Download Math 018 Review Sheet v.3

Math 018 Review Sheet v.3
Tyrone Crisp
Spring 2007
1.1 - Slopes and Equations of Lines
Slopes:
• Find slopes of lines using the slope formula m =
y2 − y1
.
x2 − x1
• Positive slope % the line slopes up to the right. Negative slope & the line slopes down to
the right. Slope equals zero → horizontal line. Slope undefined ↑ vertical line.
• Large slope (in absolute value) ⇒ steep line. Eg, a line with slope 5 is steeper than a line
with slope 21 . A line with slope −6 is steeper than a line with slope −2.
Equations:
• Slope-intercept form: The equation of the line with slope m and y-intercept b is y = mx+b.
• Point-Slope form: The equation of the line through the point (x1 , y1 ) and with slope m is
y − y1 = m(x − x1 ).
• Line through two points: to find the equation of the line through points (x1 , y1 ) and (x2 , y2 ),
first find the slope using the slope formula, then apply the point-slope formula to find the
equation.
• Horizontal lines: The equation of a horizontal line has the form y = b, where b is the
y-intercept.
• Vertical lines: The equation of a vertical line has the form x = a, where a is the x-intercept.
Parallel and Perpendicular lines:
• If two lines are parallel, they have the same slope (or they are both vertical).
• If two lines are perpendicular, the product of their slopes is equal to −1 (or one is horizontal
and the other vertical).
• The condition for perpendicular lines can be rephrased as “If two lines are perpendicular,
then their slopes are negative reciprocals of one another.” (To find the negative reciprocal of
a number, write it as a fraction, turn the fraction upside-down and multiply by −1.)
Graphs:
• To find the y-intercept of a line, put x = 0 in its equation and solve for y.
• To find the x-intercept, put y = 0 in its equation and solve for x.
• To draw the graph of a line, find its x- and y-intercepts, plot these two points on the plane,
and draw a straight line through them.
• If the x- and y-intercepts are both zero, you’ll need to find another point before you can draw
the line. Choose any value for x (1 is usually a good choice) and plug that value into the
equation. Solve for y. This tells you that the point (x1 , y1 ) lies on the line, where x1 is the x
value you chose, and y1 is the resulting y value. Plot this point, and join the dots.
• An equation like y = b gives a horizontal line; to graph it, just draw a horizontal line with
y-intercept b.
• An equation like x = a gives a vertical line; to graph it, just draw a vertical line with
x-intercept a.
1.2 - Linear Functions and Applications
Linear Functions:
• If y and x are two quantities, then saying that “y is a linear function of x” means that the
value of y depends entirely on the value of x, and that when you draw a graph of y against x
you get a straight line.
• In this situation, we call y the “dependent variable” (because it depends on x). We call x the
“independent variable”.
• To emphasise that a certain quantity is a function of x, we often write it as f(x) rather than
y.
• Given a formula for a function f(x), you can find (for example) f(4) by substituting 4 for x
in the formula.
• To find x such that f(x) = 7 (for example), replace f(x) by 7 in the given formula and solve
for x.
Cost and Break-Even Analysis:
• The cost of producing any product can be broken down into the “fixed cost” (eg. for equipping a factory, training staff, etc.) and a “marginal cost” (the cost in materials and labour
required to produce one unit of the product).
• The cost function for producing x units of a certain product is given by the formula C(x) =
mx + b, where m is the marginal cost and b is the fixed cost.
• If the product is sold at a price of p dollars per unit, then the revenue function for selling x
units is R(x) = px.
• The profit function for producing and selling x units of a product is given by the formula
P(x) = R(x) − C(x): “Profit = Revenue − Cost”.
• “Negative profit” means “loss”.
• The “break-even point” is the number of units one would need to produce and sell in order
to make zero profit. To find this point, find the formula for the profit function P(x) and then
solve P(x) = 0.
Temperature:
• We knew that 0◦ C = 32◦ F and 100◦ C = 212◦ F. We then found a formula for converting
between Celsius and Fahrenheit by finding the equation of the straight line through (0, 32)
and (100, 212).
• The actual formula was F = 95 C+32 (but you won’t need to commit this equation to memory
for the exam).
• To convert from one temperature scale to the other, plug in the temperature you have for the
appropriate variable, and solve for the other (eg. to convert 40◦ C into Fahrenheit, plug in
C = 40 and solve for F).
