Download Linear Algebra Review Sheet

Linear Algebra Midterm II Review Sheet
Prepared by your generous peer, Jonah Palmer, April 2017 (my comments are in red – Rob Kusner ;-)
3.1:
Easy 3x3 determinant equation:
𝑎
|𝑑
𝑔
𝑏
𝑒
ℎ
𝑐
𝑓 | = 𝑎𝑒𝑖 − 𝑐𝑒𝑔 + 𝑏𝑓𝑔 − 𝑎𝑓ℎ + 𝑐𝑑ℎ
𝑖
Computing the determinant using cofactor expansion across rows / columns:
4
|5
0
0 2
2 2|
2 −1
𝐶𝑖𝑗 = (−1)𝑖+𝑗 det(𝐴𝑖𝑗 )
det(𝐴) = 𝑎11 𝐶11 + 𝑎12 𝐶12 + 𝑎13 𝐶13 → det(𝐴) = (−1)1+1 4𝐶11 + 0𝐶12 + (−1)1+3 2𝐶13
det(𝐴) = 4𝐶11 + 2𝐶13
2
det(𝐴) = 4 |
2

2
5 2
|+ 2|
| = 4(−6) + 2(10) = −4
−1
0 2
Note: You should get the same answer for every row or column you use cofactor expansion on.
Remember the checkerboard of ± signs (−1)𝑖+𝑗 in the cofactors 𝐶𝑖𝑗 = (−1)𝑖+𝑗 det(𝐴𝑖𝑗 )
Here 𝐴𝑖𝑗 is the minor matrix resulting from removing the row i and col j from 𝐴
Row operations and the effect on the determinant:
𝑎
[
𝑐
𝑏
7𝑎
]→[
𝑑
𝑐
𝑑𝑒𝑡 [
𝑎
𝑐

𝑎
[
𝑐
7𝑏
]
𝑑
𝑏
] = 𝑎𝑑 − 𝑏𝑐
𝑑
𝑎
𝑐
7𝑎
𝑐
7𝑏
] = 7(𝑎𝑑 − 𝑏𝑐)
𝑑
It multiplied the determinant by 7.
𝑏
𝑎 + 𝑘𝑐
]→[
𝑑
𝑐
𝑑𝑒𝑡 [
𝑑𝑒𝑡 [
𝑏 + 𝑘𝑑
]
𝑑
𝑏
] = 𝑎𝑑 − 𝑏𝑐
𝑑
𝑑𝑒𝑡 [
𝑎 + 𝑘𝑐
𝑐
𝑏 + 𝑘𝑑
] = 𝑑(𝑎 + 𝑘𝑐) − 𝑐(𝑏 + 𝑘𝑑)
𝑑

It does not change the determinant.
1
1
1
𝑘
𝑘
𝑘
[−9 4 −7] → [−9 4 −7]
6 −8 7
6 −8 7
1
1
1
𝑑𝑒𝑡 [−9 4 −7] = 41
6 −8 7

𝑘
𝑘
𝑘
𝑑𝑒𝑡 [−9 4 −7] = 41𝑘
6 −8 7
The determinant is multiplied by 𝑘.
Finding a formula relating 𝒅𝒆𝒕(𝒌𝑨) to 𝒌 and 𝒅𝒆𝒕(𝑨):
𝐴=[
𝑎
𝑐
𝑏
] , 𝒌 𝑖𝑠 𝑎 𝑠𝑐𝑎𝑙𝑎𝑟
𝑑
det(𝐴) = 𝑎𝑑 − 𝑏𝑐

