} \)
(ie. \( \begin{pmatrix} x \\ y \end{pmatrix} \) is the pre-image of \( \begin{pmatrix} 4 \\ 9 \end{pmatrix} \) under \( T \).)
(III) Linear Transformation of Vectors
\[\begin{pmatrix} x \\ y \end{pmatrix} \xrightarrow{T} T\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 2x - y \\ x + 3y \end{pmatrix} = \begin{pmatrix} p \\ q \end{pmatrix}\]
Figure 4
Two coordinate planes drawn side by side,
each with a blue vertical axis labeled y
(left) and q (right) and a horizontal axis.
On the left plane a black arrow points from
the origin to a red dot labeled with the
column vector (x, y). A large pink curved
arrow labeled T points from the left plane
to the right plane. On the right plane a
black arrow points from the origin to a red
dot, labeled in red with \( T\begin{pmatrix}
x \\ y \end{pmatrix} = \begin{pmatrix} 4 \\
9 \end{pmatrix} \). The diagram illustrates
the linear transformation T mapping the
input vector to its image.
Example in \( \mathbb{R}^3 \)
3 × 3 system: (*)
\[\begin{cases} x + 2y + z = 1 &\text{(1)} \\ 3x + y + 4z = 0 &\text{(2)} \\ 2x + 2y + 3z = 2 &\text{(3)} \end{cases}\]Note
Each equation represents a plane in \( \mathbb{R}^3 \)
Figure 5
A three-dimensional coordinate system drawn
in blue with the z-axis pointing up, the
y-axis pointing to the lower right, and the
x-axis pointing to the lower left. A red
parallelogram represents a tilted plane
cutting across the axes. A red curved arrow
points to the plane with the label 'a plane
in R cubed', illustrating that each equation
of the system corresponds to a plane in
three-dimensional space.
(I) Solving (*) is to find the intersection between 3 planes in \( \mathbb{R}^3 \)
Figure 6
A hand-drawn three-dimensional sketch
showing three flat planes intersecting one
another, drawn as large quadrilaterals
crossing at various angles. Blue coordinate
axes point up, to the lower-left, and to the
lower-right. A red dot near the center marks
the common intersection point, labelled in
red text solution with an arrow pointing to
it. Red annotation text reads 3 planes (in
general) meet a point. Two of the planes are
labelled with circled numbers 2 (upper
right) and 1 (right side). Dashed lines
indicate hidden edges behind the planes.
(II) Writing Vector as linear Combination
\[ \begin{gathered} \begin{pmatrix} x + 2y + z = 1 \\ 3x + y + 4z = 0 \\ 2x + 2y + 3z = 2 \end{pmatrix} \\[0.5em] \begin{pmatrix} x \\ 3x \\ 2x \end{pmatrix} + \begin{pmatrix} 2y \\ y \\ 2y \end{pmatrix} + \begin{pmatrix} z \\ 4z \\ 3z \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} \\[0.5em] x \begin{pmatrix} 1 \\ 3 \\ 2 \end{pmatrix} + y \begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix} + z \begin{pmatrix} 1 \\ 4 \\ 3 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} \end{gathered} \](II) Writing Vector as linear Combination
Figure 7
A 3D coordinate diagram drawn by hand. Blue
axes are labelled z-axis (pointing up),
y-axis (pointing right and down), and x-axis
(pointing lower-left). From the common
origin, several vectors are drawn. Three
black vectors point upward-right and are
labelled with column vectors: (1,4,3),
(2,1,2), and (1,3,2). A red vector points up
and to the right, labelled with the column
vector (1,0,2), representing the target
vector expressed as a linear combination of
the three black vectors.
Find \( x, y, z \):
\[x \begin{pmatrix} 1 \\ 3 \\ 2 \end{pmatrix} + y \begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix} + z \begin{pmatrix} 1 \\ 4 \\ 3 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}\](III) Linear Transformation
Figure 8
Two three-dimensional coordinate systems
drawn side by side. The left system has blue
axes labelled x, y, z in red, with a black
vector from the origin reaching a point
labelled with the column vector (x, y, z) in
red. A large pink curved arrow points from
the left diagram to the right diagram,
representing the linear transformation T.
The right system has blue axes labelled p,
q, r in red, with a black vector from the
origin reaching a point. To the right of the
second system is the equation showing the
column vector (p, q, r) equals T applied to
(x, y, z), which equals the column vector
with entries x plus 2y plus z, 3x plus y
plus 4z, and 2x plus 2y plus 3z.
Find \( \begin{pmatrix} x \\ y \\ z \end{pmatrix} \) such that \( T\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} \)
Solution of (*) (Using elimination & substitution)
\[\begin{cases} x + 2y + z = 1 & (1) \\ 3x + y + 4z = 0 & (2) \\ 2x + 2y + 3z = 2 & (3) \end{cases}\](1) \(\Rightarrow\) \( x = 1 - 2y - z \)
(2) \(\Rightarrow\) \( 3(1 - 2y - z) + y + 4z = 0 \Rightarrow -5y + z = -3 \)
(3) \(\Rightarrow\) \( 2(1 - 2y - z) + 2y + 3z = 2 \Rightarrow -2y + z = 0 \)
\[ \begin{gathered} y = 1 \qquad z = 2 \\[0.5em] x = 1 - 2(1) - 2 = -3 \end{gathered} \]Solution: \( \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} -3 \\ 1 \\ 2 \end{pmatrix} \)
"Just to make sure":
(I):
\[\begin{cases} (-3) + 2(1) + (2) = 1 \\ 3(-3) + (1) + 4(2) = 0 \\ 2(-3) + 2(1) + 3(2) = 2 \end{cases}\](II):
\[-3 \begin{pmatrix} 1 \\ 3 \\ 2 \end{pmatrix} + 1 \begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix} + 2 \begin{pmatrix} 1 \\ 4 \\ 3 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}\]