Linear Algebra Quiz: Computing Matrix Inverses
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Computing Matrix InversesQuestion 1 of 18

To begin finding the inverse of matrix A=(2413)A = \begin{pmatrix} 2 & 4 \\ 1 & 3 \end{pmatrix} using Gauss-Jordan elimination, a student first swaps rows 1 and 2, and then performs the operation R2→R2−2R1R_2 \rightarrow R_2 - 2R_1. What is the resulting matrix?

(13∣010−2∣1−2)\begin{pmatrix} 1 & 3 & | & 0 & 1 \\ 0 & -2 & | & 1 & -2 \end{pmatrix}
(13∣010−2∣10)\begin{pmatrix} 1 & 3 & | & 0 & 1 \\ 0 & -2 & | & 1 & 0 \end{pmatrix}
(13∣0102∣12)\begin{pmatrix} 1 & 3 & | & 0 & 1 \\ 0 & 2 & | & 1 & 2 \end{pmatrix}
(13∣01−1−2∣1−1)\begin{pmatrix} 1 & 3 & | & 0 & 1 \\ -1 & -2 & | & 1 & -1 \end{pmatrix}
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Linear Algebra Quiz

Linear Algebra Quiz: Computing Matrix Inverses

Practice Computing Matrix Inverses in Linear Algebra with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Computing Matrix Inverses, giving you a quick way to practice the rules, question types, and explanations that matter most for Linear Algebra.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

To begin finding the inverse of matrix A=(2413)A = \begin{pmatrix} 2 & 4 \\ 1 & 3 \end{pmatrix} using Gauss-Jordan elimination, a student first swaps rows 1 and 2, and then performs the operation R2→R2−2R1R_2 \rightarrow R_2 - 2R_1. What is the resulting matrix?

  1. (13∣010−2∣1−2)\begin{pmatrix} 1 & 3 & | & 0 & 1 \\ 0 & -2 & | & 1 & -2 \end{pmatrix} (correct answer)
  2. (13∣010−2∣10)\begin{pmatrix} 1 & 3 & | & 0 & 1 \\ 0 & -2 & | & 1 & 0 \end{pmatrix}
  3. (13∣0102∣12)\begin{pmatrix} 1 & 3 & | & 0 & 1 \\ 0 & 2 & | & 1 & 2 \end{pmatrix}
  4. (13∣01−1−2∣1−1)\begin{pmatrix} 1 & 3 & | & 0 & 1 \\ -1 & -2 & | & 1 & -1 \end{pmatrix}
Explanation: The process starts with the augmented matrix [A∣I]=(24∣1013∣01)[A|I] = \begin{pmatrix} 2 & 4 & | & 1 & 0 \\ 1 & 3 & | & 0 & 1 \end{pmatrix}. First, swapping the rows (R1↔R2R_1 \leftrightarrow R_2) gives (13∣0124∣10)\begin{pmatrix} 1 & 3 & | & 0 & 1 \\ 2 & 4 & | & 1 & 0 \end{pmatrix}. Next, the operation R2→R2−2R1R_2 \rightarrow R_2 - 2R_1 is applied to the new second row: [2,4,1,0]−2×[1,3,0,1]=[2−2,4−6,1−0,0−2]=[0,−2,1,−2][2, 4, 1, 0] - 2 \times [1, 3, 0, 1] = [2-2, 4-6, 1-0, 0-2] = [0, -2, 1, -2]. This yields the final matrix (13∣010−2∣1−2)\begin{pmatrix} 1 & 3 & | & 0 & 1 \\ 0 & -2 & | & 1 & -2 \end{pmatrix}.

Question 2

A student's work to find the inverse of A=(1235)A = \begin{pmatrix} 1 & 2 \\ 3 & 5 \end{pmatrix} is shown below.

