Skip to content
NumeriveMATH
Make learning comfortable
Changing style, motion, or reading speed keeps your current question, answers, homework, and saved progress. The school band still chooses your task layout. Saved to your learner profile across devices. All lessons and help stay available.
← Linear Algebra levels

Linear Algebra / LEVEL 1 · DIFFICULTY 4/5

Linear Algebra

Read the prerequisite, study a worked example and justify each step in these selected university topics.

3 stages · 24 practice problems · two 6-question assessment forms

STAGE ULA 1.1 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Matrices, Elimination & Inverses

Useful preparation: Two Equations, Two Unknowns

Open this topic in the university library →

For a long equation, swipe sideways inside its formula box to see every term.

Meaning & notation

A matrix records a linear rule as well as a collection of equations. Each column tells where one coordinate direction goes; multiplying by an input vector combines those columns. Elimination changes the way the equations are written while preserving exactly which inputs produce the requested output. This makes pivots evidence about solvability, rather than merely convenient arithmetic.

Matrix and linear system
An m by n matrix A has m rows and n columns. In Ax=b, the unknown vector x has n entries and the output b has m entries.
Pivot
A pivot is a leading nonzero entry used to eliminate entries in its column. In reduced row echelon form, each pivot is 1 and is the only nonzero entry in its column.
Inverse
For a square matrix A, an inverse is a matrix B satisfying AB=BA=I\displaystyle AB = BA = I. It reverses the transformation for every possible input.

Conditions before calculation

  • Row swaps, multiplication of a row by a nonzero scalar, and adding a multiple of one row to another are reversible; apply them to the entire augmented row.
  • A square n by n matrix is invertible exactly when it has n pivots, equivalently a nonzero determinant. A rectangular matrix has no two-sided inverse.
  • A zero coefficient row with a nonzero right side makes a system inconsistent. A free variable creates infinitely many solutions only when the system is consistent.

Why the method works

1. Preserve the equations

Reversibility explains why row operations preserve the solution set. Scaling by zero is excluded because it can erase a restriction.

2. Read the result

First check for a contradictory row. Then identify pivot variables and free variables; back substitution expresses pivots in terms of any free choices.

3. Verify the input

Substitute the candidate into the original equations. A correct solution must match every component of b, including equations that became redundant during elimination.

Ax=bAx=b

Worked examples

EXAMPLE 1

Solve x+2 y=5\displaystyle x + 2\, y = 5 and 3 x+5 y=13\displaystyle 3\, x + 5\, y = 13.

  1. Write coefficients and constants in matching row order.

    [1253513]\left[\begin{array}{cc|c}1&2&5\\3&5&13\end{array}\right]

  2. Subtract three times row one from row two; the x coefficient cancels.

    R2←R2−3R1:−y=−2R_2\leftarrow R_2-3R_1:\quad -y=-2

  3. Thus y=2\displaystyle y = 2 and the first equation gives x=1\displaystyle x = 1. Substitution checks both equations.

    1+2(2)=5,3(1)+5(2)=131+2(2)=5,\qquad 3(1)+5(2)=13

Conclusion: The unique solution is (1,2); two pivots leave no free variable.

EXAMPLE 2

Classify x+2 y=3\displaystyle x + 2\, y = 3 and 2 x+4 y=6\displaystyle 2\, x + 4\, y = 6, then change the last constant to 7.

  1. Eliminate the second x coefficient; with constant 6, the entire second row becomes zero.

  2. Choose y=t\displaystyle y = t freely and recover x from the first row.

    (x,y)=(3−2t,t),t∈R(x,y)=(3-2t,t),\quad t\in\mathbb R

  3. With constant 7, the same operation produces a contradiction.

    0x+0y=10x+0y=1

Conclusion: The first system has infinitely many solutions; the altered system has none. The shared coefficient matrix is singular.

Pitfalls & counterexamples

Tempting claim: A zero determinant means the system has no solution.

