Elementary Matrices - Linear Algebra

1. A square matrix $A$ is transformed into a matrix $B$ by a single elementary row operation. If matrix $B$ is singular, which of the following statements must be true?
   Answer: The original matrix $A$ must also be singular. Let the elementary row operation be represented by the elementary matrix $E$. Then $B = EA$. The determinant of $B$ is $\det(B) = \det(E)\det(A)$. Since $B$ is singular, $\det(B) = 0$. This gives $\det(E)\det(A) = 0$. All elementary matrices are invertible, so their determinants are never zero ($\det(E) \neq 0$). For the product to be zero, it must be that $\det(A) = 0$. A matrix with a determinant of zero is singular. Therefore, the original matrix $A$ must also be singular.
2. If matrices $A$ and $B$ are row equivalent, and $E_1, E_2, \ldots, E_k$ are elementary matrices such that $E_k E_{k-1} \cdots E_1 A = B$, which statement about the matrix $M = E_k E_{k-1} \cdots E_1$ is always true?
   Answer: $M$ is invertible and $M^{-1} = E_1^{-1} E_2^{-1} \cdots E_k^{-1}$ When you encounter problems about row equivalent matrices and elementary operations, you're dealing with the fundamental relationship between matrix multiplication and row operations. The key insight is that performing a sequence of elementary row operations is equivalent to multiplying by a product of elementary matrices. Since $E_k E_{k-1} \cdots E_1 A = B$, the matrix $M = E_k E_{k-1} \cdots E_1$ represents the composition of all these elementary operations. Each elementary matrix is invertible (you can always "undo" a row operation), and the product of invertible matrices is invertible. To find $M^{-1}$, you reverse both the order and each individual operation: $M^{-1} = (E_k E_{k-1} \cdots E_1)^{-1} = E_1^{-1} E_2^{-1} \cdots E_k^{-1}$. This makes choice C correct. Choice A is wrong because $M$ is a product of multiple elementary matrices, not a single elementary matrix. Choice B is incorrect because while each elementary matrix has determinant $\pm 1$ or a positive constant, their product could have any nonzero determinant. For example, if you scale a row by 2 twice, you get $\det(M) = 4$. Choice D fails because $M$ represents multiple row operations combined, not a single operation. Remember: elementary matrices are the building blocks of all invertible matrices. When you see a product of elementary matrices, focus on the invertibility property and how inverses reverse both the order of operations and each individual operation.
3. Let $E$ be the elementary matrix corresponding to the row operation $R_1 + 7R_3 \to R_1$ for $3 \times 3$ matrices. Which matrix represents $E^{-1}$?
   Answer: The elementary matrix for the operation $R_1 - 7R_3 \to R_1$. The inverse of an elementary matrix corresponds to the inverse elementary row operation. The operation is adding 7 times row 3 to row 1. The inverse operation that would undo this is subtracting 7 times row 3 from row 1. This corresponds to the row operation $R_1 - 7R_3 \to R_1$. Distractor A incorrectly uses the rule for the inverse of a scaling operation. Distractor B confuses the roles of the rows. Distractor D incorrectly suggests the matrix is its own inverse, which is only true for row-swapping matrices.
4. A matrix $A$ is transformed into a matrix $B$ by the following sequence of operations on $A$'s rows: 1. First, scale row 2 by a factor of 3. 2. Second, add the new row 2 to row 1. Which single matrix $P$ satisfies the equation $PA=B$?
   Answer: $\begin{pmatrix} 1 & 3 \\ 0 & 3 \end{pmatrix}$ Let $E_1$ be the matrix for the first operation ($3R_2 \to R_2$) and $E_2$ be the matrix for the second operation ($R_1 + R_2 \to R_1$). For $2 \times 2$ matrices, $E_1 = \begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix}$ and $E_2 = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$. The operations are applied sequentially, so $B = E_2(E_1 A) = (E_2 E_1)A$. The matrix $P$ is the product $E_2 E_1$. $P = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix} = \begin{pmatrix} 1(1)+1(0) & 1(0)+1(3) \\ 0(1)+1(0) & 0(0)+1(3) \end{pmatrix} = \begin{pmatrix} 1 & 3 \\ 0 & 3 \end{pmatrix}$. The reversed product $E_1 E_2$ would give the incorrect matrix in choice A.