Step 1 (Setup): (12∣1035∣01)\begin{pmatrix} 1 & 2 & | & 1 & 0 \\ 3 & 5 & | & 0 & 1 \end{pmatrix}

Step 2 (R2→R2−3R1R_2 \rightarrow R_2 - 3R_1): (12∣100−1∣01)\begin{pmatrix} 1 & 2 & | & 1 & 0 \\ 0 & -1 & | & 0 & 1 \end{pmatrix}

Step 3 (R2→−R2R_2 \rightarrow -R_2): (12∣1001∣0−1)\begin{pmatrix} 1 & 2 & | & 1 & 0 \\ 0 & 1 & | & 0 & -1 \end{pmatrix}

Step 4 (R1→R1−2R2R_1 \rightarrow R_1 - 2R_2): (10∣1201∣0−1)\begin{pmatrix} 1 & 0 & | & 1 & 2 \\ 0 & 1 & | & 0 & -1 \end{pmatrix}

In which step did the student make the first error?

  1. Step 1
  2. Step 2 (correct answer)
  3. Step 3
  4. Step 4
Explanation: The error occurs in Step 2. The row operation R2→R2−3R1R_2 \rightarrow R_2 - 3R_1 must be applied to the entire second row, including the part on the right side of the augmentation bar. The correct calculation for the new second row is [3,5,0,1]−3×[1,2,1,0]=[3−3,5−6,0−3,1−0]=[0,−1,−3,1][3, 5, 0, 1] - 3 \times [1, 2, 1, 0] = [3-3, 5-6, 0-3, 1-0] = [0, -1, -3, 1]. The student failed to apply the operation to the right side of the bar, incorrectly leaving it as [0,1][0, 1].

Question 3

When using the Gauss-Jordan elimination method to find the inverse of an n×nn \times n matrix AA, which of the following outcomes definitively indicates that AA is singular?

  1. The process requires a row swap operation.
  2. The process results in a row of all zeros on the left side of the augmented matrix. (correct answer)
  3. The process results in a matrix on the right side with a determinant of 0.
  4. The process results in a row of all zeros on the right side of the augmented matrix.
Explanation: If the row reduction process on the left side (the AA portion) of the augmented matrix [A∣I][A|I] results in a row containing only zeros, it is impossible to transform AA into the identity matrix II. This indicates that the original matrix AA is singular and does not have an inverse. The other options are not definitive indicators of singularity.

Question 4

To find the inverse of a non-singular 3×33 \times 3 matrix AA using the row reduction method, one must form an augmented matrix. If II is the 3×33 \times 3 identity matrix and 00 is the 3×33 \times 3 zero matrix, which is the correct initial setup?

  1. [A∣0][A | 0]
  2. [I∣A][I | A]
  3. [A∣I][A | I] (correct answer)
  4. [A∣A][A | A]
Explanation: When finding a matrix inverse using row reduction, you're essentially solving the matrix equation AX=IAX = I for the unknown matrix XX, which will be A−1A^{-1}. The row reduction method transforms this problem into an augmented matrix that you can solve systematically. The correct setup is [A∣I][A | I] because you want to perform row operations that transform the left side (matrix AA) into the identity matrix II. When you apply these same operations to the right side (which starts as II), it becomes A−1A^{-1}. Think of it as: "Start with AA on the left, end with II on the left, and whatever happens to the right side is your inverse." Option A [A∣0][A | 0] is incorrect because starting with the zero matrix on the right would give you the zero matrix after any sequence of row operations—you'd never obtain the inverse. Option B [I∣A][I | A] reverses the setup. If you reduced the left side to the identity matrix, you'd be starting with II and trying to get to II, which means no row operations are needed. The right side would remain AA, giving you AA, not A−1A^{-1}. Option D [A∣A][A | A] would give you A−1A=IA^{-1}A = I on the right side after reduction, not the inverse matrix itself. Study tip: Remember the phrase "A becomes I, I becomes A-inverse" to recall that you start with [A∣I][A | I] and end with [I∣A−1][I | A^{-1}].

Question 5

To find the inverse of A=(123014560)A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix}, the augmented matrix [A∣I][A|I] is created. After performing the row operations necessary to create zeros in the first column below the main diagonal, what is the third row of the resulting matrix?