2.1 - Systems of Linear Equations - The Echelon Method
Linear Equations:
• A linear equation in the variables x1 , x2 , . . . , xn is an equation which can be written in the
form a1 x1 + a2 x2 + . . . + an xn = k, where a1 , a2 , . . . , an , k are numbers.
• A solution to such an equation is a list (s1 , s2 , . . . , sn ) of numbers which make the equation
correct when you substitute x1 = s2 , x2 = s2 , . . ., xn = sn into the equation.
• A system of linear equations is a collection of linear equations in the same variables.
• A solution to a system of linear equations is a solution which works for every equation in the
system.
• The above four points are just terminology. You don’t need to memorise it word-for-word,
just understand what it means.
• There are three possibilities for the number of solutions to a given system:
1. Exactly one solution (“Unique Solution”);
2. No solutions (“Inconsistent”);
3. Infinitely many solutions (“Dependent”).
• In the case of systems of two equations in two variables, you can determine which of the
above possibilities applies by graphing the two lines and seeing whether they intersect at a
single point (unique solution), not at all (inconsistent), or at infinitely many points, i.e. they
are the same line (dependent).
Solving Linear Systems - Echelon Method:
• The following “transformations” can be applied to a system without changing its solutions:
1. Interchange two equations.
2. Multiply one equation by a nonzero constant.
3. Replace one equation by a nonzero multiple of itself plus a multiple of another equation.
• The strategy for solving a system of two equations in two variables is as follows:
1. Eliminate the first variable (usually called x) from the second equation.
2. Solve the second equation for the second variable (usually called y).
3. Substitute this y-value into the first equation and solve for x.
• If you get an equation like 0 = b, where b is some nonzero number, then the system is
inconsistent, and there is nothing more to be done.
• If you get an equation 0 = 0, the system is dependent. You can then find the general solution
by solving the first equation for x in terms of y.
2.2 - The Gauss-Jordan Method
Matrices:
• We can replace a system of equations by an “augmented matrix”.
• Each row of the augmented matrix corresponds to an equation in the system.
• Each column, except for the last, corresponds to a variable.
• The last column corresponds to the constants, and is usually separated from the others by a
vertical line.
• To write a system of equations as a matrix:
1. Rearrange all of the equations so that all variables are on the left; all constants are on
the right; and all variables are in the same order.
2. Then working from left to right, top to bottom, put the numbers from the equations
into a matrix. If a variable is missing from a particular equation, put a zero as the
corresponding entry of the matrix.
• The following three “row operations” may be performed on an augmented matrix without
changing its solutions:
1. Interchange two rows.
2. Multiply all the entries in one row by a nonzero constant.
3. Replace one row by a nonzero multiple of itself plus a multiple of another row.
The Gauss-Jordan Method:
• To solve a system of equations, do the following:
1. Rearrange the equations into standard form.
2. Write the system as an augmented matrix. If there is a common factor in any row,
divide it out.
3. In the first column, eliminate all entries except the first. (“Eliminate” means “Use row
operations to make equal to zero”.) If there is now a common factor in any row, divide
it out.
4. In the second column, eliminate all entries except the second. Divide out common
factors.
5. In the third column, eliminate all entries except the third. Divide out common factors.
6. Keep going until you’ve dealt with every column.
7. Translate back into equations; the system should be solved.
• In the above procedure, it may be necessary to interchange two rows to get a nonzero entry
in a diagonal position.
• If some row of your matrix becomes [0 0 . . . 0 | b] where b 6= 0, the system is inconsistent.
Game over.
• If you get a row consisting entirely of zeros, the system is dependent. To find the general
solution, translate the matrix back into equations and solve in terms of a parameter.
Word Problems:
• First, you should check to see if the question is asking about a special situation which we’ve
studied in detail, eg. graphs, break-even analysis or temperature conversion. In this case,
apply the methods appropriate to that kind of question.
• Otherwise, here is a general strategy:
1. Identify the two or three (or very rarely, four) different quantities whose values you’re
being asked to find.
2. Assign a different letter to each of those quantities. Choose the letters so that it’s
obvious what they represent.
3. Use the information you’re given to find some equations. You need to find as many
equations as you have variables.