𝑘𝑎
det(𝑘𝐴) = [
𝑘𝑐
𝑘𝑏
] = 𝑘 2 (𝑎𝑑 − 𝑏𝑐)
𝑘𝑑
Note: Notice the example at the top of the page. When one row is filled with k’s, the
determinant is multiplied by 𝑘. However, when two rows are multiplied by k’s, then the
determinant is 𝑘 2.
3.2:
Properties of Determinants:
1.
2.
3.
4.
5.
A multiple of one row of 𝐴 is added to another row to produce matrix 𝐵, then: |𝐴| = |𝐵|
If two rows are interchanged to produce a matrix 𝐵, then: |𝐵| = −|𝐴|
If one row is multiplied by 𝑘 to produce a matrix 𝐵, then: |𝐵| = 𝑘|𝐴|
If 𝐴 & 𝐵 are both 𝑛 × 𝑛 matricies then: |𝐴𝐵| = |𝐴| × |𝐵|
|𝐴𝑇 | = |𝐴|

Note: |𝐴| = det(𝐴) , |𝐵| = det(𝐵)
True / False Statements:
1) A row replacement operation does not affect the determinant of a matrix. i.e. adding some row to
another row
a) TRUE
2) The determinant of A is the product of the pivots in any echelon form U of A, multiplied by (−1)𝑟 ,
where r is the number of row interchanges made during row reduction from A to U.
a) FALSE
i) Reduction to an echelon form may also include scaling a row by a nonzero constant, which
will change the value of the determinant.
3) If the columns of A are linearly dependent, then det(𝐴) = 0.
a) TRUE
i) If columns of A are linearly dependent, A is not invertible.
4) det(𝐴 + 𝐵) = det(𝐴) + det(𝐵).
a) FALSE
i) Not one of the properties!
Showing equivalent equations using determinants:
𝑆ℎ𝑜𝑤 𝑡ℎ𝑎𝑡 det(𝑃𝐴𝑃−1 ) = det(𝐴)
det(𝑃𝐴𝑃−1 ) = (det(𝑃))(det(𝐴))(det(𝑃−1 )) → det(𝑃−1 ) = (det(𝑃))−1
det(𝑃) ∗ det(𝑃)−1 = 1 → 1 ∗ det(𝐴) = det(𝐴)
3.3:
Finding Area using determinants:
Find the area of the parallelogram whose vertices are listed: (0,0), (7,5), (11,6), (18,11)


Note: The area of the parallelogram determined by the columns of 𝐴 is |det(𝐴)|.
Note: We can use any two vertices here; we’ll use (7,5) & (11,6).
7 11
7 11
[ ],[ ] → 𝐴 = [
]
5
6
5 6
det(𝐴) = −13 → |det(𝐴)| = 13
The area of the parallelogram is 13 square units. Note that the sign of the determinant gives info about
the “right” or “left” handedness (orientation) of the parallelotope.
Finding volume using determinants:
Find the volume of the parallelepiped with one vertex at the origin and adjacent vertices at
(3,0, −4), (1,5,6), & (6,2,0).

Note: If 𝐴 is a 3x3 matrix, the volume of the parallelepiped determined by the columns of 𝐴 is
|det(𝐴)|.
3
1 6
3 1
[ 0 ] , [5] , [2] → 𝐴 = [ 0 5
−4 6 0
−4 6
|det(𝐴)| = 76
The volume of the parallelepiped is 76.
6
2]
0
Computing the area of an image of 𝑺 under the mapping 𝒙 → 𝑨𝒙:
−2
−2
4 −7
] and 𝑏2 = [ ], and let 𝐴 = [
].
6
9
−2 7
Compute the area of the image of S under the mapping 𝑥 → 𝐴𝑥.
Let S be the parallelogram determined by the vectors 𝑏1 = [
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑇(𝑆) = |det(𝐴)| ∙ {𝑎𝑟𝑒𝑎 𝑜𝑓 𝑆}
|det(𝑏1 𝑏2 )| = |−2 −2 | = |−6| = 6 = 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑆
6
9
|det(𝐴)| = | 4 −7| = 14
−2 7
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑇(𝑆) = 14 ∙ 6 = 84
4.1:
Sets of vectors with inclusion and exclusion:
Let V be the set of vectors shown below.
𝑥
𝑉 = {[𝑦] : 𝑥 ≥ 1, 𝑦 > 1}
It helps to draw pictures of these sets to see what’s going on, especially when looking for a counterexample to show a property fails.
If u and v are in V, is u + v in V? Why?
𝑎
𝑐
𝑎+𝑐
𝑢 = [ ] , and 𝑣 = [ ], then 𝑢 + 𝑣 = [
]
𝑏
𝑑
𝑏+𝑑
Since 𝑎 ≥ 1 and 𝑐 ≥ 1, then 𝑎 + 𝑐 ≥ 2, which implies 𝑎 + 1 ≥ 1
Since 𝑏 > 1 and 𝑑 > 1, then 𝑏 + 𝑑 > 2, which implies 𝑏 + 𝑑 > 1
Since 𝑎 + 𝑐 ≥ 1 & 𝑏 + 𝑑 > 1, this means that u + v is in V.
Find a specific vector u in V and a specific scalar c such that cu is not in V.