  1. [0415∣50−1][0 \quad 4 \quad 15 \quad | \quad 5 \quad 0 \quad -1]
  2. [0−4−15∣001][0 \quad -4 \quad -15 \quad | \quad 0 \quad 0 \quad 1]
  3. [560∣001][5 \quad 6 \quad 0 \quad | \quad 0 \quad 0 \quad 1]
  4. [0−4−15∣−501][0 \quad -4 \quad -15 \quad | \quad -5 \quad 0 \quad 1] (correct answer)
Explanation: When finding a matrix inverse using the Gauss-Jordan method, you create an augmented matrix [A∣I][A|I] and perform row operations to transform the left side into the identity matrix. The key is systematically creating zeros below each pivot element. Starting with [A∣I]=(123∣100014∣010560∣001)[A|I] = \begin{pmatrix} 1 & 2 & 3 & | & 1 & 0 & 0 \\ 0 & 1 & 4 & | & 0 & 1 & 0 \\ 5 & 6 & 0 & | & 0 & 0 & 1 \end{pmatrix}, you need to eliminate the 5 in position (3,1) to create a zero below the first pivot. The first column already has a zero in position (2,1), so you only need one operation: R3−5R1→R3R_3 - 5R_1 \rightarrow R_3. This gives you:
  • New third row: (5,6,0,0,0,1)−5(1,2,3,1,0,0)=(0,−4,−15,−5,0,1)(5, 6, 0, 0, 0, 1) - 5(1, 2, 3, 1, 0, 0) = (0, -4, -15, -5, 0, 1)
The resulting third row is [0−4−15∣−501][0 \quad -4 \quad -15 \quad | \quad -5 \quad 0 \quad 1], which is answer choice D. Choice A has incorrect signs and the wrong augmented portion. Choice B has the correct left side but shows [001][0 \quad 0 \quad 1] in the augmented portion, which would result from a different operation entirely. Choice C is the original third row with no operations performed. Study tip: When eliminating entries below a pivot, always use the operation Ri−(entry to eliminate)×(pivot row)→RiR_i - (\text{entry to eliminate}) \times (\text{pivot row}) \rightarrow R_i. Track both sides of the augmented matrix carefully—the identity matrix side records your row operations and becomes the inverse.

Question 6

A student attempts to find the inverse of matrix A=(121013134)A = \begin{pmatrix} 1 & 2 & 1 \\ 0 & 1 & 3 \\ 1 & 3 & 4 \end{pmatrix} using row reduction. Which of the following matrices represents a correct intermediate step in this process?

  1. (100∣100010∣010001∣001)\begin{pmatrix} 1 & 0 & 0 & | & 1 & 0 & 0 \\ 0 & 1 & 0 & | & 0 & 1 & 0 \\ 0 & 0 & 1 & | & 0 & 0 & 1 \end{pmatrix}
  2. (100∣115010∣33−3001∣−1−11)\begin{pmatrix} 1 & 0 & 0 & | & 1 & 1 & 5 \\ 0 & 1 & 0 & | & 3 & 3 & -3 \\ 0 & 0 & 1 & | & -1 & -1 & 1 \end{pmatrix}
  3. (121∣100013∣010013∣001)\begin{pmatrix} 1 & 2 & 1 & | & 1 & 0 & 0 \\ 0 & 1 & 3 & | & 0 & 1 & 0 \\ 0 & 1 & 3 & | & 0 & 0 & 1 \end{pmatrix}
  4. (121∣100013∣010000∣−1−11)\begin{pmatrix} 1 & 2 & 1 & | & 1 & 0 & 0 \\ 0 & 1 & 3 & | & 0 & 1 & 0 \\ 0 & 0 & 0 & | & -1 & -1 & 1 \end{pmatrix} (correct answer)
Explanation: When finding a matrix inverse using row reduction, you're performing operations on the augmented matrix [A∣I][A|I] to transform it into [I∣A−1][I|A^{-1}]. The key insight is recognizing what constitutes a valid intermediate step versus an impossible outcome. Let's trace through the process. You start with (121∣100013∣010134∣001)\begin{pmatrix} 1 & 2 & 1 & | & 1 & 0 & 0 \\ 0 & 1 & 3 & | & 0 & 1 & 0 \\ 1 & 3 & 4 & | & 0 & 0 & 1 \end{pmatrix}. After eliminating the first column's lower entry (subtracting row 1 from row 3), you get $$ \begin{pmatrix} 1 & 2 & 1 & | & 1 & 0 & 0 \ 0 & 1 & 3 & | & 0 & 1 & 0 \ 0 & 1 & 3 & | & -1 & 0 & 1 \end{pmatrix} Option A shows the final result, not an intermediate step. Option B claims to be the final inverse, but this would only be valid if the matrix were invertible. Option C shows an impossible intermediate step—the third row should become $$ \begin{pmatrix} 0 & 1 & 3 & | & -1 & 0 & 1 \end{pmatrix} $$ after the first elimination, not what's shown. Most importantly, option D reveals that this matrix is actually singular (non-invertible) because the left side has a zero row. This is the correct realization a student should reach during the process. **Study tip:** When row reducing for inverse, if you ever get a zero row on the left side, stop immediately—the matrix has no inverse. Don't assume every matrix in a problem is invertible.

