LU Decomposition - Linear Algebra

To begin the LU decomposition of matrix $A = \begin{pmatrix} 2 & 3 & 1 \\ -4 & -1 & 2 \\ 6 & 13 & 8 \end{pmatrix}$ , the row operation applied to row 2 to create a zero in the first column is $R_2 \to R_2 + 2R_1$ . What is the entry $l_{21}$ in the resulting unit lower triangular matrix $L$ ?
Answer: $-2$ The entries $l_{ij}$ (for $i>j$ ) in the matrix $L$ are the multipliers used in Gaussian elimination. The elimination step is defined as $R_i \to R_i - l_{ij}R_j$ . The question states the operation is $R_2 \to R_2 + 2R_1$ . Comparing this to the definition, we have $R_2 \to R_2 - l_{21}R_1$ . Thus, $-l_{21} = 2$ , which means $l_{21} = -2$ . Alternatively, the multiplier is always calculated as $l_{21} = a_{21}/a_{11} = -4/2 = -2$ .
The LU decomposition of a matrix $A$ is given by $L = \begin{pmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 3 & -1 & 1 \end{pmatrix}$ and $U = \begin{pmatrix} 2 & 1 & 4 \\ 0 & 1 & 3 \\ 0 & 0 & -1 \end{pmatrix}$ . To solve the system $Ax=b$ for $b = \begin{pmatrix} 2 \\ -1 \\ 7 \end{pmatrix}$ , one first solves $Ly=b$ . What is the intermediate vector $y$ ?
Answer: $\begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix}$ We solve $Ly=b$ using forward substitution. From the first row, $1y_1 = 2 \implies y_1=2$ . From the second row, $-2y_1 + 1y_2 = -1 \implies -2(2) + y_2 = -1 \implies y_2 = 3$ . From the third row, $3y_1 - 1y_2 + 1y_3 = 7 \implies 3(2) - 3 + y_3 = 7 \implies 3 + y_3 = 7 \implies y_3 = 4$ . So, $y = \begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix}$ .
A matrix $A$ is decomposed as $A = LU$ where both $L$ and $U$ are $3 \times 3$ . If we modify the decomposition by multiplying $L$ by a diagonal matrix $D = \text{diag}(2, \frac{1}{2}, 3)$ on the right and $U$ by $D^{-1}$ on the left to get $\tilde{L} = LD$ and $\tilde{U} = D^{-1}U$ , what is the diagonal element $\tilde{u}_{22}$ in terms of the original $u_{22}$ ?
Answer: $\tilde{u}_{22} = 2u_{22}$ Matrix decomposition transformations test your understanding of how scaling operations affect matrix products. When you see a question about modifying an existing decomposition, focus on how the transformations interact at each position. Starting with $A = LU$ , the modified decomposition gives us $\tilde{L} = LD$ and $\tilde{U} = D^{-1}U$ . Let's verify this preserves the original matrix: $\tilde{L}\tilde{U} = (LD)(D^{-1}U) = LDD^{-1}U = LU = A$ ✓ To find $\tilde{u}_{22}$ , we need to understand how $D^{-1}$ affects $U$ . Since $D = \text{diag}(2, \frac{1}{2}, 3)$ , we have $D^{-1} = \text{diag}(\frac{1}{2}, 2, \frac{1}{3})$ . When we multiply $D^{-1}U$ , the diagonal matrix $D^{-1}$ scales each row of $U$ by the corresponding diagonal element. Specifically, row 2 of $U$ gets multiplied by the (2,2) entry of $D^{-1}$ , which is 2. Therefore: $\tilde{u}_{22} = 2u_{22}$ . Option A ( $\frac{u_{22}}{2}$ ) incorrectly uses the original diagonal element from $D$ instead of $D^{-1}$ . Option B ( $u_{22}$ ) assumes the transformation doesn't affect the diagonal, missing that $D^{-1} \neq I$ . Option D ( $\frac{u_{22}}{3}$ ) confuses the (2,2) position with the (3,3) position of $D^{-1}$ . The correct answer is C: $\tilde{u}_{22} = 2u_{22}$ . Strategy tip: In matrix scaling problems, always track which matrix acts on rows versus columns, and remember that $D^{-1}$ inverts each diagonal element individually.