Note: Any vector for u will work, make one up.
2
2𝑐
𝑢 = [ ], then 𝑐𝑢 = [ ]
2
2𝑐

Note: cu will not be in V if 2𝑐 < 1. So, find a c where this is true. Ex. 𝑐 = −1.
2
If 𝑢 = [ ], and 𝑐 = −1, then cu will not be in V.
2
A subspace of a vector space V is a subset H of V properties:
1) The zero vector of V is in H. This is equivalent to V being nonempty.
2) H is closed under vector addition. That is, for each u and v in H, the sum u + v is in H.
3) H is closed under scalar multiplication. This is, for each u in H and each scalar c, the vector cu is in H.
Showing H is not a subspace of ℝ𝟐 :
𝑥
𝐻 = {[𝑦] : 16𝑥 2 + 9𝑦 2 ≤ 1}
Find two vectors and a vector and a scalar to show that H is not a subspace of ℝ2 . Picture! (Ellipse!)

𝑢=[

Note: For two vectors to show that H is not a subspace of ℝ2 , they must show that H is not
closed under addition.
1
0
( )
4 ] , 𝑣 = [ 1 ] , 𝑢 + 𝑣 is not in H.
(3)
0
Note: For a vector and a scalar to show that H is not a subspace of ℝ2 , they must show that H is
not closed under scalar multiplication.
𝑥
𝑥
𝑐𝑥
𝑢 = [𝑦] is in H. The vector cu is equal to 𝑐 [𝑦] = [𝑐𝑦]
Say, 𝑐 = 5. Find a vector u such that cu is not in H. 16(𝑐𝑥)2 + 9(𝑐𝑦)2 > 1
0.17
𝑢=[
]
−0.22

Note: 16(0.17)2 + 9(−0.22)2 = 0.898 ≤ 1
𝑐𝑢 = 5 [

0.17
0.85
]=[
]
−0.22
−1.1
Note: 16(0.85)2 + 9(−1.1)2 = 22.450 > 1
Spanning and vectors:
5𝑡
Let H be the set of all vectors of the form [−𝑡]. Find a vector v in ℝ3 such that 𝐻 = 𝑆𝑝𝑎𝑛{𝑣}.
0
5
𝑣 = [−1]
0

Note: Due to the theorem; The Span of any nonempty set is always a subspace!
If 𝑣1 , … , 𝑣𝑝 are in a vector space V, then 𝑆𝑝𝑎𝑛{𝑣1 , … , 𝑣𝑝 } is a subspace of V.
We can say because v is in ℝ3 and 𝐻 = 𝑆𝑝𝑎𝑛{𝑣}, H is a subspace of ℝ3
Determining subspace of vectors:
1
10
6
5
𝑣1 = [ 0 ] , 𝑣2 = [1] , 𝑣3 = [ 2 ] , 𝑎𝑛𝑑 𝑤 = [1]
−1
12
5
6
Is w in the subspace spanned by {𝑣1 , 𝑣2 , 𝑣3 }?