Correction: It means the coefficient matrix is not invertible; the right side determines whether solutions are absent or nonunique.

Test the claim: The zero matrix with zero right side accepts every input vector.

Check your understanding

Can swapping two equations change the solution set?

Write a reason before opening the explanation. This reflection is not graded or saved.

Compare your reasoning

No.

Both restrictions are still imposed, and swapping them back reverses the operation.

Connect this idea

Elimination reveals rank and nullity. In computation, solving by pivoted elimination is generally preferable to explicitly forming an inverse; small pivots also signal the need to consider numerical accuracy.

Practise this topic →Review concept flashcards

Reference reading

Use these texts to study the underlying definitions and theorems in greater depth. The explanations and examples above are written for Numerive.

Preview the eight practice prompts
  1. Problem 1

    Solve the system x+y=7\displaystyle x + y = 7, x+2 y=12\displaystyle x + 2\, y = 12. Find y.

    Practise this exact problem →
  2. Problem 2

    Find the determinant [[2,1],[2,3]]. Find its determinant.

    Practise this exact problem →
  3. Problem 3

    For A=[[2,0],[0,2]]\displaystyle A = \left[\left[2, 0\right], \left[0, 2\right]\right], what is the top-left entry of A inverse?

    Practise this exact problem →
  4. Problem 4

    Multiply A=[[1,2],[0,1]]\displaystyle A = \left[\left[1, 2\right], \left[0, 1\right]\right] and B=[[2,0],[3,1]]\displaystyle B = \left[\left[2, 0\right], \left[3, 1\right]\right]. What is the top-left entry of AB?

    Practise this exact problem →
  5. Problem 5New context

    Two measured totals satisfy x+y=9\displaystyle x + y = 9, x+2 y=15\displaystyle x + 2\, y = 15. Find y.

    Practise this exact problem →
  6. Problem 6New context

    A transformation has matrix [[3,1],[2,3]]. Find its determinant.

    Practise this exact problem →
  7. Problem 7New context

    A coordinate transform uses A=[[3,0],[0,2]]\displaystyle A = \left[\left[3, 0\right], \left[0, 2\right]\right], what is the top-left entry of A inverse?

    Practise this exact problem →
  8. Problem 8New context

    A two-stage transformation uses A=[[1,3],[0,1]]\displaystyle A = \left[\left[1, 3\right], \left[0, 1\right]\right] and B=[[2,0],[3,1]]\displaystyle B = \left[\left[2, 0\right], \left[3, 1\right]\right]. What is the top-left entry of AB?

    Practise this exact problem →
Open stage ULA 1.1 in the student workspace →

STAGE ULA 1.2 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Span, Basis, Rank & Nullity

Useful preparation: Matrices, Elimination & Inverses

Open this topic in the university library →

For a long equation, swipe sideways inside its formula box to see every term.

Meaning & notation

A span describes all outputs that a collection of directions can build. A basis removes redundant directions while retaining every possible output. Rank counts directions that survive a linear transformation, whereas nullity counts independent input directions that become invisible. These are dimensions of spaces, not simply counts of nonzero rows in the original display.

Subspace and span
A subspace contains zero and is closed under addition and scalar multiplication. The span of vectors is the set of all their linear combinations.
Basis and dimension
A basis is a linearly independent spanning list. Independence means that a zero linear combination forces every coefficient to be zero. Dimension is the number of vectors in a basis.
Rank and nullity
For A with n columns, rank is the dimension of its column space; nullity is the dimension of its nullspace, the set of x satisfying Ax=0.

Conditions before calculation

  • Use one fixed scalar field, here the real numbers. For a nonempty subset of a vector space, closure under all linear combinations proves it is a subspace.
  • Rank plus nullity equals the number of input coordinates n, not necessarily the number of output coordinates m.
  • Row operations preserve the nullspace and dependencies among columns. To obtain a basis for the original column space, select pivot columns from the original matrix.

