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Solution Manual for Introduction to Linear Algebra, 7th Edition by Gilbert Strang | Complete Answer Key | Latest 2025/2026 Verified Edition

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This complete solution manual and answer key corresponds to Introduction to Linear Algebra (7th Edition) by Gilbert Strang. It provides fully worked-out solutions to all exercises, ensuring clear understanding of each concept and step. Topics covered include vector spaces, linear transformations, matrix algebra, determinants, eigenvalues and eigenvectors, least squares problems, orthogonality, and applications in engineering and data science. Covers All Chapters – Complete Answer Key Step-by-Step Verified Solutions Ideal for Self-Study and Exam Preparation Aligned with the 2025/2026 Revised Edition | A+ ResourceThis complete solution manual and answer key corresponds to Introduction to Linear Algebra (7th Edition) by Gilbert Strang. It provides fully worked-out solutions to all exercises, ensuring clear understanding of each concept and step. Topics covered include vector spaces, linear transformations, matrix algebra, determinants, eigenvalues and eigenvectors, least squares problems, orthogonality, and applications in engineering and data science. Covers All Chapters – Complete Answer Key Step-by-Step Verified Solutions Ideal for Self-Study and Exam Preparation Aligned with the 2025/2026 Revised Edition | A+ ResourceThis complete solution manual and answer key corresponds to Introduction to Linear Algebra (7th Edition) by Gilbert Strang. It provides fully worked-out solutions to all exercises, ensuring clear understanding of each concept and step. Topics covered include vector spaces, linear transformations, matrix algebra, determinants, eigenvalues and eigenvectors, least squares problems, orthogonality, and applications in engineering and data science. Covers All Chapters – Complete Answer Key Step-by-Step Verified Solutions Ideal for Self-Study and Exam Preparation Aligned with the 2025/2026 Revised Edition | A+ ResourceThis complete solution manual and answer key corresponds to Introduction to Linear Algebra (7th Edition) by Gilbert Strang. It provides fully worked-out solutions to all exercises, ensuring clear understanding of each concept and step. Topics covered include vector spaces, linear transformations, matrix algebra, determinants, eigenvalues and eigenvectors, least squares problems, orthogonality, and applications in engineering and data science. Covers All Chapters – Complete Answer Key Step-by-Step Verified Solutions Ideal for Self-Study and Exam Preparation Aligned with the 2025/2026 Revised Edition | A+ Resource

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Institution
Introduction To Linear Algebra
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Introduction to Linear Algebra











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Institution
Introduction to Linear Algebra
Course
Introduction to Linear Algebra

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Uploaded on
October 11, 2025
Number of pages
238
Written in
2025/2026
Type
Exam (elaborations)
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ALL 10 CHAPTERS COVERED
zd zd zd zd zd




zd
zd
zd




zd




zd




zd




Answer key
zd

,2 SolutionstoExercises

Problem Set 1.1, page 8 zd zd zd zd




1
The combinations give (a) a line in R3 (b) a plane in R3 (c) all of R3.
zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd




2
v + w = (2,3) and v − w = (6,−1) will be the diagonals of the parallelogram with v and
zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd




zd w as two sides going out from (0,0).
zd zd zd zd zd zd zd




3
This problem gives the diagonals v + w and v − w of the parallelogram and asks for the
zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd




zd sides: The opposite of Problem 2. In this example v = (3,3) and w = (2,−2).
zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd




4
3v + w = (7,5) and cv + dw = (2c + d,c + 2d).
zd zd zd zd zd zd zd zd zd zd zd zd zd zd




5
u+v = (−2,3,1) and u+v+w = (0,0,0) and 2u+2v+w = ( add first answers) = (−2,3,1).
zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd




zd The vectors u,v,w are in the same plane because a combination gives (0,0,0). Stated
zd zd zd zd zd zd zd zd zd zd zd zd zd




zd another way: u = −v − w is in the plane of v and w.
zd zd zd zd zd zd zd zd zd zd zd zd zd zd




6
The components of every cv + dw add to zero because the components of v and of w
zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd




add to zero. c = 3 and d = 9 give (3,3,−6). There is no solution to cv+dw = (3,3,6)
zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd




zd because 3 + 3 + 6 is not zero. zd zd zd zd zd zd zd zd




7
The nine combinations c(2,1) + d(0,1) with c = 0,1,2 and d = (0,1,2) will lie on a
zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd




zd lattice. If we took all whole numbers c and d, the lattice would lie over the whole
zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd




zd plane.

8
The other diagonal is v − w (or else w − v). Adding diagonals gives 2v (or 2w).
zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd




9
The fourth corner can be (4,4) or (4,0) or (−2,2). Three possible parallelograms!
zd zd zd zd zd zd zd zd zd zd zd zd




10
i − j = (1,1,0) is in the base (x-y plane). i + j + k = (1,1,1) is the opposite corner from
zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd




zd (0,0,0). Points in the cube have 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1.
zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd

,SolutionstoExercises 3

11
Four more corners (1,1,0),(1,0,1),(0,1,1),(1,1,1). The center point is
zd zd zd zd zd zd zd zd .
Centers of faces are zd zd zd zd .