Note: To determine if w is in the subspace spanned by {𝑣1 , 𝑣2 , 𝑣3 }, we must determine if w is a
linear combination of 𝑣1 , 𝑣2 𝑎𝑛𝑑 𝑣3 .
1 5 10 6
[𝑣1 𝑣2 𝑣3 𝑤] = [ 0 1 2 1]
−1 6 12 5
1
1 5 10 6
[ 0 1 2 1] ~ [0
0
−1 6 12 5

0 0 1
1 2 1]
0 0 0
Note: No row has the form [0 … 0 𝑏] where b is nonzero. Therefore, w is a linear combination of
𝑣1 , 𝑣2 , 𝑎𝑛𝑑 𝑣3
Spanning Sets:
Find a set S of vectors that spans W.
3𝑎 − 6𝑏
7𝑏 + 3𝑐
𝑊=[
]
6𝑐 − 4𝑎
8𝑏
𝑎
𝑏
𝑐
3
−6 0
0
7
3
𝑆𝑝𝑎𝑛𝑛𝑖𝑛𝑔 𝑠𝑒𝑡 𝑆 = {[ ] , [ ] , [ ]}
−4
0
6
0
8
0
Vector Space Axioms:
1)
2)
3)
4)
The sum u + v is in V Some consider this to be a “given” rather than an “axiom”….
u+v=v+u
(u + v) + w = u + (v + w)
V has a vector – such that u + 0 = u
5)
6)
7)
8)
9)
10)
For each u in V, there is a vector -u in V such that u + (-u) = 0
The scalar multiple cu is in V Some consider this to be a “given” rather than an “axiom”….
c(u + v) = cu + cv
(c + d)u = cu + du
c(du) = (cd)u
1u = u
3), 4) and 5) are the axioms of a “group” and including 2) makes it a commutative (abelian) group
7) and 8) are distributive laws relating scalar “+” (one of these) with vector “+” (three of these)
10) keeps things from collapsing (without it, the “trivial” scalar multiplication rule “cv=0” for all scalars c
and vectors v would satisfy the other axioms)
4.2:
Null space of an m x n matrix A:
2
6 −6 3
Determine if 𝑤 = [4] is in 𝑁𝑢𝑙(𝐴), where 𝐴 = [ 7 −3 8 ]
1
−2 8 11
Also called the “kernel” 𝑘𝑒𝑟(𝐴)
𝑁𝑢𝑙(𝐴) = {𝑥: 𝑥 𝑖𝑠 𝑖𝑛 ℝ𝑛 & 𝐴𝑥 = 0}

Note: To test if w satisfies 𝐴𝑤 = 0, computer Aw. If the product Aw is 0, then w is in 𝑁𝑢𝑙(𝐴).
6 −6 3 2
−9
𝐴𝑤 = [ 7 −3 8 ] [4] = [ 10 ]
−2 8 11 1
39
𝐴𝑤 ≠ 0, so w is not in 𝑁𝑢𝑙(𝐴).
Explicit Description of Nul A by Listing Vectors that span the null space:
𝐴=[

1 6
0 0
−7 0
]
8 0
Note: Need to find the general solution of Ax = 0 in terms of free variables.
[𝐴 0] = [1 6 −7 0
0 0 8 0
𝑥1 + 6𝑥2 = 0,
𝑥1 = −6𝑥2 ,
0
1
]~[
0
0
6 0
0 1
𝑥3 = 0
𝑥3 = 0
0 0
]
0 0
𝑥1
−6𝑥2
−6
0
𝑥2
𝑥2
1
0
[𝑥 ] = [
] = 𝑥2 [ ] + 𝑥4 [ ]
0
0
0
3
𝑥4
𝑥4
0
1
−6 0
1
0
Spanning set for Nul A = {[ ] , [ ]}
0
0
0
1
Show a given set is a vector space:
𝑝
𝑞
𝑊 = {[ ] ∶ 2𝑝 + 𝑞 = −2𝑠,
𝑟
𝑠
−2𝑝 = −3𝑠 + 3𝑟}
2𝑝 + 𝑞 + 0𝑟 + 2𝑠 = 0
−2𝑝 + 0𝑞 − 3𝑟 + 3𝑠 = 0
2 1
𝐴𝑥 = [
−2 0