5. Consider the matrix equation $XA = B$ where $A = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}$ and $B = \begin{pmatrix} 2 & 1 \\ 6 & 3 \end{pmatrix}$. If $X$ must be an elementary matrix, which row operation does $X$ represent?
   Answer: Add $3$ times row $1$ to row $2$ We need $X = BA^{-1}$. First, $A^{-1} = \frac{1}{\det(A)}\begin{pmatrix} 3 & -1 \\ 0 & 2 \end{pmatrix} = \frac{1}{6}\begin{pmatrix} 3 & -1 \\ 0 & 2 \end{pmatrix}$. Then $X = \begin{pmatrix} 2 & 1 \\ 6 & 3 \end{pmatrix} \cdot \frac{1}{6}\begin{pmatrix} 3 & -1 \\ 0 & 2 \end{pmatrix} = \frac{1}{6}\begin{pmatrix} 6 & 0 \\ 18 & 6 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix}$. This elementary matrix adds 3 times row 1 to row 2, since it has 1's on the diagonal and a 3 in position (2,1).
6. Matrix $A$ can be reduced to matrix $B$ using exactly three elementary row operations: first multiply row $2$ by $\frac{1}{3}$, then add $-2$ times row $1$ to row $3$, then swap rows $1$ and $2$. If $\det(A) = 12$, what is $\det(B)$?
   Answer: $-4$ When you encounter questions about determinants and elementary row operations, remember that each type of row operation affects the determinant in a predictable way. This is a fundamental relationship in linear algebra that connects matrix transformations to determinant values. Let's trace through each operation applied to matrix $A$ to find $\det(B)$. Starting with $\det(A) = 12$: Operation 1: Multiply row 2 by $\frac{1}{3}$. When you multiply a row by a scalar $k$, the determinant gets multiplied by $k$. So $\det = 12 \times \frac{1}{3} = 4$. Operation 2: Add $-2$ times row 1 to row 3. Adding a multiple of one row to another row leaves the determinant unchanged. So $\det = 4$. Operation 3: Swap rows 1 and 2. Swapping any two rows multiplies the determinant by $-1$. So $\det(B) = 4 \times (-1) = -4$. Looking at the wrong answers: (A) 4 forgets the final row swap that changes the sign. (B) 12 incorrectly assumes row operations don't change the determinant at all. (D) -12 likely applies the sign change from the row swap but forgets that multiplying row 2 by $\frac{1}{3}$ reduces the determinant. Key strategy: Memorize the three determinant rules for row operations: scaling a row multiplies the determinant by that scalar, adding multiples of rows doesn't change it, and swapping rows multiplies by $-1$. Apply them sequentially in the order given.
7. If $P$ and $Q$ are elementary matrices such that $PA = QA$ for some invertible matrix $A$, which statement must be true?
   Answer: $PQ^{-1} = I$ and therefore $P = Q$ From $PA = QA$, we get $(P - Q)A = 0$. Since $A$ is invertible, we can multiply on the right by $A^{-1}$ to get $P - Q = 0$, so $P = Q$. Equivalently, $PA = QA$ implies $PQ^{-1}QA = QA$, so $PQ^{-1} = I$, giving $P = Q$. Choice A is wrong because we need $P = Q$, not just equivalence. Choice B states the conclusion but not the reasoning. Choice C incorrectly restricts the type of elementary matrix.
8. Let $E$ be a $3 \times 3$ elementary matrix that corresponds to subtracting 5 times row 1 from row 3 ($R_3 - 5R_1 \to R_3$). What is the entry in the third row, first column of the matrix $E^2$?
   Answer: $-10$ The elementary matrix for the operation $R_3 - 5R_1 \to R_3$ is $E = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -5 & 0 & 1 \end{pmatrix}$. To find $E^2$, we multiply $E$ by itself: $E^2 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -5 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -5 & 0 & 1 \end{pmatrix}$. The entry in the third row, first column of the product is found by taking the dot product of the third row of the first matrix and the first column of the second matrix: $(-5)(1) + (0)(0) + (1)(-5) = -5 - 5 = -10$. Conceptually, applying the operation twice means the total effect is $R_3 - 10R_1 \to R_3$.
9. Let $E$ be the $3 \times 3$ elementary matrix that performs the row operation $R_2 - 3R_3 \to R_2$. If $A$ is any $3 \times n$ matrix, which of the following represents the matrix product $EA$?