4. Solve these equations.
5. Interpret your solution back into words.
2.3 Matrices - Addition and Subtraction
Matrices—Generalities:
• A matrix (not augmented) is just a rectangular array of numbers. Each number in the matrix
is called an entry. Each horizontal line of entries is called a row. Each vertical line of entries
is called a column.
• A matrix with m rows and n columns is called an m × n matrix. For example,
1 2 3
4 5 6
is a 2 × 3 matrix.
• A matrix with just a single row is called a row matrix. A matrix with just a single column is
called a column matrix. A matrix with just one row and just one column is more commonly
known as a number. A matrix with the same number of rows as columns is called a square
matrix.
• Two matrices are equal if they are of the same size, and each entry in one is equal to the
corresponding entry in the other.
Adding and Subtracting Matrices:
• You can only add or subtract matrices that have the same size. If two matrices have different
sizes, their sum is undefined.
• If A and B are two matrices of the same size, then their sum A + B is another matrix of the
same size, each of whose entries is equal to the sum of the corresponding entries of A and
B.
• Similarly, if A and B have the same size, then A − B also has the same size as A and B, and
the entries of A − B are obtained by subtracting the entries of B from those of A.
• Basically, adding and subtracting matrices is defined in the obvious way. Just be careful with
minus signs.
• The matrix whose entries are all 0 is called the zero matrix. There is a zero matrix of each
size: eg the 2 × 2 zero matrix, the 5 × 7 zero matrix, etc.
• The zero matrix behaves like the number 0, in the sense that A + 0 = 0 for every matrix A
(the 0 in this equation denotes the zero matrix, not the number zero - you can’t add a matrix
and a number).
• If A is a matrix, then the matrix you get by negating (i.e. changing the sign in front of)
each entry of A is called the negative of A, denoted −A. This matrix has the property that
A + (−A) = 0, where 0 here means the zero matrix.
2.4 Matrix Multiplication
Scalar multiplication:
• “Scalar” means “number”.
• When you multiply a matrix by a number, you get a matrix of the same size as the one you
started with.
• To multiply a matrix by a number, just multiply each entry of the matrix by that number. Be
careful with negative signs.
Row × column:
• If R is a 1 × n row matrix, and C is an n × 1 column matrix, then RC is just a number (or in
other words, a 1 × 1 matrix).
• To work out which number it is, multiply the first entry of R by the first entry of C; add the
product of the second entries; add the product of the third entries; and so on.
• In symbols,
r 1 r2 r3
 
c1
 c2 


 
. . . rn  c3  = r1 c1 + r2 c2 + r3 c3 + . . . + rn cn .
 .. 
.
cn
• If the row and the column don’t have the same number of entries, their product is undefined.
• The product CR, (i.e. in the opposite order to what we’ve been talking about), is not the
same as RC (in fact, the product CR will be an n × n matrix, not a number).
Matrix × matrix:
• Warning: The product of two matrices is not defined in the obvious way: i.e. you don’t just
multiply corresponding entries together.
• If A is an m × n matrix, and B is a p × q matrix, then the product AB is only defined if
n = p; if this is the case, the product is an m × q matrix.
• The way to remember this is: if you want to multiply A times B (in that order), write each
matrix with its size underneath:
A B
m×n p×q
The product AB is defined only if the inner numbers match (i.e. n = p), and then the size
of the product is given by the outside numbers (m × q).
• Once you know the size of the product, calculate each entry using the formula:
Entry in i’th row, j’th column of AB = (i’th row of A) · (j’th column of B),
where the product on the right is a product row×column as discussed above.
• Warning: Even if AB and BA are both defined, in most cases they will not be equal; they
might even be of different sizes. So you need to be careful about the order in which you
multiply.
• Matrix multiplication satisfies the following properties (assume in each case that the matrices
are of the appropriate sizes, so that all of the products are defined) :
1. (AB)C = A(BC)
2. A(B + C) = AB + AC
3. (A + B)C = AC + BC
4. 0A = A0 = 0, where 0 denotes the zero matrix.
• A square (n × n) matrix which has 1’s down the main diagonal and zeros everywhere else,
is called the n × n identity matrix, denoted In or just I. This matrix has the property that
IA = AI = A for every n × n matrix A. That is, in the world of n × n matrices, the identity
matrix I acts just like the number 1.
2.5 Matrix Inverses
Inverses—Generalities:
• A square (n × n) matrix is called invertible if there is another n × n matrix B with the
property that AB = BA = In : i.e. when you multiply A by B on either side, you get the
identity matrix.