Consider the block matrix $A = \begin{pmatrix} B & C \\ 0 & D \end{pmatrix}$ where $B$ and $D$ are square matrices of sizes $2 \times 2$ and $3 \times 3$ respectively. If $B$ has LU decomposition $B = L_1 U_1$ and $D$ has LU decomposition $D = L_2 U_2$ , what is the structure of the $L$ matrix in the LU decomposition of $A$ ?
Answer: $L = \begin{pmatrix} L_1 & 0 \\ 0 & L_2 \end{pmatrix}$ For a block upper triangular matrix like $A$ , the LU decomposition preserves the block structure. Since the lower-left block is already zero, no elimination is needed there, and $L$ becomes block diagonal: $L = \begin{pmatrix} L_1 & 0 \\ 0 & L_2 \end{pmatrix}$ and $U = \begin{pmatrix} U_1 & L_1^{-1}C \\ 0 & U_2 \end{pmatrix}$ . Choice B incorrectly places the off-diagonal term in $L$ . Choice C violates the structure of $L$ . Choice D would apply if $A$ were block lower triangular instead.
A student performs LU decomposition on a $4 \times 4$ matrix $A$ and obtains $L$ and $U$ . Later, they discover that $A$ was actually the transpose of the intended matrix. If the intended matrix $A^T$ has LU decomposition $A^T = L'U'$ , what is the relationship between the decompositions?
Answer: $(L')^T = U$ and $(U')^T = L$ If $A = LU$ , then $A^T = (LU)^T = U^T L^T$ . Since $A^T = L'U'$ where $L'$ is lower triangular with unit diagonal and $U'$ is upper triangular, we need $L'U' = U^T L^T$ . Since $U^T$ is lower triangular and $L^T$ is upper triangular with unit diagonal, we have $L' = U^T$ and $U' = L^T$ , which means $(L')^T = U$ and $(U')^T = L$ . Choices B and C represent common misconceptions about matrix transposes and decomposition properties.
If matrix $A$ has LU decomposition $A = LU$ where $L$ has unit diagonal, and we know that $\det(A) = -12$ and $u_{11} = 2$ , $u_{22} = -3$ , what is $u_{33}$ ?
Answer: $2$ Since $\det(A) = \det(L) \cdot \det(U)$ and $L$ has unit diagonal elements, $\det(L) = 1$ . Therefore, $\det(A) = \det(U) = u_{11} \cdot u_{22} \cdot u_{33}$ . Substituting known values: $-12 = 2 \cdot (-3) \cdot u_{33} = -6u_{33}$ . Solving: $u_{33} = \frac{-12}{-6} = 2$ . Choice A would result from forgetting the negative sign in $u_{22}$ . Choices C and D would result from errors in the determinant relationship or arithmetic mistakes.
For which of the following matrices does a unique LU decomposition (where $L$ is unit lower triangular) not exist without row interchanges?
Answer: $\begin{pmatrix} 0 & 1 \\ 2 & 3 \end{pmatrix}$ An LU decomposition without pivoting exists if and only if all leading principal minors of the matrix are non-zero. For matrix (B), the first leading principal minor is the top-left element, which is $0$ . Since this minor is zero, the LU decomposition cannot be formed without a row interchange. The other matrices have all non-zero leading principal minors.
Let $A$ be an invertible $n \times n$ matrix with a unique LU decomposition $A=LU$ , where $L$ is a unit lower triangular matrix and $U$ is an upper triangular matrix. Which of the following statements is necessarily false?
Answer: The matrix $U$ can be singular. The determinant of a product of matrices is the product of their determinants: $\det(A) = \det(L)\det(U)$ . Since $L$ is unit lower triangular, its diagonal entries are all $1$ , so $\det(L) = 1$ . This means $\det(A) = \det(U)$ . The problem states that $A$ is invertible, so $\det(A) \neq 0$ . Consequently, $\det(U)$ must also be non-zero, which means $U$ must be invertible (non-singular). Therefore, the statement that $U$ can be singular is false.