Question 7

During the process of finding the inverse of a matrix AA by applying row operations to [A∣I][A|I], the augmented matrix is reduced to (1−2∣1000∣31)\begin{pmatrix} 1 & -2 & | & 1 & 0 \\ 0 & 0 & | & 3 & 1 \end{pmatrix}. What can be concluded about the inverse of matrix AA?

  1. The inverse of AA does not exist. (correct answer)
  2. The inverse of AA is (1031)\begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix}.
  3. The inverse of AA is (1−200)\begin{pmatrix} 1 & -2 \\ 0 & 0 \end{pmatrix}.
  4. The inverse of AA is (1301)\begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix}.
Explanation: The row reduction process reveals that the original matrix AA is singular (i.e., it does not have an inverse) if it is impossible to obtain the identity matrix on the left side of the augmented matrix. The presence of a row of all zeros on the left side, [0,0][0, 0], indicates that the matrix is singular. Therefore, the inverse of AA does not exist.

Question 8

A student is finding the inverse of A=(4253)A = \begin{pmatrix} 4 & 2 \\ 5 & 3 \end{pmatrix}. As a first step, the student applies the row operation R1→−R1+R2R_1 \rightarrow -R_1 + R_2. What is the entire first row of the resulting augmented matrix?

  1. [−1−1∣1−1][-1 \quad -1 \quad | \quad 1 \quad -1]
  2. [11∣10][1 \quad 1 \quad | \quad 1 \quad 0]
  3. [11∣11][1 \quad 1 \quad | \quad 1 \quad 1]
  4. [11∣−11][1 \quad 1 \quad | \quad -1 \quad 1] (correct answer)
Explanation: When finding a matrix inverse using row operations, you set up an augmented matrix [A∣I][A|I] where the identity matrix is on the right. Your goal is to transform the left side into the identity matrix through row operations, which simultaneously transforms the right side into A−1A^{-1}. Starting with $$A = \begin{pmatrix} 4 & 2 \ 5 & 3 \end{pmatrix} \begin{pmatrix} 4 & 2 & | & 1 & 0 \ 5 & 3 & | & 0 & 1 \end{pmatrix} The row operation $$R_1 \rightarrow -R_1 + R_2$$ means you replace row 1 with the result of $$-(\text{row 1}) + (\text{row 2})$$. Let's calculate each element: - First element: $$-4 + 5 = 1$$ - Second element: $$-2 + 3 = 1$$ - Third element: $$-1 + 0 = -1$$ - Fourth element: $$-0 + 1 = 1$$ This gives us the new first row: $$[1 \quad 1 \quad | \quad -1 \quad 1]$$, which is answer D. Looking at the wrong answers: A gives $$[-1, -1]$$ for the left side, which would result from $$R_1 - R_2$$ instead. B has the correct left side but shows $$[1, 0]$$ on the right, ignoring the sign change in the identity portion. C shows $$[1, 1]$$ on the right, which would happen if you forgot to apply the negative sign to the first row entirely. Remember: row operations must be applied to the entire row, including both the original matrix portion and the identity matrix portion of the augmented matrix.