𝑝
0 2 𝑞
0
][ ] = [ ]
𝑟
−3 3
0
𝑠
Note: this given set represents the set of all solutions to the homogeneous system of equations.
o Therefore; the set 𝑊 = 𝑁𝑢𝑙(𝐴)
Note: The null space of an m x n matrix A is a subspace of ℝ𝑛 . Equivalently, the set of all
solutions to a system 𝐴𝑥 = 0 of m homogeneous linear equations in n unknowns is a subspace
of ℝ𝑛 .
Nonzero Vectors in Nul A and Col A:
−35 −40
−7
−8
𝐴=[
]
−7
−8
14
16
First find the general solution of Ax = 0 in terms of free variables.
−35 −40 0
1
−7
−8
0
[𝐴 0] = [
]~ 0
−7
−8 0
0
14
16 0 [
0
8
7
0
0
0
0
0
0
0]
8
8
𝑥1 + 𝑥2 = 0 → 𝑥1 = − 𝑥2
7
7

Note: To find a nonzero vector in Nul A, choose a nonzero value for the free variable, say
𝑥2 = −7, and substitute into the general solution.
8
𝑥1 = − (−7) = 8
7
𝑁𝑢𝑙(𝐴) = 𝑆𝑝𝑎𝑛([

8
])
−7
Note: To find a nonzero vector in Col A, choose any column of A as the nonzero vector.
−35
−7
𝐶𝑜𝑙(𝐴) = 𝑆𝑝𝑎𝑛([
])
−7
14
Determine if a vector is in Col(A) / Nul(A):
−10 −8 −6
1
𝐴=[ 2
4
2 ] and 𝑤 = [ 1 ]. Determine if w is in Col(A) / Nul(A)?
6
0
2
−3
𝐶𝑜𝑙(𝐴) = {𝑏 ∶ 𝑏 = 𝐴𝑥 𝑓𝑜𝑟 𝑠𝑜𝑚𝑒 𝑥 𝑖𝑛 ℝ}
Also called the “image” 𝑖𝑚(𝐴)

Note: If the equation Ax = w is consistent, then w is in Col(A).
1
1 0
−10 −8 −6 1
3
[𝐴 𝑤] = [ 2
4
2
1 ]~
1
0 1
6
0
2 −3
3
[0 0 0

−
1
2
1
0 ]
Note: Since the echelon form of the augmented matrix has no row of the form [0 … 0 𝑏] the
equation is consistent. Thus, w is in Col(A).

Note: If the equation Aw = 0 is true, then w is in Nul(A).
−10 −8 −6 1
0
[ 2
4
2 ] [ 1 ] = [ 0]
6
0
2 −3
0
w is in Nul(A).
True / False Statements:
1) A null space is a vector space.
a) TRUE
2) The column space of an m x n matrix is in ℝ𝑚
a) TRUE
3) The column space of A, Col(A), is the set of all solutions of Ax = b.
a) FALSE
i) 𝐶𝑜𝑙(𝐴) = {𝑏 ∶ 𝑏 = 𝐴𝑥 𝑓𝑜𝑟 𝑠𝑜𝑚𝑒 𝑥 𝑖𝑛 ℝ𝑛 }
4) The null space of A, Nul(A), is the kernel of the mapping 𝑥 → 𝐴𝑥.
a) TRUE
i) The kernel of a linear transformation T, from a vector space V to a vector space W, is the set
of all u in V such that 𝑇(𝑢) = 0. Thus the kernel of a matrix transformation 𝑇(𝑥) = 𝐴𝑥 is the
null space of A.
5) The range of a linear transformation is a vector space. Again, most folks call this the “image”.
a) TRUE
i) The range of a linear transformation T, from a vector space V to a vector space W, is a
subspace of W.
6) The set of all solutions of a homogeneous linear differential equation is the kernel of a linear
transformation.
a) TRUE
i) The linear transformation maps each function f to a linear combination of f and at least one
of its derivatives, exactly as these appear in the homogeneous linear differential equation.
4.3:
Invertible Matrix Theorem:
Let A be a square n x n matrix. Then the following statements are equivalent.
1) The matrix A is an invertible matrix.
2) The matrix A is row equivalent to the n x n identity matrix.
3)
4)
5)
6)
The matrix A has n pivot positions.
The equation Ax = 0 has only the trivial solution.
The columns of A form a linearly independent set.
The columns of A span ℝ𝑛
Determining if a set is a basis:
1 1 1
Determine whether the set {[1] , [0] , [1]} is a basis for ℝ3 .
0 0 1
1 1
𝐴 = [1 0
0 0