• If such a matrix B exists, it is called the inverse of A, and is denoted by A−1 .
• Not every matrix is invertible. For example, the zero matrix is never invertible; there are
other (nonzero) matrices which are also not invertible.
Determinants:
• The determinant of a square matrix A is a number, denoted by det(A).
• The matrix A is invertible if and only if det(A) does not equal zero.
• 2×2: For a 2 × 2 matrix, the determinant is given by the formula
a b
det
= ad − bc.
c d
• 3×3: To find the determinant of a 3 × 3 matrix, copy the first two columns of the matrix to
the right of the matrix. You then have three “downward” diagonals (each consisting of three
numbers), and three “upward” diagonals (also consisting of three numbers each). Let d1 , d2
and d3 denote the products of the downward diagonals; let u1 , u2 and u3 denote the product
of the upward diagonals. Then the determinant is given by
det(A) = (d1 + d2 + d3 ) − (u1 + u2 + u3 ).
Calculating inverses:
• Before you try to calculate the inverse of a matrix, check that it is invertible by calculating
the determinant.
• 2×2: There is a formula—swap the main diagonal, negate the off-diagonals and divide by
the determinant:
1
a b
d −b
−1
If A =
then A =
c d
det(A) −c a
• 3×3: There is no formula (or there is, but it’s horrible). To find the inverse of A, apply the
Gauss-Jordan method to the augmented matrix [A | I3 ]. When you’ve got the left-hand side
to look like I3 , the right-hand side will be A−1 . Check your answer by multiplying A−1 by
A and making sure you get I3 .
Matrix Equations:
• Any system of linear equations can be written as a single matrix equation.
• The matrix equation will be of the form AX = B, where A is the coefficient matrix, X is the
variable matrix, and B is the constant matrix for the system.
• The system has a unique solution if and only if A is invertible; in this case, the solution is
given by X = A−1 B.
3.1 Graphing linear inequalities
Working with inequalities: An inequality can be rearranged just like an equation, except that
whenever you multiply or divide both sides by a negative number, the inequality changes direction.
Graphing a single linear inequality:
1. Draw the boundary line: replace the inequality by an =, find the x- and y- intercepts, and
then draw the line:
• If the inequality is strict (i.e. < or >), draw the boundary as a broken line—this indicates that the boundary is not included in the region.
• If the inequality is non-strict (≤ or ≥), use a solid line to indicate that the boundary is
included in the region.
2. If the boundary is not vertical, decide if the feasible region is above or below the boundary.
To do this, solve the inequality for y;
• If the result involves “y ≥” or “y >”, then y is allowed to be large, so the feasible
region is above the boundary.
• If the result is “y ≤” or “y <”, then y is allowed to be small, so the region is below the
boundary.
3. If the boundary is vertical (i.e. its equation is of the form x = c), then decide whether the
feasible region is to the left or to the right of the boundary, using the rule
• “x >” ⇒ x is large ⇒ feasible region is to the right.
• “x <” ⇒ x is small ⇒ feasible region is to the left.
4. Shade the appropriate region.
Graphing a system of linear inequalities:
1. For each inequality, draw the boundary line and decide whether the feasible region is above
or below (or left/right for a vertical boundary). Put some arrows on each boundary line to
indicate which side the feasible region lies on.
2. Once you’ve drawn all the boundaries, the feasible region is the region where all the arrows
point. Shade it.
3.2 Solving linear programming problems graphically
Constraints and objectives: A linear programming problem involves:
• Variables: For us, there are always two variables, x and y. They represent two quantities
whose values we can choose, and for which we seek the “best” choice.
• Constraints: Not all choices for the values of the variables will be possible; this is expressed
by the constraints of the problem. The constraints are nothing but linear inequalities involving x and y. In any problem, there will usually be several constraints.
• Objective: The objective is what tells us which choices of x and y are better than others. In
practice, there will be some quantity that we wish to maximise or minimise; this quantity
will depend on (i.e. be a function of) x and y. We usually denote this function by the letter
z; this is the objective function. In any problem, there will be only one objective function.
How to solve linear programming problems:
1. Graph the feasible region determined by the constraints.
2. Find the coordinates of each corner of the feasible region. To do this, notice that each
corner is the point of intersection of two boundary lines; solve the equations of those lines
simultaneously to find the coordinates of the corner point.