Question 9

The process of finding the inverse of a 2×22 \times 2 matrix AA via row reduction yields the final augmented matrix (10∣3−101∣−21)\begin{pmatrix} 1 & 0 & | & 3 & -1 \\ 0 & 1 & | & -2 & 1 \end{pmatrix}. What is the sum of the elements of the original matrix AA?

  1. 1
  2. 7 (correct answer)
  3. -7
  4. 0
Explanation: The final augmented matrix is [I∣A−1][I|A^{-1}], so we are given A−1=(3−1−21)A^{-1} = \begin{pmatrix} 3 & -1 \\ -2 & 1 \end{pmatrix}. To find the original matrix AA, we must compute the inverse of A−1A^{-1}, since A=(A−1)−1A = (A^{-1})^{-1}. Using the formula for the inverse of a 2×22 \times 2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix}, which is 1ad−bc(d−b−ca)\frac{1}{ad-bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}, we find the determinant of A−1A^{-1} is (3)(1)−(−1)(−2)=3−2=1(3)(1) - (-1)(-2) = 3 - 2 = 1. Thus, A=11(1−(−1)−(−2)3)=(1123)A = \frac{1}{1}\begin{pmatrix} 1 & -(-1) \\ -(-2) & 3 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 2 & 3 \end{pmatrix}. The sum of the elements of AA is 1+1+2+3=71+1+2+3=7.

Question 10

A student attempts to find the inverse of matrix B=(12−1013211)B = \begin{pmatrix} 1 & 2 & -1 \\ 0 & 1 & 3 \\ 2 & 1 & 1 \end{pmatrix} using row reduction. After performing the row operation R3←R3−2R1R_3 \leftarrow R_3 - 2R_1, what should be the next strategic row operation to continue toward reduced row echelon form?

  1. R3←R3+3R2R_3 \leftarrow R_3 + 3R_2 to eliminate the entry in position (3,2) (correct answer)
  2. R1←R1−2R2R_1 \leftarrow R_1 - 2R_2 to eliminate the entry in position (1,2)
  3. R3←13R3R_3 \leftarrow \frac{1}{3}R_3 to create a leading 1 in row 3
  4. R2←R2−3R3R_2 \leftarrow R_2 - 3R_3 to eliminate the entry in position (2,3)
Explanation: After the operation R₃ ← R₃ - 2R₁, the matrix becomes [[1,2,-1],[0,1,3],[0,-3,3]]. The next logical step in Gaussian elimination is to eliminate the -3 in position (3,2) by adding 3 times row 2 to row 3. Choice B is premature (done during back-substitution), Choice C should wait until after eliminating the (3,2) entry, and Choice D references an operation that doesn't make sense given the current matrix state.

Question 11

Consider the augmented matrix [10210001−1010000ab1]\left[\begin{array}{ccc|ccc} 1 & 0 & 2 & 1 & 0 & 0 \\ 0 & 1 & -1 & 0 & 1 & 0 \\ 0 & 0 & 0 & a & b & 1 \end{array}\right] that resulted from attempting to find the inverse of a 3×3 matrix using row reduction. What can be concluded about the original matrix and the values of aa and bb?