   Answer: Matrix $A$ with its second row replaced by the difference of its second row and 3 times its third row. The elementary matrix $E$ corresponding to the row operation $R_i + kR_j \to R_i$ is the identity matrix with an additional entry $k$ at position $(i, j)$. For the operation $R_2 - 3R_3 \to R_2$ on a $3 \times 3$ matrix, $i=2, j=3, k=-3$. The matrix is $E = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & -3 \\ 0 & 0 & 1 \end{pmatrix}$. Left-multiplying a matrix $A$ by $E$ (i.e., forming $EA$) performs this exact row operation on $A$. Therefore, the product $EA$ results in a new matrix where the second row of $A$ has been replaced by the original second row of $A$ minus 3 times the third row of $A$.
10. Let $E_1$ be the elementary matrix for the operation $R_1 \leftrightarrow R_2$, and $E_2$ be the matrix for $R_2 - 4R_1 \to R_2$. If $B = E_2 E_1 A$, which sequence of operations transforms matrix $A$ into matrix $B$?
    Answer: First, swap rows 1 and 2, then subtract 4 times the new row 1 from the new row 2. Matrix multiplication acts from right to left. In the product $E_2 E_1 A$, the matrix $E_1$ acts on $A$ first, and then the matrix $E_2$ acts on the result $(E_1 A)$. Therefore, the first operation is the one corresponding to $E_1$, which is swapping rows 1 and 2. The second operation is the one corresponding to $E_2$, which is subtracting 4 times row 1 from row 2. This second operation is applied to the matrix after the first operation has been completed. Thus, it's 4 times the new row 1 that is subtracted from the new row 2.
11. An invertible matrix $A$ can be row-reduced to the identity matrix $I$ by a sequence of elementary row operations represented by elementary matrices $E_1, E_2, \dots, E_k$. If $E_k \dots E_2 E_1 A = I$, which expression correctly represents $A$?
    Answer: $A = E_1^{-1} E_2^{-1} \dots E_k^{-1}$ To solve for $A$ in the equation $E_k \dots E_2 E_1 A = I$, we must multiply both sides by the inverses of the elementary matrices. To isolate $A$, we must 'undo' the multiplications in the reverse order. We start by left-multiplying by $E_k^{-1}$, then $E_{k-1}^{-1}$, and so on, until we multiply by $E_1^{-1}$. This gives $A = E_1^{-1} E_2^{-1} \dots E_k^{-1} I = E_1^{-1} E_2^{-1} \dots E_k^{-1}$. This relies on the property that for invertible matrices $P, Q$, $(PQ)^{-1} = Q^{-1}P^{-1}$.
12. Which of the following $3 \times 3$ matrices is NOT an elementary matrix?
    Answer: $\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{pmatrix}$ An elementary matrix is obtained by performing a single elementary row operation on an identity matrix. All elementary matrices are invertible. Choice A is obtained from the identity matrix by swapping rows 2 and 3 ($R_2 \leftrightarrow R_3$). Choice B is obtained by the operation $R_3 - 2R_2 \to R_3$. Choice C is obtained by scaling row 2 by 5 ($5R_2 \to R_2$). Choice D has a row of zeros. Its determinant is 0, so it is not invertible. No single elementary row operation performed on an identity matrix can create a row of zeros. Therefore, it is not an elementary matrix.
13. A $3 \times 3$ matrix $A$ with $\det(A) = 4$ is transformed into a matrix $B$ by the following sequence of row operations: $R_1 \leftrightarrow R_3$, then $2R_2 \to R_2$, then $R_3 + 5R_2 \to R_3$. What is the determinant of $B$?
    Answer: $-8$ The determinant of the resulting matrix is the product of the determinant of the original matrix and the determinants of the elementary matrices corresponding to the row operations. 1. $R_1 \leftrightarrow R_3$: The determinant of this elementary matrix is $-1$. 2. $2R_2 \to R_2$: The determinant of this elementary matrix is $2$. 3. $R_3 + 5R_2 \to R_3$: The determinant of a row replacement elementary matrix is always $1$. Therefore, $\det(B) = (\det(E_3))(\det(E_2))(\det(E_1))(\det(A)) = (1)(2)(-1)(4) = -8$.