A matrix $A$ is factored into a unit lower triangular matrix $L$ and an upper triangular matrix $U$ , where $L = \begin{pmatrix} 1 & 0 & 0 \\ -3 & 1 & 0 \\ 2 & 4 & 1 \end{pmatrix} \quad \text{and} \quad U = \begin{pmatrix} 5 & 1 & -2 \\ 0 & -1 & 3 \\ 0 & 0 & 2 \end{pmatrix}$ What is the entry $a_{32}$ of the original matrix $A$ ?
Answer: $-2$ Since $A=LU$ , the entry $a_{ij}$ is the dot product of the $i$ -th row of $L$ and the $j$ -th column of $U$ . To find $a_{32}$ , we take the dot product of the third row of $L$ and the second column of $U$ . \\ Third row of $L$ : $\begin{pmatrix} 2 & 4 & 1 \end{pmatrix}$ . \\ Second column of $U$ : $\begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}$ . \\ The dot product is $a_{32} = (2)(1) + (4)(-1) + (1)(0) = 2 - 4 + 0 = -2$ .
The LU decomposition for a matrix $A$ is $L = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ -2 & 0 & 1 \end{pmatrix}$ and $U = \begin{pmatrix} 2 & -1 & 1 \\ 0 & 3 & -2 \\ 0 & 0 & 4 \end{pmatrix}$ . For $b = \begin{pmatrix} 2 \\ 5 \\ -2 \end{pmatrix}$ , what is the value of $x_2$ in the solution to $Ax=b$ ?
Answer: $4/3$ First, solve $Ly=b$ . $y_1 = 2$ . $y_1 + y_2 = 5 \implies 2 + y_2 = 5 \implies y_2 = 3$ . $-2y_1 + y_3 = -2 \implies -4 + y_3 = -2 \implies y_3=2$ . So $y = \begin{pmatrix} 2 \\ 3 \\ 2 \end{pmatrix}$ . Next, solve $Ux=y$ . We only need $x_2$ . We find $x_3$ first from the last row: $4x_3 = 2 \implies x_3 = 1/2$ . Then use the second row: $3x_2 - 2x_3 = y_2 \implies 3x_2 - 2(1/2) = 3 \implies 3x_2 - 1 = 3 \implies 3x_2 = 4 \implies x_2 = 4/3$ .
A $3 \times 3$ matrix $A$ has the property that during LU decomposition without pivoting, the second leading principal minor equals zero while the first leading principal minor is nonzero. Which statement about the LU decomposition is correct?
Answer: The decomposition fails because we encounter division by zero during elimination When the second leading principal minor is zero but the first is nonzero, we have $a_{11} \neq 0$ but $\det\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} = 0$ . During elimination, after the first step, the pivot position $(2,2)$ becomes zero, causing division by zero when trying to eliminate below the second pivot. Choice A is wrong because $L$ has unit diagonal. Choice C describes what we'd need to do to fix it, not what happens without pivoting. Choice D is wrong because we can't even reach the point of having $U$ fully formed.
A research team needs to solve the matrix equation $Ax=b$ for a large, fixed $n \times n$ matrix $A$ and over one thousand different right-hand side vectors $b_1, b_2, \ldots, b_{1000}$ . Which statement best explains the computational advantage of using an LU decomposition of $A$ ?
Answer: The LU decomposition needs to be computed only once, and then solving for each vector $b_k$ requires only computationally inexpensive substitutions. The main advantage of LU decomposition is that the computationally expensive part (Gaussian elimination on $A$ , with complexity $O(n^3)$ ) is done only once to find $L$ and $U$ . Solving the system for any new vector $b$ then involves a forward substitution ( $Ly=b$ ) and a backward substitution ( $Ux=y$ ), both of which are much faster (complexity $O(n^2)$ ). This is far more efficient than re-running elimination for each of the thousand vectors or computing the matrix inverse.
Given the matrix $A = \begin{pmatrix} 2 & 1 & -1 \\ 4 & 5 & -1 \\ -2 & 5 & 2 \end{pmatrix}$ . In the LU decomposition $A=LU$ , where $L$ is a unit lower triangular matrix, what is the value of the entry $l_{31}$ ?
