Linear Algebra Quiz: Lu Decomposition

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Lu DecompositionQuestion 1 of 19

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 $x$ ?

\begin{pmatrix} 4/3 \\ 2 \\ 3 \end{pmatrix}

\begin{pmatrix} 35/6 \\ 7/2 \\ -3 \end{pmatrix}

\begin{pmatrix} 8/3 \\ -2/3 \\ 3 \end{pmatrix}

\begin{pmatrix} 5/6 \\ 1/2 \\ 3 \end{pmatrix}

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Linear Algebra Quiz

Linear Algebra Quiz: Lu Decomposition

Practice Lu Decomposition 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 Lu Decomposition, giving you a quick way to practice the rules, question types, and explanations that matter most for Linear Algebra.

How to use this quiz

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.

All questions

Question 1

$\begin{pmatrix} 4/3 \\ 2 \\ 3 \end{pmatrix}$
$\begin{pmatrix} 35/6 \\ 7/2 \\ -3 \end{pmatrix}$
$\begin{pmatrix} 8/3 \\ -2/3 \\ 3 \end{pmatrix}$
$\begin{pmatrix} 5/6 \\ 1/2 \\ 3 \end{pmatrix}$ (correct answer)

Explanation: 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}

Question 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 $Ax=b$ ?

$3$
$4/3$ (correct answer)
$7/3$
$2/3$

Explanation: 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

Question 3

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 $y$ ?

$\begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix}$ (correct answer)
$\begin{pmatrix} 2 \\ -5 \\ -4 \end{pmatrix}$
$\begin{pmatrix} 5 \\ 20 \\ -7 \end{pmatrix}$
$\begin{pmatrix} 2 \\ -1 \\ 7 \end{pmatrix}$

Explanation: 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}

Question 4

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?

The matrix $L$ is always invertible.
The matrix $U$ can be singular. (correct answer)
The determinant of $L$ is always $1$ .
The determinant of $A$ is equal to the product of the diagonal entries of $U$ .

Explanation: 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.

Question 5

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 $L$ ?

$2$
$-1/2$
$-4$
$-2$ (correct answer)

Explanation: 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

Question 6

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 $l_{31}$ ?

$1$
$-1$ (correct answer)
$2$
$-2$

Explanation: 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

Question 7

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 $A$ .

$-7$
$-3$
$1$ (correct answer)
$5$

Explanation: 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

Question 8

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 $U$ ?

$\begin{pmatrix} 1 & 1 & 3 \\ 0 & 2 & 1 \\ 0 & 0 & 4 \end{pmatrix}$ (correct answer)
$\begin{pmatrix} 1 & 1 & 3 \\ 0 & 2 & 1 \\ 0 & 6 & 7 \end{pmatrix}$
$\begin{pmatrix} 1 & 1 & 3 \\ 0 & 2 & 1 \\ -2 & 4 & 1 \end{pmatrix}$
$\begin{pmatrix} 1 & 1 & 3 \\ 0 & 2 & 1 \\ 0 & 0 & 10 \end{pmatrix}$

Explanation: 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

\begin{pmatrix} 1&1&3\\0&2&1\\0&0&4 \end{pmatrix}

Question 9

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 $A$ ?

$9$
$-1$
$-2$ (correct answer)
$4$

Explanation: 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

Question 10

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 $A$ ?

Calculating the inverse $A^{-1}$ is faster than finding the LU decomposition, so $x=A^{-1}b$ should be used for each vector.
The LU decomposition needs to be computed only once, and then solving for each vector $b_k$ requires only computationally inexpensive substitutions. (correct answer)
For each new vector $b_k$ , the entire Gaussian elimination process for the augmented matrix $[A|b_k]$ must be repeated from scratch.
The LU decomposition directly provides the solution vector $x$ without the need for forward or backward substitution.

Explanation: 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.

Question 11

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 $u_{33}$ ?

$-2$
$2$ (correct answer)
$4$
$-4$

Explanation: 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.

Question 12

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?

The decomposition exists and is unique, but $L$ contains a zero diagonal element
The decomposition fails because we encounter division by zero during elimination (correct answer)
The decomposition exists but requires partial pivoting to complete successfully
The decomposition produces a singular matrix $U$ with two zero diagonal entries

Explanation: 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.

Question 13

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?

$(L')^T = U$ and $(U')^T = L$ (correct answer)
$(L')^T = L$ and $(U')^T = U$
$L' = U^T$ and $U' = L^T$
No simple relationship exists between the decompositions

Explanation: 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.

Question 14

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 $a_{21}$ ?

$1$
$3$
$2$ (correct answer)
$4$

Explanation: 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.

Question 15

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 $u_{22}$ ?

$\tilde{u}_{22} = \frac{u_{22}}{2}$
$\tilde{u}_{22} = u_{22}$
$\tilde{u}_{22} = 2u_{22}$ (correct answer)
$\tilde{u}_{22} = \frac{u_{22}}{3}$

Explanation: 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.

Question 16

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 $A$ ?

(correct answer)

Explanation: 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.

Question 17

For which of the following matrices does a unique LU decomposition (where $L$ is unit lower triangular) not exist without row interchanges?

$\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$
$\begin{pmatrix} 0 & 1 \\ 2 & 3 \end{pmatrix}$ (correct answer)
$\begin{pmatrix} 2 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 2 \end{pmatrix}$
$\begin{pmatrix} 3 & 0 & 0 \\ 1 & 3 & 0 \\ 2 & 1 & 3 \end{pmatrix}$

Explanation: 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.

Question 18

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 $A$ ?

$3$
$24$
$-23$
$-24$ (correct answer)

Explanation: 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

Question 19

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 $(2,1)$ ?

The elimination proceeds normally, and the multiplier $l_{21} = 2$ is stored in $L$
The elimination fails immediately because the first pivot is too small for numerical stability
The elimination proceeds, but the resulting $U$ matrix becomes singular in the second step
The elimination creates a zero in position $(2,2)$ , preventing further progress without pivoting (correct answer)

Explanation: 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.