14. The matrix $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ is transformed into $B = \begin{pmatrix} 1 & 2 \\ 0 & -2 \end{pmatrix}$ by left-multiplication with an elementary matrix $E$. What is the matrix $E$?
    Answer: $\begin{pmatrix} 1 & 0 \\ -3 & 1 \end{pmatrix}$ To transform matrix $A$ into matrix $B$, we need to find the row operation that achieves this. The first row remains unchanged. The second row of $B$ can be obtained from the second row of $A$ by the operation $R_2 - 3R_1 \to R_2$. Specifically, $(3, 4) - 3 \times (1, 2) = (3-3, 4-6) = (0, -2)$. The elementary matrix $E$ for this operation is found by applying the same operation to the $2 \times 2$ identity matrix $I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$. Applying $R_2 - 3R_1 \to R_2$ to $I$ yields $E = \begin{pmatrix} 1 & 0 \\ -3 & 1 \end{pmatrix}$.
15. Let $E$ be an elementary matrix of size $4 \times 4$. If $\det(E) = -1$ and $E^2 = I$, what can be concluded about the row operation represented by $E$?
    Answer: $E$ swaps two distinct rows Elementary matrices have determinant: (1) $\det = 1$ for row addition, (2) $\det = c$ for multiplying a row by $c \neq 0$, (3) $\det = -1$ for row swaps. Since $\det(E) = -1$, $E$ is a row swap matrix. Row swap matrices satisfy $E^2 = I$, confirming this. Choice A would require $\det(E) = -1$ and $E^2 = I$, but multiplying by $-1$ gives $\det = -1$ yet $E^2 \neq I$. Choice C gives $\det = 1$. Choice D ignores the constraints.
16. Let $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ and $E = \begin{pmatrix} 1 & 5 \\ 0 & 1 \end{pmatrix}$. The matrix product $AE$ results in which transformation on matrix $A$?
    Answer: Replacing column 2 with the sum of column 2 and 5 times column 1. When an elementary matrix $E$ multiplies a matrix $A$ on the right (forming $AE$), it performs column operations on $A$. The product is $AE = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 1 & 5 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} a & 5a+b \\ c & 5c+d \end{pmatrix}$. The first column is unchanged. The new second column is the original second column plus 5 times the original first column. This corresponds to the column operation $C_2 + 5C_1 \to C_2$. Left-multiplication ($EA$) results in row operations.
17. Let $A$ be a $3 \times 3$ matrix with $\det(A) = 5$. After applying the sequence of row operations: (i) add $3$ times row $1$ to row $2$, (ii) multiply row $3$ by $-2$, (iii) swap rows $2$ and $3$, the resulting matrix is $B$. If $E$ is the single elementary matrix equivalent to this sequence, what is $\det(E^{-1})$?
    Answer: $\frac{1}{2}$ Let $E_1, E_2, E_3$ be the elementary matrices for operations (i), (ii), (iii) respectively. Then $E = E_3E_2E_1$ and $B = EA$. The determinants are: $\det(E_1) = 1$ (row addition), $\det(E_2) = -2$ (row scaling), $\det(E_3) = -1$ (row swap). So $\det(E) = \det(E_3)\det(E_2)\det(E_1) = (-1)(-2)(1) = 2$. Therefore $\det(E^{-1}) = \frac{1}{\det(E)} = \frac{1}{2}$. Choice B has the wrong sign. Choices C and D give $\det(E)$ instead of $\det(E^{-1})$.
18. A $2 \times 2$ matrix $A$ is transformed by the row operation $R_1 - 2R_2 \to R_1$ to produce the matrix $B = \begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix}$. What is the original matrix $A$?
    Answer: $\begin{pmatrix} 7 & 2 \\ 3 & 1 \end{pmatrix}$ The row operation corresponds to the equation $EA=B$, where $E$ is the elementary matrix for $R_1 - 2R_2 \to R_1$, which is $E = \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix}$. To find $A$, we must compute $A = E^{-1}B$. The inverse operation is $R_1 + 2R_2 \to R_1$, so the inverse matrix is $E^{-1} = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$. Thus, $A = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix} = \begin{pmatrix} 1(1)+2(3) & 1(0)+2(1) \\ 0(1)+1(3) & 0(0)+1(1) \end{pmatrix} = \begin{pmatrix} 7 & 2 \\ 3 & 1 \end{pmatrix}$.

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