  1. The original matrix is singular, and the values of aa and bb indicate the specific linear dependence among the original rows
  2. The original matrix has an inverse, but aa and bb must both be zero to complete the process successfully
  3. The original matrix is singular, and aa and bb must both be zero for the system to be consistent (correct answer)
  4. The original matrix has an inverse if and only if aa and bb are both nonzero values
Explanation: The presence of a zero row [0 0 0] in the left portion of the augmented matrix indicates that the original matrix is singular (not invertible). For the system to be consistent during the row reduction process, the corresponding entries on the right side must also form a zero row, meaning a = b = 0. If a or b were nonzero, it would represent an inconsistent system. Choice A is partially correct about singularity but wrong about the interpretation of a and b. Choice B incorrectly claims the matrix has an inverse. Choice D incorrectly suggests nonzero a and b would make the matrix invertible.

Question 12

Consider a 3×3 upper triangular matrix H=(2310−14003)H = \begin{pmatrix} 2 & 3 & 1 \\ 0 & -1 & 4 \\ 0 & 0 & 3 \end{pmatrix} . When using row reduction to find H−1H^{-1}, which statement best describes the computational advantage of this structure?

  1. The row reduction process is identical to that of any other matrix, so there is no computational advantage
  2. The determinant is immediately visible as the product of diagonal entries, allowing for quick verification of invertibility before starting
  3. Upper triangular matrices always have simpler inverses that can be written down directly without row operations
  4. No elimination below the diagonal is needed, so the process begins directly with back-substitution operations to eliminate above the diagonal (correct answer)
Explanation: When you encounter matrix inversion problems involving special matrix structures like upper triangular matrices, the key insight is recognizing how their structure affects the row reduction algorithm. To find H−1H^{-1}, you set up the augmented matrix [H∣I][H|I] and use row operations to transform it to [I∣H−1][I|H^{-1}]. For a general matrix, this involves two phases: forward elimination (creating zeros below the diagonal) and back-substitution (creating zeros above the diagonal). However, since HH is already upper triangular, all entries below the diagonal are already zero. This means you can skip the entire forward elimination phase and jump directly to back-substitution operations, where you systematically eliminate entries above the diagonal from right to left. Answer choice A is incorrect because the computational advantage is significant—you're essentially cutting the algorithm in half. Choice B, while true that the determinant 2×(−1)×3=−62 \times (-1) \times 3 = -6 is easily computed, describes a verification step rather than the main computational advantage during inversion. Choice C is false because you still need row operations; upper triangular matrices don't have inverses that can simply be "written down" without calculation. The correct answer is D because it captures the essential computational savings: no elimination below the diagonal is needed, allowing you to begin immediately with back-substitution. Study tip: When you see structured matrices (triangular, diagonal, etc.) in inversion problems, always consider how their existing zero patterns can shortcut the standard row reduction process.

Question 13

A student correctly reduces the augmented matrix [35102401]\left[\begin{array}{cc|cc} 3 & 5 & 1 & 0 \\ 2 & 4 & 0 & 1 \end{array}\right] to [102−5201−132]\left[\begin{array}{cc|cc} 1 & 0 & 2 & -\frac{5}{2} \\ 0 & 1 & -1 & \frac{3}{2} \end{array}\right]. If the student wants to verify this result, which calculation would be most efficient?

  1. Check that each column of the computed inverse satisfies the original system when substituted as a solution vector
  2. Recalculate the inverse using the formula $$\frac{1}{\text{det}(A)} \begin{pmatrix} d & -b \ -c & a \end{pmatrix}
  3. Verify that the determinant of the original matrix equals the reciprocal of the determinant of the computed inverse
  4. Multiply $$ \begin{pmatrix} 3 & 5 \ 2 & 4 \end{pmatrix} \begin{pmatrix} 2 & -\frac{5}{2} \ -1 & \frac{3}{2} \end{pmatrix} (correct answer)
Explanation: When you see an augmented matrix being reduced to find an inverse, you're looking at the process [A∣I]→[I∣A−1][A|I] \rightarrow [I|A^{-1}]. The most fundamental way to verify any matrix inverse is to check that AA−1=IAA^{-1} = I. Option D gives you exactly this verification. When you multiply (3524)(2−52−132)\begin{pmatrix} 3 & 5 \\ 2 & 4 \end{pmatrix} \begin{pmatrix} 2 & -\frac{5}{2} \\ -1 & \frac{3}{2} \end{pmatrix}, you get $$ \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} Option A is unnecessarily complicated. While each column of the inverse does solve $$A\mathbf{x} = \mathbf{e}_i$$ (where $$\mathbf{e}_i$$ are standard basis vectors), checking this requires multiple system solutions rather than one simple multiplication. Option B involves recalculating the entire inverse using the $$2 \times 2$$ formula, which defeats the purpose of verification. You'd essentially be redoing the original work rather than checking it. Option C misses the point entirely. For any invertible matrix, $$\det(A^{-1}) = \frac{1}{\det(A)}$$ is always true, but this doesn't verify that your computed matrix is actually $$A^{-1}$$. You could have the right determinant but completely wrong entries. Remember: when verifying a matrix inverse, always use the definition $$AA^{-1} = I$$. It's the most direct and reliable check, requiring minimal computation while giving you complete confidence in your result.