1
1 0 0
1] ~ [ 0 1 0]
1
0 0 1
Note: The columns of A form a linearly independent set and they span ℝ3
Note: The set is a basis if the set is linearly independent and they span ℝ3
Basis of a null space of a matrix:
1
[0
3
0 −3 4
1 −5 7 ] Find the basis for the null space of the matrix.
−2 1 −2
1 0 −3 4 0
1 0
[𝐴 0] = [0 1 −5 7 0] ~ [0 1
3 −2 1 −2 0
0 0
−3 4
−5 7
0 0
0
0]
0
𝑥1 − 3𝑥3 + 4𝑥4 = 0
𝑥2 − 5𝑥3 + 7𝑥4 = 0
𝑥1 = 3𝑥3 − 4𝑥4
𝑥2 = 5𝑥3 − 7𝑥4
𝑥1
3
−4
𝑥2
5
−7
𝑥 = [𝑥 ] = 𝑥3 [ ] + 𝑥4 [ ]
1
0
3
𝑥4
0
1

3
−4
5
−7
Note: The basis for the null space of the matrix will only be the basis if [ ] and [ ] are linearly
1
0
0
1
independent. They are independent however, because neither is a multiple of the other.
True / False Statements:
1) A linearly independent set in a subspace H is a basis for H.
a) FALSE
i) The subspace spanned by the set must also coincide with H. One can extend to a maximal
linearly independent set – that extended set will be a basis – this is how one shows any
vector space has a basis.
2) If a finite set S of nonzero vectors spans a vector space V, then some subset of S is a basis for V.
Dually, a minimal spanning set is a basis, but this works easily only in the finite-dimensional case.
a) TRUE
i) Spanning Set Theorem
3) A basis is a linearly independent set that is as large as possible. (see above)
a) TRUE
4) The standard method for producing a spanning set for Nul A sometimes fails to produce a basis for
Nul A.
a) FALSE
i) The method always produces an independent set.
5) If B is an echelon form of a matrix A, then the pivot columns of B form a basis for Col A.
a) FALSE
i) The columns of an echelon form B of A are not necessarily in the column space of A.
4.4:
Finding vector x using a given basis and a coordinate vector [𝒙]𝑩 :
Find vector x determined by the given coordinate vector [𝑥]𝐵 and the given basis 𝐵.
−7 −5
8
𝐵 = {[ ] , [ ]} , [𝑥]𝐵 = [ ]
3
3
6
𝑥 = 𝑐1 𝑏1 + 𝑐2 𝑏2
𝑥 = (8) [
𝑐
[𝑥]𝐵 = [𝑐1 ]
2
𝐵 = {𝑏1 , 𝑏2 }
−7
−71
−5
] + (3) [ ] = [
]
3
42
6
Finding the coordinate vector [𝒙]𝑩 given vector x and basis 𝑩:
Find the coordinate vector [𝑥]𝐵 of x relative to the given basis 𝐵 = {𝑏1 , 𝑏2 }.
10
−6
54
],𝑏 = [ ],𝑥 = [
]
−7 2
5
−41
𝑐1
Note: We are looking for the vector [𝑥]𝐵 = [𝑐 ] here.
2
𝑏1 = [

