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Linear Algebra Proofs ACTUAL EXAM AND PRACTICE EXAM COMPLETE QUESTIONS AND CORRECT DETAILED ANSWERS (VERIFIED ANSWERS) |ALREADY GRADED A+ $15.49   Add to cart

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Linear Algebra Proofs ACTUAL EXAM AND PRACTICE EXAM COMPLETE QUESTIONS AND CORRECT DETAILED ANSWERS (VERIFIED ANSWERS) |ALREADY GRADED A+

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  • Linear Algebra A
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  • Linear Algebra A

Linear Algebra Proofs ACTUAL EXAM AND PRACTICE EXAM COMPLETE QUESTIONS AND CORRECT DETAILED ANSWERS (VERIFIED ANSWERS) |ALREADY GRADED A+

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  • September 16, 2024
  • 127
  • 2024/2025
  • Exam (elaborations)
  • Questions & answers
  • Linear Algebra A
  • Linear Algebra A
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Linear Algebra Proofs ACTUAL EXAM AND
PRACTICE EXAM COMPLETE
QUESTIONS AND CORRECT DETAILED
ANSWERS (VERIFIED ANSWERS)
|ALREADY GRADED A+



Let A, B , C be n x n matrices such that AB =
BA = I and AC = CA = I.
Prove that B = C. - ✔✔ANSWER✔✔-Proof.
Suppose AB = BA = I and AC = CA = I.
Then, B = BI = B(AC) = (BA)C = CI = C

Prove that if 𝐴 is an invertible 𝑛 × 𝑛 matrix, then
for each vector 𝑏 ∈ ℝ^𝑛 , 𝐴𝑥⃑ = vector 𝑏 has the
unique solution 𝑥⃑ = 𝐴^−1 vector 𝑏. -
✔✔ANSWER✔✔-Proof.
Assume Z is an invertible 𝑛 × 𝑛 matrix.
Then, 𝐴^−1 exists. For each vector 𝑏 ∈ ℝ^𝑛 ,
𝐴𝑥⃑ = vector 𝑏 → 𝐴^−1(𝐴𝑥⃑ ) =𝐴^−1 vector 𝑏
→ (𝐴^−1𝐴)𝑥⃑ = 𝐴^−1 vector 𝑏
→ I𝑛𝑥⃑ = 𝐴^−1 vector 𝑏
→ 𝑥⃑ = 𝐴^−1 vector 𝑏

,Since 𝐴^−1 is unique, the solution of 𝑥⃑ = 𝐴^−1
vector 𝑏 for 𝐴𝑥⃑ = vector 𝑏 is unique.

Suppose P is invertible and A = PBP^-1. Solve
for B in terms of A. - ✔✔ANSWER✔✔-Proof.
Assume P is invertible. Then P^-1 exists. Let A
= PBP^-1.
Then, P^-1 AP = P^-1(PBP^-1)P
= (P^-1P)B(P^-1P)
= IBI
=B

Use matrix algebra to show that if A is invertible
and D satisfies AD=I, then D=A−1. -
✔✔ANSWER✔✔-Proof.
Suppose A is invertible and AD = I.
Then A^-1 exists.
A^-1(AD) = A^-1I
(A^-1A)D = A^-1
ID = A^-1
D = A^-1

Suppose A and B are an 𝑛 × 𝑛 matrices, B is
invertible, and AB is invertible. SHow that A is
invertible. - ✔✔ANSWER✔✔-Proof.

,Suppose A and B are 𝑛 × 𝑛 matrices, B and AB
are invertible.
Since B is invertible, B^-1 exists. Let C = AB.
CB^-1 = (AB)B^-1 = A(BB^-1) = AI = A
B is invertible → B^-1 is also invertible.
Since B^-1, C = AB are invertible,
A = CB^-1 is also invertible,
because product of invertible matrices is
invertible.
1.1
10. a.
Is the statement "Every elementary row
operation is reversible" true or false? Explain. -
✔✔ANSWER✔✔-True, because replacement,
interchanging, and scaling are all reversible.

1.1
10. b.
Is the statement A 5×6 matrix has six rows" true
or false? Explain. - ✔✔ANSWER✔✔-False,
because a 5×6 matrix has five rows and six
columns.

1.1

, 10. c.
Is the statement "The solution set of a linear
system involving variables x1, ..., xn is a list of
numbers (s1, ..., sn) that makes each equation
in the system a true statement when the values
s1, ..., sn are substituted for x1, ..., xn,
respectively" true or false? Explain. -
✔✔ANSWER✔✔-False, because the
description applies to a single solution. The
solution set consists of all possible solutions.

1.1
10. d.

Is the statement "Two fundamental questions
about a linear system involve existence and
uniqueness" true or false? Explain. -
✔✔ANSWER✔✔-True, because two
fundamental questions address whether the
solution exists and whether there is only one
solution.

1.1
11. a.

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