Question 14

A 3×3 matrix EE requires exactly 8 row operations to reduce [E∣I][E|I] to [I∣E−1][I|E^{-1}] using the Gauss-Jordan method. If the same matrix were reduced using only Gaussian elimination to reach row echelon form and then back-substitution, approximately how many operations would be needed?

  1. Exactly 8 operations, since both methods require the same number of steps for any given matrix
  2. Approximately 5-6 operations, since Gaussian elimination requires fewer operations than Gauss-Jordan reduction (correct answer)
  3. Approximately 10-12 operations, since back-substitution typically requires more steps than direct elimination
  4. The number cannot be determined without knowing the specific entries of matrix EE
Explanation: Gaussian elimination to row echelon form typically requires fewer operations than full Gauss-Jordan reduction to reduced row echelon form. The Gauss-Jordan method eliminates entries both above and below the diagonal simultaneously, while Gaussian elimination only eliminates below the diagonal initially. For a 3×3 matrix requiring 8 operations in Gauss-Jordan, roughly 5-6 operations would achieve row echelon form. Choice A incorrectly assumes equal operation counts. Choice C incorrectly suggests more operations are needed. Choice D is wrong because the relative efficiency of methods is generally predictable regardless of specific entries.

Question 15

Two students are finding the inverse of G=(2132)G = \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix} using row reduction. Student A begins with row operation R1←12R1R_1 \leftarrow \frac{1}{2}R_1, while Student B begins with R2←R2−32R1R_2 \leftarrow R_2 - \frac{3}{2}R_1. Assuming both students complete the process correctly, how will their final answers compare?

  1. Student A will get the correct inverse, but Student B's approach will lead to computational errors due to the fractional coefficients introduced early
  2. Both students will obtain identical final answers, but Student A's method will require fewer arithmetic operations overall
  3. Both students will obtain identical final answers, but Student B's method avoids fractions longer and may be less error-prone (correct answer)
  4. The students will obtain different but equivalent forms of the inverse that differ by a scalar multiple
Explanation: Both approaches are valid and will yield the same final inverse matrix G^(-1) = [[2,-1],[-3,2]]. Student A's approach introduces fractions early (making calculations potentially more error-prone), while Student B's approach eliminates the (2,1) position first while keeping integer arithmetic longer. Student B's method is often preferred in practice because it delays the introduction of fractions. Choice A incorrectly suggests one approach leads to errors. Choice B incorrectly claims A requires fewer operations. Choice D incorrectly suggests different final results.

Question 16

What is the entry in the first row, second column of the inverse of the matrix A=(3122)A = \begin{pmatrix} 3 & 1 \\ 2 & 2 \end{pmatrix}?