Why the method works

1. Separate generation from uniqueness

Spanning asks whether every target can be generated. Independence asks whether its coordinates in the proposed basis are unique; a nonzero dependency allows different coordinate lists for the same vector.

2. Count inputs systematically

In echelon form, each pivot determines one variable and each free variable contributes one nullspace direction.

rank⁡(A)+nullity⁡(A)=n\operatorname{rank}(A)+\operatorname{nullity}(A)=n

3. Describe every solution

If one particular solution exists, adding a nullspace vector changes the input without changing the output. Conversely, subtracting any two solutions gives a nullspace vector.

{x:Ax=b}=xp+ker⁡A\{x:Ax=b\}=x_p+\ker A

Worked examples

EXAMPLE 1

Find rank and nullity for A with rows (1,2,0) and (0,0,1).

  1. Columns one and three are independent; column two is twice column one. Thus the image has dimension two.

  2. Solve the homogeneous equations and choose the free variable x 2=t\displaystyle x\, 2 = t.

    x1+2x2=0,x3=0x_1+2x_2=0,\quad x_3=0

  3. Write all solutions as multiples of one nonzero vector.

    x=t(−2,1,0),ker⁡A=span⁡{(−2,1,0)}x=t(-2,1,0),\quad \ker A=\operatorname{span}\{(-2,1,0)\}

Conclusion: Rank is 2, nullity is 1, and their sum is the three input coordinates.

EXAMPLE 2

Find a basis for the plane x+y+z=0\displaystyle x + y + z = 0 in real three-dimensional space.

  1. Set y=s\displaystyle y = s and z=t\displaystyle z = t because the equation then determines x.

  2. Separate the two independent choices.

    (x,y,z)=s(−1,1,0)+t(−1,0,1)(x,y,z)=s(-1,1,0)+t(-1,0,1)

  3. If this combination is zero, its second and third coordinates force s=t=0\displaystyle s = t = 0. Both vectors also satisfy the plane equation.

Conclusion: The displayed vectors form a basis; the plane has dimension two.

Pitfalls & counterexamples

Tempting claim: Any plane in three-dimensional space is a vector subspace.

Correction: A subspace must include the origin and remain closed under linear combinations.

Test the claim: The plane x+y+z=1\displaystyle x + y + z = 1 omits the zero vector and is not a subspace.

Check your understanding

Can four vectors be linearly independent in real three-dimensional space?

Write a reason before opening the explanation. This reflection is not graded or saved.

Compare your reasoning

No.

A three-row matrix has at most three pivots, so four columns leave a free variable in the homogeneous system.

Connect this idea

Basis coordinates make abstract vector spaces computable. The same ideas apply to polynomials, functions, and solution spaces of homogeneous linear differential equations.

Practise this topic →Review concept flashcards

Reference reading

Use these texts to study the underlying definitions and theorems in greater depth. The explanations and examples above are written for Numerive.

Preview the eight practice prompts
  1. Problem 1

    A matrix has 5 columns and rank 2. Find its nullity.

    Practise this exact problem →
  2. Problem 2

    In R2\displaystyle R^{2} are (1,2) and (2,4). Find the dimension of their span.

    Practise this exact problem →
  3. Problem 3

    A map T (x,y)=(x+1,y)\displaystyle T\, \left(x, y\right) = \left(x + 1, y\right) is proposed to be linear. Is it linear? Enter 1 for yes, 0 for no.

    Practise this exact problem →
  4. Problem 4

    For a linear map T (1,0)=(2,1)\displaystyle T\, \left(1, 0\right) = \left(2, 1\right) and T (0,1)=(2,3)\displaystyle T\, \left(0, 1\right) = \left(2, 3\right). Find the first coordinate of T(2,3).

    Practise this exact problem →
  5. Problem 5New context

    A linear model matrix has 6 columns and rank 3. Find its nullity.