3. The maximal and minimal values of the objective function occur at one of the corners. Draw
a table with two columns, “Corner” and “Value of the objective function”. For each corner,
plug the x and y coordinates in to the formula for the objective function to get the value of
the function at that point. Enter the numbers into your table.
4. Look for the largest/smallest value of z (depending on whether the question asked you for a
maximum or minimum, respectively).
5. Answer the question. For example, if you were asked to minimise z, and you found that the
smallest value of z was 6 for the corner (5, 7), you would answer as follows: “The minimal
value of z is 6, which occurs at the point (5, 7).”
3.3 Applications of linear programming (a.k.a. word problems)
Linear programming can be used to solve real-world problems. These problems typically involve
choosing the best combination of two quantities, subject to some constraints. To solve such a
problem,
1. Read through the whole problem before you start writing anything.
2. Assign your variables: there will be two quantities whose values are to be chosen; call one
of these x and the other y.
3. Determine the constraints: read through the question and look for any information that will
give you an inequality. Key words to look out for are “at least”, “at most”, “no more than”,
etc. Translate each such piece of information into an inequality involving x and y. Remember:
• In many cases, x and y will denote quantities of real-world objects. Such a quantity
can never be negative; in such cases, you should add “x ≥ 0” and “y ≥ 0” to your list
of constraints, even if this is not specifically mentioned in the problem.
• There will always be some information in the question that you won’t use at this step;
this comes up in the next step.
4. Find the objective function: The problem will always involve either maximising or minimising some quantity (eg. maximise revenue, minimise cost, etc.) Find an expression for this
quantity in terms of x and y, and call it z. This is the objective function.
5. The problem has now been translated into algebra, and can be solved as in the previous
section: graph the feasible region, find the corner points, make a table, find the maximum/minimum.
6. When you give your answer, you should translate back into words. For example, if x represents the number of product A produced, and y represents the number of product B produced,
and the objective was the maximise profit, then you should give your final answer in the form
“The maximal profit is $15000, which can be attained by producing 300 of product A and
700 of product B”, and not something like “The maximal value of z is 15000, which occurs
at the point (300, 700).”
4 Linear transformations
Note: This material is not covered in the textbook. In particular, it has nothing to do with Chapter
4 in your text. It will be covered on the exam. For more details, consult the notes you took in class.
Basic notions and definitions:
• Recall from Section 1.2 that a linear function can be thought of as an object which takes a
number x and transforms it into a new number, f(x).
• A linear transformation is similar, but instead
of transforming one number
into
another
x
x
number, these transform one 2 × 1 matrix
into another 2 × 1 matrix, T
.
y
y
• Each linear transformation T is associated with a 2 × 2 matrix A, in the sense that
x
x
T
=A
,
y
y
where the right-hand side is the usual product of a 2 × 2 matrix with a 2 × 1 matrix.
• In other words, we transform one 2 × 1 matrix into another, by multiplying the original
matrix by some given 2 × 2 matrix A.
x
x
• The matrix T
is called the image of
under T .
y
y
3 −4
Examples:(1) Let A =
, and let T be the associated linear transformation. Find the
−1
3
−1
image of
under T .
2
−1
−1
Solution: “The image of
under T ” means T
. By definition, we compute this image
2
2
−1
by multiplying
by the matrix A:
2
−1
3 −4 −1
−11
T
=
=
.
2
−1
3
2
7
−1
−11
So the image of
under T is
.
2
7
1 −2
x
(2) Let A =
, and let T be the corresponding linear transformation. Find a matrix
5 −1 y
0
whose image under T is
.
9
x
0
Solution: We want to solve T
=
, or in other words
y
9
1 −2 x
0
=
.
5 −1 y
9
Rewrite this matrix equation as a system of two linear equations: x−2y = 0 and 5x−y = 9. Solve
this system: subtract five times the first equation from the second, to get 9y = 9. This implies that
y = 1. Now plug this in to the first equation to get x − 2 = 0, or in other words x = 2. So the
solution is
x
2
=
.
y
1
You can check that
this is correct by calculating the image of this point under T , and making sure
0
that it really is
.
