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Algebra 2 Flashcards: Solving Systems Using Matrix Inverses

Study Solving Systems Using Matrix Inverses in Algebra 2 with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

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What this deck covers

This deck focuses on Solving Systems Using Matrix Inverses, giving you a quick way to review the definitions, rules, and examples that matter most for Algebra 2.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Algebra 2 Flashcards: Solving Systems Using Matrix Inverses

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QUESTION

Solve using an inverse: (1002)x⃗=(56)\begin{pmatrix}1&0\\0&2\end{pmatrix}\vec{x}=\begin{pmatrix}5\\6\end{pmatrix}(10​02​)x=(56​).

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ANSWER

x⃗=(53)\vec{x}=\begin{pmatrix}5\\3\end{pmatrix}x=(53​). Multiplying the inverse matrix by the vector yields the solution.

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Flashcard 1: Solve using an inverse: (1002)x⃗=(56)\begin{pmatrix}1&0\\0&2\end{pmatrix}\vec{x}=\begin{pmatrix}5\\6\end{pmatrix}(10​02​)x=(56​).

Answer: x⃗=(53)\vec{x}=\begin{pmatrix}5\\3\end{pmatrix}x=(53​). Multiplying the inverse matrix by the vector yields the solution.

Flashcard 2: What is the matrix size of AAA in Ax⃗=b⃗A\vec{x}=\vec{b}Ax=b for a system of nnn equations in nnn unknowns?

Answer: AAA is n×nn\times nn×n. A square n×nn\times nn×n matrix corresponds to nnn equations in nnn variables.

Flashcard 3: What is det⁡ ⁣((3124))\det\!\left(\begin{pmatrix}3&1\\2&4\end{pmatrix}\right)det((32​14​))?

Answer: 101010. Computed using the 2x2 determinant formula ad−bcad-bcad−bc.

Flashcard 4: Which matrix must have the same size as AAA to form [A∣I][A\mid I][A∣I] for an n×nn\times nn×n matrix?

Answer: The identity matrix InI_nIn​. InI_nIn​ is square n×nn\times nn×n to match AAA's dimensions for augmentation.

Flashcard 5: Identify whether A=(100002030)A=\begin{pmatrix}1&0&0\\0&0&2\\0&3&0\end{pmatrix}A=​100​003​020​​ is invertible using det⁡(A)\det(A)det(A).

Answer: Invertible because det⁡(A)=−6≠0\det(A)=-6\neq 0det(A)=−6=0. The non-zero determinant confirms the matrix is invertible.

Flashcard 6: Find ((02−31))−1\left(\begin{pmatrix}0&2\\-3&1\end{pmatrix}\right)^{-1}((0−3​21​))−1.

Answer: (16−13frac120)\begin{pmatrix}\frac{1}{6}&-\frac{1}{3}\\frac{1}{2}&0\end{pmatrix}(61​frac12​−31​0​). The inverse is scaled by 1/det⁡(A)1/\det(A)1/det(A) using the adjugate matrix.

Flashcard 7: What does Ax⃗=b⃗A\vec{x}=\vec{b}Ax=b represent when AAA is 3×33\times 33×3?

Answer: A system of 333 linear equations in 333 unknowns. The dimensions of AAA determine the number of equations and unknowns.

Flashcard 8: Identify the matrix equation you should solve to find A−1A^{-1}A−1 using row reduction.

Answer: Row-reduce [A∣In][A\mid I_n][A∣In​] to [In∣A−1][I_n\mid A^{-1}][In​∣A−1]. Row reduction of the augmented matrix yields the inverse on the right.

Flashcard 9: Identify the correct criterion: AAA is invertible iff det⁡(A)=0\det(A)=0det(A)=0 or det⁡(A)≠0\det(A)\neq 0det(A)=0?

Answer: AAA is invertible iff det⁡(A)≠0\det(A)\neq 0det(A)=0. Invertibility requires a non-zero determinant.

Flashcard 10: Compute (3A)−1(3A)^{-1}(3A)−1 if A−1=MA^{-1}=MA−1=M and 3≠03\neq 03=0.

Answer: (3A)−1=13M(3A)^{-1}=\frac{1}{3}M(3A)−1=31​M. The scalar 3 requires dividing the inverse by 3.

Flashcard 11: What equation defines the inverse matrix A−1A^{-1}A−1 of a square matrix AAA?

Answer: AA−1=IAA^{-1}=IAA−1=I (equivalently A−1A=IA^{-1}A=IA−1A=I). The inverse satisfies the multiplicative identity property when multiplied on the left or right by the original matrix.

Flashcard 12: What is A−1A^{-1}A−1 for A=(1000−20004)A=\begin{pmatrix}1&0&0\\0&-2&0\\0&0&4\end{pmatrix}A=​100​0−20​004​​?

Answer: A−1=(1000−1200014)A^{-1}=\begin{pmatrix}1&0&0\\0&-\frac{1}{2}&0\\0&0&\frac{1}{4}\end{pmatrix}A−1=​100​0−21​0​0041​​​. Reciprocate each diagonal entry to form the inverse.

Flashcard 13: What is the inverse of a 2×22\times 22×2 matrix A=(abcd)A=\begin{pmatrix}a&b\\c&d\end{pmatrix}A=(ac​bd​) when it exists?

Answer: A−1=1ad−bc(d−b−ca)A^{-1}=\frac{1}{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}A−1=ad−bc1​(d−c​−ba​). The formula scales the adjugate matrix by the reciprocal of the determinant to satisfy AA−1=IAA^{-1}=IAA−1=I.

Flashcard 14: What is the determinant of the identity matrix InI_nIn​?