    Practise this exact problem →
  6. Problem 6New context

    Two proposed basis vectors in R2\displaystyle R^{2} are (1,3) and (2,6). Find the dimension of their span.

    Practise this exact problem →
  7. Problem 7New context

    A coordinate map T (x,y)=(x+2,y)\displaystyle T\, \left(x, y\right) = \left(x + 2, y\right) is proposed to be linear. Is it linear? Enter 1 for yes, 0 for no.

    Practise this exact problem →
  8. Problem 8New context

    A linear transformation satisfies T (1,0)=(3,1)\displaystyle T\, \left(1, 0\right) = \left(3, 1\right) and T (0,1)=(2,3)\displaystyle T\, \left(0, 1\right) = \left(2, 3\right). Find the first coordinate of T(2,3).

    Practise this exact problem →
Open stage ULA 1.2 in the student workspace →

STAGE ULA 1.3 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Eigenvalues & Orthogonal Projection

Useful preparation: Span, Basis, Rank & Nullity

Open this topic in the university library →

For a long equation, swipe sideways inside its formula box to see every term.

Meaning & notation

Most vectors change direction under a matrix. Eigenvectors identify exceptional directions on which the whole transformation acts like multiplication by one number. Projection is a complementary idea: retain the component inside a chosen subspace and discard the perpendicular component. A projection therefore acts with eigenvalue one on its target and zero on the perpendicular directions.

Eigenpair
For a square matrix A, a nonzero vector v and scalar lambda form an eigenpair when Av=lambda v. The scalar may be zero even though v cannot be zero.
Diagonalization
A diagonalization factors A as P times D times the inverse of P, using a basis of eigenvectors in the columns of P and matching eigenvalues in the diagonal matrix D.
Orthogonal projection
The orthogonal projection of b onto a subspace is the vector p in that subspace for which b−p\displaystyle b - p is perpendicular to every vector in the subspace.

Conditions before calculation

  • Eigenvalues require a square matrix. Real matrices can have nonreal eigenvalues, so specify whether calculations are over the real or complex numbers.
  • An n by n matrix is diagonalizable exactly when it has n independent eigenvectors. Distinct eigenvalues guarantee independence; repeated eigenvalues alone do not settle the question.
  • For least squares, the normal equations always hold at a minimizer. The inverse of A transpose A exists only when A has full column rank; real symmetric matrices admit an orthonormal eigenbasis.

Why the method works

1. Find preserved directions

A nonzero solution of the homogeneous eigenvector equation exists exactly when its coefficient matrix is singular.

(A−λI)v=0,det⁡(A−λI)=0(A-\lambda I)v=0,\qquad \det(A-\lambda I)=0

2. Check multiplicities

For each eigenvalue, solve its nullspace problem. Count independent eigenvectors, not just roots of the characteristic polynomial, before using a diagonalization.

3. Minimize perpendicular error

Moving within the column space cannot shorten a perpendicular residual. Algebraically, its dot product with each column vanishes, giving the normal equations.

AT(b−Ax^)=0A^{\mathsf T}(b-A\widehat x)=0

Worked examples

EXAMPLE 1

Diagonalize the matrix with rows (3,1) and (0,2).

  1. The triangular determinant gives two distinct roots.

    det⁡(A−λI)=(3−λ)(2−λ)\det(A-\lambda I)=(3-\lambda)(2-\lambda)

  2. Solving each nullspace gives directions (1,0) and (-1,1). Direct multiplication verifies eigenvalues 3 and 2.

  3. Use those independent vectors as columns; their determinant is 1.

    P=(1−101),D=(3002),A=PDP−1P=\begin{pmatrix}1&-1\\0&1\end{pmatrix},\quad D=\begin{pmatrix}3&0\\0&2\end{pmatrix},\quad A=PDP^{-1}

Conclusion: The two eigenvectors form a basis, so powers of A reduce to powers of 3 and 2.