12
The combinations of i = (1,0,0) and i + j = (1,1,0) fill the xy plane in xyz space.
zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd




13
Sum = zero vector. Sum = −2:00 vector = 8:00 vector. 2:00 is 30◦ from horizontal
zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd




.
14
Moving the origin to 6:00 adds j = (0,1) to every vector. So the sum of twelve vectors
zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd




zd changes from 0 to 12j = (0,12). zd zd zd zd zd zd




15
The point v + w is three-fourths of the way to v starting from w. The vector v + w
zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd




is halfway to u = v + w. The vector v + w is 2u (the far corner of the
zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd




parallelogram).

16
All combinations with c + d = 1 are on the line that passes through v and w. The point V
zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd




= −v + 2w is on that line but it is beyond w.
zd zd zd zd zd zd zd zd zd zd zd zd zd




17
All vectors cv + cw are on the line passing through (0,0) and u
zd zd zd zd zd zd zd zd zd zd zd zd zd zd w. That line
zd zd




zd continues out beyond v + w and back beyond (0,0). With c ≥ 0, half of this line is
zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd




zd removed, leaving a ray that starts at (0,0). zd zd zd zd zd zd zd




18
The combinations cv + dw with 0 ≤ c ≤ 1 and 0 ≤ d ≤ 1 fill the parallelogram with
zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd




zd sides v and w. For example, if v = (1,0) and w = (0,1) then cv + dw fills the unit
zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd




zd square. But when v = (a,0) and w = (b,0) these combinations only fill a segment of
zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd




a line.
zd




19
With c ≥ 0 and d ≥ 0 we get the infinite “cone” or “wedge” between v and w. For
zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd




zd example, if v = (1,0) and w = (0,1), then the cone is the whole quadrant x ≥ 0, y ≥ 0.
zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd




zd Question: What if w = −v? The cone opens to a half-space. But the combinations of v =
zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd




zd (1,0) and w = (−1,0) only fill a line.
zd zd zd zd zd zd zd zd

, 2 SolutionstoExercises

20
(a) u + v + w is the center of the triangle between u,v and w;
zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd w lies between
zd zd




zd u and w (b) To fill the triangle keep c≥0, d≥0, e≥0, and c+d+e = 1.
zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd




21
The sum is (v−u)+(w−v)+(u−w) = zero vector. Those three sides of a triangle are in
zd zd zd zd zd zd zd zd zd zd zd zd zd zd




zd the same plane!
zd zd




22
The vector (u +v +w) is outside the pyramid because
zd zd zd zd zd zd zd zd zd .
23
All vectors are combinations of u,v,w as drawn (not in the same plane). Start by seeing
zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd




zd that cu + dv fills a plane, then adding ew fills all of R3.
zd zd zd zd zd zd zd zd zd zd zd zd zd




24
The combinations of u and v fill one plane. The combinations of v and w fill another
zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd




zd plane. Those planes meet in a line: only the vectors cv are in both planes.
zd zd zd zd zd zd zd zd zd zd zd zd zd zd




25
(a) For a line, choose u = v = w = any nonzero vector (b) For a plane, choose u and v in
zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd




zd different directions. A combination like w = u + v is in the same plane.
zd zd zd zd zd zd zd zd zd zd zd zd zd zd




26
Two equations come from the two components: c + 3d = 14 and 2c + d = 8. The
zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd




zd solution is c = 2 and d = 4. Then 2(1,2) + 4(3,1) = (14,8).
zd zd zd zd zd zd zd zd zd zd zd zd zd zd




27
A four-dimensional cube has 24 = 16 corners and 2 · 4 = 8 three-dimensional faces and
zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd




zd 24 two-dimensional faces and 32 edges in Worked Example 2.4 A.
zd zd zd zd zd zd zd zd zd zd




28
There are 6 unknown numbers v1,v2,v3,w1,w2,w3. The six equations come from the
zd zd zd zd zd zd zd zd zd zd zd




zd components of v + w = (4,5,6) and v − w = (2,5,8). Add to find 2v = (6,10,14) so v =
zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd




zd (3,5,7) and w = (1,0,−1). zd zd zd zd




29
Fact: For any three vectors u,v,w in the plane, some combination cu + dv + ew is the
zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd




zd zero vector (beyond the obvious c = d = e = 0). So if there is one combination
zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd zd




zd Cu+Dv+Ew that produces b, there will be many more—just add c,d,e or 2c,2d,2e
zd zd zd zd zd zd zd zd zd zd zd zd




to the particular solution C,D,E.
zd zd zd zd

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