9
x
Geometric description of linear transformations: We can think of a 2 × 1 matrix
as correy
sponding to the point (x, y) in the plane. Consequently, we can think of a linear transformation as
a function which takes points in the plane and moves them to different points in the plane. There
are several possibilities:
• Dilation: Every point (except the origin itself) is moved away from the origin in a straight
line.
• Contraction: Every point (except the origin itself) is moved toward the origin along a straight
line.
• Rotation: The points are rotated around the origin, but don’t move toward or away from the
origin.
• Reflection: All of the points on some straight line are fixed, while other points are moved to
the opposite side of the line.
• A combination: Most linear transformations don’t fall neatly into one of the above categories, but are a combination of the above.
0 1
Example: Let T be the linear transformation corresponding to the matrix
. Which of the
1 0
following best describes T ? (a) Dilation; (b) Contraction; (c) Rotation; (d) Reflection; (e) Some
combination of the other options.
1
0
−1
0
1
Solution: Choose a few points, say
,
,
,
,
, and calculate their images under
0
1
0
−1
1
T . For example,
1
0 1 1
0
T
=
=
,
0
1 0 0
1
1
0
so the image of
is
. Similarly, we get
0
1
0
0
−1
1
1
0
1
−1
T
=
, T
=
, T
=
, T
=
.
1
0
0
−1
−1
0
1
1
Draw a picture which shows each point, along with its image under T and an arrow from each point
to its image. You’ll see that T moves each point to the point on the opposite side of the line y = x,
and leaves the points on this line where they are. So T is a reflection.
Linear transformations and lines:
• Fact: If T is a linear transformation, and L is a straight line, then the image of L under
T is either a single point, or another straight line. (This is why they are called “linear”
transformations.)
• To find the image of a given line:
1. Find two points on the original line: usually the x- and y-intercepts will do.
2. Calculate the images of these two points.
3. If these two image points are different, find the equation of the straight line which
passes through those points.
4. If the two points you found in step 2 are the same, then the image of the entire line is
that same point.
1 0
Example: Let T be the linear transformation associated to the matrix
. Find the image of
1 2
the line 2y = x + 2 under T .
Solution:
1. Find two points on the original
line: the y-intercept is found by putting x= 0; we get y = 1,
0
−2
so the line passes through
. Similarly, we find that the x-intercept is
.
0
1
2. Calculate the images of these two points: we have
0
1 0 0
0
T
=
=
,
1
1 2 1
2
−2
−2
1 0 −2
T
=
=
.
0
1 2
0
−2
0
−2
3. We know that the image of the given line passes through the points
and
. Find the
2
−2
equation of the line through these two points in the usual way: the slope is
m=
y2 − y1
−2 − 2
−4
=
=
= 2,
x2 − x1
−2 − 0
−2
and we now use the point-slope formula,
y − y1 = m(x − x1 ) ⇒ y − 2 = 2(x − 0) ⇒ y = 2x + 2.
4. So the image of the line 2y = x + 2 is the line y = 2x + 2.
Linear transformations and regions:
• Fact: If T is a linear transformation and R is a region in the plane, then the image T (R)
is either a region, a line segment, or a point. (The latter only happens when T is the zero
matrix).
• To find the image of a region:
1. Find the image of each corner point.
2. Join these images in the same order as the original corners.
3. Shade the interior.
• Another fact: If T corresponds to the matrix A, then we have
Area(T (R)) = | det(A)| Area(R),
where | det(A)| is the absolute value of the determinant of A: that is, it’s just det(A) made
positive - ignore the minus sign if there is one.
−3 2
Examples: (1) Let T be the linear transformation associated to the matrix A =
, and let
−4 1
R be a region with area 3. Find the area of T (R).
Solution: We have det(A) = −3 − (−8) = 5, and Area(R) = 3, so by the usual formula,
Area(T (R)) = 5 · 3 = 15.
−3 0
(2) Let R be the region shown below. Use the linear transformation T with matrix A =
1 1
to compute the area of R.
0
0
Solution: Apply T to this region. The image is a rectangle with corners at the points
,
,
0
2
−3
−3
, and
. The sides of this rectangle have length 3 and 2, so we have Area(T (R)) = 6.
0
2
The matrix A has determinant det(A) = −3, so | det(A)| = 3. Thus the formula gives us
Area(T (R)) = | det(A)| Area(R) ⇒ 6 = 3 Area(R) ⇒ Area(R) = 2.
So the area of R is 2.