  1. −14-\frac{1}{4} (correct answer)
  2. 14\frac{1}{4}
  3. 12\frac{1}{2}
  4. −12-\frac{1}{2}
Explanation: To find the inverse, we row reduce [A∣I]=(31∣1022∣01)[A|I] = \begin{pmatrix} 3 & 1 & | & 1 & 0 \\ 2 & 2 & | & 0 & 1 \end{pmatrix}.
  1. R1→13R1R_1 \rightarrow \frac{1}{3}R_1: (11/3∣1/3022∣01)\begin{pmatrix} 1 & 1/3 & | & 1/3 & 0 \\ 2 & 2 & | & 0 & 1 \end{pmatrix}.
  2. R2→R2−2R1R_2 \rightarrow R_2 - 2R_1: (11/3∣1/3004/3∣−2/31)\begin{pmatrix} 1 & 1/3 & | & 1/3 & 0 \\ 0 & 4/3 & | & -2/3 & 1 \end{pmatrix}.
  3. R2→34R2R_2 \rightarrow \frac{3}{4}R_2: (11/3∣1/3001∣−1/23/4)\begin{pmatrix} 1 & 1/3 & | & 1/3 & 0 \\ 0 & 1 & | & -1/2 & 3/4 \end{pmatrix}.
  4. R1→R1−13R2R_1 \rightarrow R_1 - \frac{1}{3}R_2: (10∣1/2−1/401∣−1/23/4)\begin{pmatrix} 1 & 0 & | & 1/2 & -1/4 \\ 0 & 1 & | & -1/2 & 3/4 \end{pmatrix}. The inverse is A−1=(1/2−1/4−1/23/4)A^{-1} = \begin{pmatrix} 1/2 & -1/4 \\ -1/2 & 3/4 \end{pmatrix}. The entry in the first row, second column is −14-\frac{1}{4}.

Question 17

What is the second row of the inverse of the matrix A=(123014560)A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix}?

  1. [−24185][-24 \quad 18 \quad 5]
  2. [−541][-5 \quad 4 \quad 1]
  3. [20−15−4][20 \quad -15 \quad -4] (correct answer)
  4. [014][0 \quad 1 \quad 4]
Explanation: By performing Gauss-Jordan elimination on the augmented matrix [A∣I][A|I], we transform the left side into the identity matrix. The right side becomes the inverse. The full inverse is A−1=(−2418520−15−4−541)A^{-1} = \begin{pmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{pmatrix}. The second row of A−1A^{-1} is [20−15−4][20 \quad -15 \quad -4]. Distractors include the first row of the inverse, the third row of the inverse, and the second row of the original matrix AA.

Question 18

After correctly applying Gauss-Jordan elimination to the augmented matrix [A∣I][A|I], a student obtains the final matrix [I∣B][I|B]. Which equation must be true based on the definition of a matrix inverse?

  1. A=BTA = B^T
  2. A=BA = B
  3. A=B−1A = B^{-1} (correct answer)
  4. A+B=IA + B = I
Explanation: When you encounter Gauss-Jordan elimination applied to an augmented matrix [A∣I][A|I], you're witnessing the systematic process of finding a matrix inverse. The key insight is understanding what this process accomplishes: it transforms the left side (matrix AA) into the identity matrix II while simultaneously applying the same row operations to the right side (initially II). The definition of matrix inverse tells us that if B=A−1B = A^{-1}, then AB=BA=IAB = BA = I. During Gauss-Jordan elimination, you're essentially multiplying both sides of [A∣I][A|I] by A−1A^{-1} through row operations. This gives you [A−1A∣A−1I]=[I∣A−1][A^{-1}A|A^{-1}I] = [I|A^{-1}]. Since the final matrix is [I∣B][I|B], we know that B=A−1B = A^{-1}, which means A=B−1A = B^{-1}. Looking at the wrong answers: Choice A (A=BTA = B^T) confuses inverse with transpose—these are completely different operations. Choice B (A=BA = B) would only be true if AA were its own inverse, which happens only for very special matrices like some reflection matrices. Choice D (A+B=IA + B = I) incorrectly suggests an additive relationship when matrix inverses involve multiplication. Remember this pattern: whenever you see Gauss-Jordan elimination on [A∣I][A|I] resulting in [I∣B][I|B], the matrix BB is always A−1A^{-1}. This is the fundamental method for computing matrix inverses, so connecting the process to the inverse definition is crucial for linear algebra success.