Answer: det⁡(In)=1\det(I_n)=1det(In​)=1. It is the product of the diagonal entries, all of which are 1.

Flashcard 15: Identify the correct solution form for Ax⃗=b⃗A\vec{x}=\vec{b}Ax=b when AAA is invertible: x⃗=Ab⃗\vec{x}=A\vec{b}x=Ab or x⃗=A−1b⃗\vec{x}=A^{-1}\vec{b}x=A−1b?

Answer: x⃗=A−1b⃗\vec{x}=A^{-1}\vec{b}x=A−1b. The correct form uses the inverse matrix multiplied on the left by b⃗\vec{b}b.

Flashcard 16: What does it mean for the system Ax⃗=b⃗A\vec{x}=\vec{b}Ax=b if AAA is not invertible?

Answer: The system has either no solution or infinitely many solutions. Non-invertible matrices lead to systems that are either inconsistent or have infinite solutions.

Flashcard 17: What is the solution of Ax⃗=0⃗A\vec{x}=\vec{0}Ax=0 when AAA is invertible?

Answer: x⃗=0⃗\vec{x}=\vec{0}x=0. Invertible AAA implies only the trivial solution for the homogeneous system.

Flashcard 18: What is the determinant of A=(ab0d)A=\begin{pmatrix}a&b\\0&d\end{pmatrix}A=(a0​bd​)?

Answer: det⁡(A)=ad\det(A)=addet(A)=ad. For upper triangular matrices, the determinant equals the product of diagonals.

Flashcard 19: What is the inverse of a diagonal matrix D=diag(d1,…,dn)D=\mathrm{diag}(d_1,\dots,d_n)D=diag(d1​,…,dn​) when all di≠0d_i\neq 0di​=0?

Answer: D−1=diag(1d1,…,1dn)D^{-1}=\mathrm{diag}(\frac{1}{d_1},\dots,\frac{1}{d_n})D−1=diag(d1​1​,…,dn​1​). Each diagonal entry is reciprocated to ensure the product is the identity matrix.

Flashcard 20: What is the determinant of a diagonal matrix D=diag(d1,…,dn)D=\mathrm{diag}(d_1,\dots,d_n)D=diag(d1​,…,dn​)?

Answer: det⁡(D)=d1d2⋯dn\det(D)=d_1d_2\cdots d_ndet(D)=d1​d2​⋯dn​. The determinant is the product of its diagonal entries.

Flashcard 21: What is the determinant of a triangular matrix in terms of its diagonal entries?

Answer: det⁡(A)\det(A)det(A) equals the product of the diagonal entries. For triangular matrices, the determinant is the product of the diagonal elements.

Flashcard 22: What does it mean for the columns of AAA if AAA is invertible?

Answer: The columns are linearly independent. Invertibility requires the columns to form a linearly independent set.

Flashcard 23: What does it mean for the system Ax⃗=b⃗A\vec{x}=\vec{b}Ax=b if AAA is not invertible?

Answer: The system has either no solution or infinitely many solutions. Non-invertible matrices lead to systems that are either inconsistent or have infinite solutions.

Flashcard 24: What is the augmented matrix form of the system Ax⃗=b⃗A\vec{x}=\vec{b}Ax=b?

Answer: [A∣b⃗][A\mid \vec{b}][A∣b]. The augmented matrix combines coefficients and constants for row reduction.

Flashcard 25: What is the matrix size of AAA in Ax⃗=b⃗A\vec{x}=\vec{b}Ax=b for a system of nnn equations in nnn unknowns?

Answer: AAA is n×nn\times nn×n. A square n×nn\times nn×n matrix corresponds to nnn equations in nnn variables.

Flashcard 26: Identify the correct solution form for Ax⃗=b⃗A\vec{x}=\vec{b}Ax=b when AAA is invertible: x⃗=Ab⃗\vec{x}=A\vec{b}x=Ab or x⃗=A−1b⃗\vec{x}=A^{-1}\vec{b}x=A−1b?

Answer: x⃗=A−1b⃗\vec{x}=A^{-1}\vec{b}x=A−1b. The correct form uses the inverse matrix multiplied on the left by b⃗\vec{b}b.

Flashcard 27: What is A−1A^{-1}A−1 for A=(2005)A=\begin{pmatrix}2&0\\0&5\end{pmatrix}A=(20​05​)?

Answer: A−1=(120015)A^{-1}=\begin{pmatrix}\frac{1}{2}&0\\0&\frac{1}{5}\end{pmatrix}A−1=(21​0​051​​). The inverse of a diagonal matrix reciprocates each diagonal entry.

Flashcard 28: What is det⁡ ⁣((3124))\det\!\left(\begin{pmatrix}3&1\\2&4\end{pmatrix}\right)det((32​14​))?

Answer: 101010. Computed using the 2x2 determinant formula ad−bcad-bcad−bc.

Flashcard 29: Find ((1234))−1\left(\begin{pmatrix}1&2\\3&4\end{pmatrix}\right)^{-1}((13​24​))−1.

Answer: (−21frac32−12)\begin{pmatrix}-2&1\\frac{3}{2}&-\frac{1}{2}\end{pmatrix}(−2frac32​1−21​​). The 2x2 inverse formula scales the adjugate by 1/det⁡(A)1/\det(A)1/det(A).

Flashcard 30: Identify whether A=(1224)A=\begin{pmatrix}1&2\\2&4\end{pmatrix}A=(12​24​) is invertible.

Answer: Not invertible because det⁡(A)=0\det(A)=0det(A)=0. A zero determinant confirms the matrix is singular.