Free Printable WorksheetMathAlgebra IIMath CurriculumALG2.231

ALG2.231 Determinants and Invertibility (Extension) Worksheet

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Worksheet Overview & Skills Covered

Core academic competencies and curriculum alignment

This Algebra II Math worksheet provides structured practice on ALG2.231 Determinants and Invertibility (Extension). Students are guided through targeted exercises that build foundational understanding, computational accuracy, and critical thinking skills.

ALG2.231 Determinants and Invertibility (Extension) - Select the correct computed value for determinants (2×2) and determine whether a matrix is invertible; connect invertibility to unique solutions. (Course extension; included in many Algebra 2 scopes such as IXL Algebra 2.)

Whether you are an educator preparing classroom handouts, a tutor reinforcing weekly topics, or a parent supporting homeschool learning, this printable resource gives students the deliberate practice required to achieve subject fluency and master state standards.

Grade

Algebra II

Subject

Math

Domain

Math Curriculum

Formats

PDF & DOCX

Sample Practice Problems

Preview the types of problems and questions covered in this worksheet

Below are sample questions representative of the exercises included on this worksheet and related ALG2.231 Determinants and Invertibility (Extension) skill assessments:

Problem 1Difficulty: Mastery
For A(p) = [[p, p + 1], [p - 1, p]], which statement is true for every real value of p?
AA(p) is invertible only when p = 0
BA(p) is invertible because det(A(p)) = 1
CA(p) is not invertible only when p = 1
DA(p) is not invertible because det(A(p)) = 0
Solution & Explanation:The determinant is p*p - (p + 1)(p - 1) = p2 - (p2 - 1) = 1. Since the determinant is always 1, A(p) is always invertible.
Problem 2Difficulty: Mastery
Matrix A = [[1, 2], [3, 6]] is changed to matrix B = [[1, 2], [3, 7]]. Which explanation best describes the change in invertibility?
AA has determinant 1, while B has determinant 0, so A is not invertible
BBoth matrices are not invertible because their first rows match
CBoth matrices are invertible because all entries are positive
DA has determinant 0, while B has determinant 1, so B is invertible
Solution & Explanation:det(A) = 1*6 - 2*3 = 0. det(B) = 1*7 - 2*3 = 1. Changing the bottom-right entry breaks the proportional row relationship and makes B invertible.
Problem 3Difficulty: Mastery
For the system 7x + ay = 4 and -2x + 6y = 1, which value of a makes the system fail to have a unique solution?
A21
B-3
C-21
D3
Solution & Explanation:The coefficient determinant is 7*6 - a*(-2) = 42 + 2a. Set 42 + 2a = 0 to get a = -21.

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Frequently Asked Questions

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QWhat grade level and standards does this worksheet align with?

This worksheet is designed specifically for Algebra II Math students. It aligns with educational standard ALG2.231 (No external standard code was listed in the curriculum source for this skill. Skill statement in this prompt cell: Select the correct computed value for determinants (2×2) and determine whether a matrix is invertible; connect invertibility to unique solutions. (Course extension; included in many Algebra 2 scopes such as IXL Algebra 2.) Use the skill code ALG2.231 and this prompt-provided wording as the supporting standard description for this workflow.).

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