𝑐1 [
10
−6
1 0
54
10 −6 54
] + 𝑐2 [ ] = [
] → [𝑏1 𝑏2 𝑥] = [
]~[
−7
5
0 1
−7 5 −41
−41
3
]
−4
𝑥1 = 3, 𝑥2 = −4 → 𝑐1 = 3, 𝑐2 = −4
𝑥 = 3𝑏1 − 4𝑏2 → [𝑥]𝐵 = [
3
]
−4
Using an inverse matrix to find [𝒙]𝑩 :
Use an inverse matrix to find [𝑥]𝐵 for the given x and 𝐵.
4
−1
−5
𝐵 = {[ ] , [ ]} , 𝑥 = [ ]
−5
3
6
Recall:
𝑃𝐵 = [𝑏1 𝑏2 … 𝑏𝑛 ]
𝐴=[
𝑎
𝑐
𝑥 = 𝑐1 𝑏1 + 𝑐2 𝑏2 + ⋯ + 𝑐𝑛 𝑏𝑛
1
𝑏
𝑑
] → 𝐴−1 = det(𝐴) [
𝑑
−𝑐
𝑥 = 𝑃𝐵 [𝑥]𝐵
𝑃𝐵−1 𝑥 = [𝑥]𝐵
−𝑏
]
𝑎
4 −5
−6 −5
𝑃𝐵 = [
] → 𝑃𝐵−1 = [
]
−5 6
−5 −4
[𝑥]𝐵 = 𝑃𝐵−1 𝑥
[𝑥]𝐵 = [−6 −5] [−1] = [−9]
−7
−5 −4 3
Finding coordinate vectors relative to a set 𝑩:
The set 𝐵 = {1 − 𝑡 2 , 𝑡 − 𝑡 2 , 1 + 2𝑡 − 𝑡 2 } is a basis for ℙ2 . Find the coordinate vector of 𝑝(𝑡) = −10 +
𝑡 + 5𝑡 2 relative to 𝐵.

Note: To find the coordinate vector, split up whole numbers, 𝑡’s of the first degree, and 𝑡 2 ’s of
the second degree.
𝑐1 (1 − 𝑡 2 ) + 𝑐2 (𝑡 − 𝑡 2 ) + 𝑐3 (1 + 2𝑡 − 𝑡 2 ) = −10 + 𝑡 + 5𝑡 2
𝑐1 − 𝑐2 𝑡 2 + 𝑐2 𝑡 − 𝑐2 𝑡 2 + 𝑐3 + 𝑐3 2𝑡 − 𝑐3 𝑡 2 = −10 + 𝑡 + 5𝑡 2
(1)𝑐1 + (1)𝑐3 = −10
- Whole numbers here!
𝑐2 + 2𝑐3 = 1
- 𝑡’s of the first degree here!
(−1)𝑐1 + (−1)𝑐2 + (−1)𝑐3 = 5
-𝑡 2 ’s of the second degree here!

Note: Now we need to take this system of equations and form an augmented matrix.
1
0
1 −10
1 0 0
[0
1
2
1 ] ~ [0 1 0
−1 −1 −1
5
0 0 1
−8
5]
−2
𝑐1 = −8, 𝑐2 = 5, 𝑐3 = −2
−8
[𝑝]𝐵 = [ 5 ]
−2
Finding matrix A using the inverse of 𝑷𝑩 :
3
−4
Let 𝐵 = {[ ] , [ ]} . Since the coordinate mapping determined by B is a linear transformation from
−5
7
ℝ2 into ℝ2 , this mapping must be implemented by some 2x2 matrix A. Find it.
𝐴 = 𝑃𝐵−1
𝑃𝐵 = [
3 −4
]
−5 7
1
𝐴−1 = det(𝐴) [
𝑑
−𝑐
−𝑏
]
𝑎
det(𝑃𝐵 ) = 1
7
𝐴 = 𝑃𝐵−1 = [
5
4
]
3