EXAMPLE 2

Project b=(3,1)\displaystyle b = \left(3, 1\right) onto the line spanned by a=(1,1)\displaystyle a = \left(1, 1\right).

  1. Every candidate on the line has form ca. Require its residual to be perpendicular to a.

    aT(b−ca)=0a^{\mathsf T}(b-ca)=0

  2. Solve the scalar equation using the squared length of a.

    c=aTbaTa=42=2c=\frac{a^{\mathsf T}b}{a^{\mathsf T}a}=\frac42=2

  3. The projection is (2,2), with residual (1,-1). Their direction test succeeds.

    (1,1)⋅(1,−1)=0(1,1)\mathbin{\cdot}(1,-1)=0

Conclusion: The closest point is (2,2); the least-squares error has squared length 2.

Pitfalls & counterexamples

Tempting claim: Every square matrix is diagonalizable.

Correction: Enough independent eigenvectors are required.

Test the claim: The matrix with rows (1,1) and (0,1) has only the eigenvalue 1 and a one-dimensional eigenspace.

Check your understanding

Can zero be an eigenvalue of an invertible matrix?

Write a reason before opening the explanation. This reflection is not graded or saved.

Compare your reasoning

No.

A zero eigenvalue would give a nonzero v with Av=0, contradicting invertibility.

Explore an interactive illustration
Projection leaves a perpendicular residual

Project b onto the line spanned by a=(1,1)\displaystyle a = \left(1, 1\right). This illustrates a one-column least-squares problem.

Follow the perpendicular projection
0%
b

Moving point: (3, 3)

c=a⋅ba⋅a=3+32=3\displaystyle c = \frac{a \cdot b}{a \cdot a} = \frac{3 + 3}{2} = 3. Projection=(3,3); residual=(0,0). Its dot product with (1,1) is 0, confirming perpendicularity.

Connect this idea

Eigenvalues govern modes in linear differential equations. Orthogonal projection produces least-squares regression and separates explained components from residuals.

Practise this topic →Review concept flashcards

Reference reading

Use these texts to study the underlying definitions and theorems in greater depth. The explanations and examples above are written for Numerive.

Preview the eight practice prompts
  1. Problem 1

    For A=[[2,0],[0,3]]\displaystyle A = \left[\left[2, 0\right], \left[0, 3\right]\right], find the eigenvalue for v=(1,0)\displaystyle v = \left(1, 0\right).

    Practise this exact problem →
  2. Problem 2

    A triangular matrix [[2,1],[0,4]]. What is the sum of its eigenvalues?

    Practise this exact problem →
  3. Problem 3

    Project b=(2,4)\displaystyle b = \left(2, 4\right) onto a=(1,1)\displaystyle a = \left(1, 1\right). If the projection is ca, find c.

    Practise this exact problem →
  4. Problem 4

    A constant fit to data 1, 3, 8 by least squares. Find the fitted constant.

    Practise this exact problem →
  5. Problem 5New context

    A stretching map has A=[[3,0],[0,3]]\displaystyle A = \left[\left[3, 0\right], \left[0, 3\right]\right], find the eigenvalue for v=(1,0)\displaystyle v = \left(1, 0\right).

    Practise this exact problem →
  6. Problem 6New context

    A linear map has matrix [[3,1],[0,5]]. What is the sum of its eigenvalues?

    Practise this exact problem →
  7. Problem 7New context

    A least-squares model projects b=(3,5)\displaystyle b = \left(3, 5\right) onto a=(1,1)\displaystyle a = \left(1, 1\right). If the projection is ca, find c.

    Practise this exact problem →
  8. Problem 8New context

    A sensor calibration fits a single constant to data 2, 4, 9 by least squares. Find the fitted constant.

    Practise this exact problem →
Open stage ULA 1.3 in the student workspace →
Browse another level →
Numerive