Answer: $-1$ To find the entries of $L$ , we perform Gaussian elimination on $A$ to get $U$ . The multipliers used become the entries of $L$ . The entry $l_{31}$ is the multiplier used to create a zero in the third row, first column. The operation is $R_3 \to R_3 - m R_1$ , where $m = a_{31}/a_{11} = -2/2 = -1$ . Thus, $l_{31} = -1$ . The operation would be $R_3 \to R_3 - (-1)R_1 = R_3 + R_1$ .
The LU decomposition of a matrix $A$ results in the upper triangular matrix $U = \begin{pmatrix} 2 & -1 & 5 \\ 0 & -3 & 6 \\ 0 & 0 & 4 \end{pmatrix}$ . What is the determinant of matrix $A$ ?
Answer: $-24$ For an LU decomposition $A=LU$ , the determinant is $\det(A) = \det(L)\det(U)$ . By convention, $L$ is a unit lower triangular matrix, meaning its diagonal entries are all 1. The determinant of any triangular matrix is the product of its diagonal entries, so $\det(L)=1$ . Therefore, $\det(A) = \det(U)$ . For the given matrix $U$ , the determinant is the product of its diagonal entries: $\det(U) = (2)(-3)(4) = -24$ . Thus, $\det(A) = -24$ .
In the LU decomposition $A = LU$ of a $3 \times 3$ matrix, suppose $l_{21} = \frac{1}{3}$ , $l_{31} = \frac{2}{3}$ , and $l_{32} = \frac{3}{4}$ . If the first pivot is $u_{11} = 6$ and the original matrix element $a_{31} = 4$ , what was the value of $a_{21}$ ?
Answer: $2$ LU decomposition questions test your understanding of how Gaussian elimination relates to matrix factorization. When you see given values for the L and U matrices, think about the fundamental relationship: each element in the original matrix A can be expressed through the row-column multiplication of L and U. In LU decomposition, the relationship $A = LU$ means that $a_{ij} = \sum_{k=1}^{\min(i,j)} l_{ik} \cdot u_{kj}$ . For element $a_{21}$ , this becomes $a_{21} = l_{21} \cdot u_{11}$ since we only need the first term in the sum. Given that $l_{21} = \frac{1}{3}$ and $u_{11} = 6$ , we can calculate: $a_{21} = l_{21} \cdot u_{11} = \frac{1}{3} \cdot 6 = 2$ We can verify this approach using the given information about $a_{31} = 4$ . Since $a_{31} = l_{31} \cdot u_{11} = \frac{2}{3} \cdot 6 = 4$ , our method is consistent. Choice A ( $1$ ) would result from incorrectly using $l_{21} = \frac{1}{6}$ instead of $\frac{1}{3}$ . Choice B ( $3$ ) represents the denominator from $l_{21}$ , showing confusion about which values to multiply. Choice D ( $4$ ) is the value of $a_{31}$ , indicating you mixed up which matrix element was being asked for. Study tip: In LU decomposition problems, always write out the matrix multiplication formula for the specific element you need. The pattern $a_{ij} = \sum l_{ik} \cdot u_{kj}$ is your reliable roadmap to avoid mixing up given values.
Consider the matrix $A = \begin{pmatrix} 0 & 2 & 1 \\ 1 & 1 & 3 \\ -2 & 4 & 1 \end{pmatrix}$ . A row interchange is necessary to proceed with LU decomposition. Let $P$ be the permutation matrix that swaps rows 1 and 2. For the factorization $PA=LU$ , what is the resulting upper triangular matrix $U$ ?
Answer: $\begin{pmatrix} 1 & 1 & 3 \\ 0 & 2 & 1 \\ 0 & 0 & 4 \end{pmatrix}$ First, apply the permutation: $PA = \begin{pmatrix} 0&1&0\\1&0&0\\0&0&1 \end{pmatrix} \begin{pmatrix} 0&2&1\\1&1&3\\-2&4&1 \end{pmatrix} = \begin{pmatrix} 1&1&3\\0&2&1\\-2&4&1 \end{pmatrix}$ . Let this be $A'$ . Now, perform elimination on $A'$ . For the first column, $R_3 \to R_3 - (-2)R_1 = R_3 + 2R_1$ , giving the new matrix $\begin{pmatrix} 1&1&3\\0&2&1\\0&6&7 \end{pmatrix}$ . For the second column, $R_3 \to R_3 - (3)R_2$ , giving the new third row $(0,6,7) - 3(0,2,1) = (0,0,4)$ . The final matrix $U$ is $\begin{pmatrix} 1&1&3\\0&2&1\\0&0&4 \end{pmatrix}$ .
Consider the matrix $A = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 7 \\ 1 & 3 & 6 \end{pmatrix}$ . During LU decomposition, what happens when we attempt to eliminate the entry in position $(2,1)$ ?
Answer: The elimination creates a zero in position $(2,2)$ , preventing further progress without pivoting When you encounter LU decomposition problems, focus on what happens at each elimination step and how it affects the matrix structure. Let's trace through the elimination process for this matrix. Starting with $A = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 7 \\ 1 & 3 & 6 \end{pmatrix}$ , the first pivot is $a_{11} = 1$ . To eliminate the $(2,1)$ entry, we calculate the multiplier $l_{21} = \frac{a_{21}}{a_{11}} = \frac{2}{1} = 2$ . We then subtract 2 times row 1 from row 2: Row 2 becomes: $(2, 4, 7) - 2(1, 2, 3) = (0, 0, 1)$ Similarly, for the $(3,1)$ entry: $l_{31} = 1$ , and row 3 becomes: $(1, 3, 6) - 1(1, 2, 3) = (0, 1, 3)$ After the first elimination step, we have: $\begin{pmatrix} 1 & 2 & 3 \\ 0 & 0 & 1 \\ 0 & 1 & 3 \end{pmatrix}$ The critical issue is that position $(2,2)$ is now zero, which means we cannot use it as the next pivot. This makes answer D correct. Let's examine why the other answers miss the mark: A incorrectly suggests normal elimination when the zero pivot prevents continuation. B is wrong because the first pivot (1) is perfectly acceptable for stability. C misidentifies when the singularity issue occurs—it's the zero pivot that's problematic, not the final $U$ matrix. Remember: In LU decomposition, always check if your next pivot is zero. When it is, you need row pivoting to continue the decomposition successfully.
Let $A = \begin{pmatrix} 1 & 2 & 3 \\ -2 & -3 & -4 \\ 3 & 7 & 10 \end{pmatrix}$ . Find the entry $u_{22}$ of the upper triangular matrix $U$ in the LU decomposition of $A$ .
Answer: $1$ To find $U$ , we use Gaussian elimination. The first step is to eliminate the entry $a_{21}$ . The multiplier is $l_{21} = -2/1 = -2$ . The row operation is $R_2 \to R_2 - (-2)R_1 = R_2 + 2R_1$ . The new second row is $(-2, -3, -4) + 2(1, 2, 3) = (-2+2, -3+4, -4+6) = (0, 1, 2)$ . The matrix becomes $\begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ ... & ... & ... \end{pmatrix}$ . The entry $u_{22}$ is the new element in the (2,2) position, which is $1$ .
In solving a linear system $Ax=b$ , the problem is factored as $A=LU$ and then solved as $Ly=b$ followed by $Ux=y$ . Given $U = \begin{pmatrix} 3 & -1 & 2 \\ 0 & 2 & 1 \\ 0 & 0 & -2 \end{pmatrix}$ and the intermediate solution $y = \begin{pmatrix} 8 \\ 4 \\ -6 \end{pmatrix}$ , what is the final solution vector $x$ ?
Answer: $\begin{pmatrix} 5/6 \\ 1/2 \\ 3 \end{pmatrix}$ We solve $Ux=y$ using backward substitution. From the last row, $-2x_3 = -6 \implies x_3 = 3$ . From the second row, $2x_2 + x_3 = 4 \implies 2x_2 + 3 = 4 \implies 2x_2 = 1 \implies x_2 = 1/2$ . From the first row, $3x_1 - x_2 + 2x_3 = 8 \implies 3x_1 - 1/2 + 2(3) = 8 \implies 3x_1 + 5.5 = 8 \implies 3x_1 = 2.5 \implies x_1 = 5/6$ . The solution is $x = \begin{pmatrix} 5/6 \\ 1/2 \\ 3 \end{pmatrix}$ .

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