Linear algebra questions with solutions are provided here for practice and to understand what is linear algebra and its application to solving problems. Linear algebra is a branch of mathematics that deals with vectors, vector spaces, and linear functions which operate on vectors and follow vector addition.

Following are the main topics under linear algebra:

• Matrices and determinants
• Vector Spaces
• System of linear equations
• Linear transformations
• Inner product spaces
• Diagonalizations and quadratic forms

We shall practice a few problems based on these topics.

Learn more about linear algebra and its applications.

Linear Algebra Questions with Solutions

Let us solve a few questions based on linear algebra.

Question 1:

Show that the matrix A is unitary matrix

Solution:

A matrix is said to be unitary if and only if AA* = A*A = I, where A* is the transpose of the conjugate of A.

Given,

Transpose of A

Now,

Similarly, we can show that A*A = I

Hence, A is a unitary matrix.

Also refer: Types of Matrices

Question 2:

Find the rank and the nullity of the following matrix:

Solution:

To find the rank and nullity of the given matrix, we transform the given matrix into a row-reduced echelon form, by performing elementary transformations.

Let

Applying R₂ → R₂ – 2R₁ and R₃ → R₃ + R₁

Applying C₂ → C₂ + 2C₁, C₃ → C₃ + C₁ and C₄ → C₄ – 4C₁

Applying R₃ → R₃ – R₂

Applying R₂ → (⅕)R₂

Applying C₄ → C₄ + (⅗)C₃

∴ number of non-zero rows of the row-reduced echelon form of A = rank of A = 2

number of zero rows of the row-reduced echelon form of A = nullity of A = 2

Learn more about rank and nullity.

Question 3:

Solve the following system of linear equations:

x + y + z = 6

x + 2y + 3z = 14

x + 4y + 7z = 30

Solution:

The given linear equations can be written in the form of a matrix equation AX = B, where

The augmented matrix [A| B] is-

We reduce the given matrix to row echelon form by applying elementary row transformations

Applying R₂ → R₂ – R₁, R₃ → R₃ – R₁

Applying R₃ → R₃ – 3R₂

Since, Rank of A = Rank of [A : B] = 2 < number of unknowns

∴ the given system of linear equations has an infinite number of solutions.

Thus, we get from the row reduced echelon form matrix

x + y + z = 6 ….(i)

y + 2z = 8

⇒ y = 8 – 2z putting this value of y in (i), we get

x + 8 – 2z + z = 6

⇒ x – z = –2

⇒ x = z – 2

Now taking different values of z will give different values of the given system of equations.

Also Read:

• Transpose of Matrix
• Determinant of a Matrix
• Matrix Multiplication
• Matrix Operations
• Special Matrices
• Vector Spaces

Question 4:

Show that the set V = {(x, y) ∈ R² | xy ≥ 0} is not a vector space of R².

Solution:

For V to be a vector space, it is required that V must be closed under addition, that is for any x and y in V, x + y ∈ V

Let ( – 1, 0) and (0, 1) ∈ V

Now, ( – 1, 0) + (0, 1) = ( –1 + 0, 0 + 1) = ( –1, 1)

But, –1 × 1 = –1 < 0 ⇒ ( –1, 1) ∉ V.

∴ V is not a vector space in R².

Question 5:

Find the eigenvalues of

Solution:

Given

The characteristic polynomial is given by

Eigenvalues of A are the roots of the above cubic equation,

𝜆³ – 8𝜆² + 17𝜆 – 4 = 0

⇒ (𝜆 – 4)(𝜆² – 4𝜆 + 1) = 0

Solving this we get,

𝜆 = 4, 𝜆 = 2 ±√3

These are the eigenvalues of A.

Also check: Eigenvalues and Eigenvectors

Question 5:

Determine whether the following vector is linearly dependent or linearly independent: (1, 2, –3, 1), (3, 7, 1, –2), (1, 3, 7, –4).

Solution:

The vectors could form the column vectors of matrix A. We shall find the rank of A by reducing it to row echelon form.

Applying R₂ → R₂ – 2R₁, R₃ → R₃ + 3R₁, and R₄ → R₄ – R₁

Applying R₃ → R₃ – 10R₂, R₄ → R₄ + 5R₂

Clearly, rank of A = 2 < number of column vectors. So, the given vectors are linearly dependent.

Question 6:

Verify whether the polynomials x³ – 5x² – 2x + 3, x³ – 1, x³ + 2x + 4 are linearly independent.

Solution:

We may construct a matrix with coefficients of x³, x², x, and constant terms.

To find the rank of A let us reduce it to row echelon form by applying elementary transformations

Applying R₂ → R₂ + 5R₁, R₃ → R₃ + 2R₁, and R₄ → R₄ – 3R₁

Applying R₂ → (⅕) R₂

Applying R₃ → R₃ – 2R₂, R₄ → R₄ + 4R₂

Applying R₄ → R₄ – (5/2)R₃

∴ rank of A = 3 = number of column vectors. So the given vectors are linearly independent.

Question 7:

Show that the following matrix is diagonalizable:

Solution:

First, we shall find the eigenvalues of A. The characteristic equation of A is given by:

⇒ (1 – 𝜆)(2 – 𝜆)(3 – 𝜆) = 0

⇒ 𝜆 = 1, 2, 3.

The eigenvector corresponding to 𝜆₁ = 1 is the non-zero solution of the following matrix equation:

(A – 1I)X = 0

Applying elementary transformation R₃ → R₃ – 2R₂ and R₂ → R₂ + R₁, we get

⇒ z = 0, x + y = 0

If we take x = 1 ⇒ y = –1

Hence, the corresponding eigen-vector X₁ = [1 –1 0]^T.

Similarly, the eigenvector corresponding to 𝜆 = 2 is given by:

Applying elementary transformation R₃ → R₃ – R₂ and R₂ → R₂ + R₁, we get

⇒ x + z = 0, x + 2y = 0

⇒ x = –2, y = 1 and z = 2 {taking y = 1}

Hence, the corresponding eigen-vector X₂ = [ –2 1 2]^T.

Finally, the eigenvector corresponding to 𝜆 = 3 is the non-zero solution of the following matrix equation:

Applying elementary transformation R₂ → R₂ + R₁ and R₃ → R₃ + 2R₂, we get

⇒ 2x + z = 0, x + y = 0

On taking x = 1, we get x = 1, y = –1 and z = –2

Hence, the corresponding eigen-vector X₃ = [ 1 –1 –2]^T.

Let us construct a matrix with these eigenvectors as its column vectors, we get

Inverse of P is

Now,

= diag(1, 2, 3)

Thus, A is diagonalizable.

Refer: Diagonalization

Question 8:

Show that the transformation T: V₂(R) → V₂(R) defined by T(a, b) = (a + b, a) ∀ a, b ∈ R is a linear transformation.

Solution:

To show that T is a linear transformation, we need to prove that,

For any x, y ∈ V₂(R)

T(x + y) = T(x) + T(y) and T(ax) = aT(x) where a is a scalar in field.

Let (x₁, y₁) and (x₂, y₂) are arbitrary elements of V₂(R)

T[(x₁, y₁) + (x₂, y₂)] = T[(x₁ + x₂, y₁ + y₂)] = (x₁ + x₂ + y₁ + y₂, x₁ + x₂) …..(i)

T(x₁, y₁) + T(x₂, y₂) = (x₁ + y₁, x₁) + (x₂ + y₂, x₂) = (x₁ + x₂ + y₁ + y₂, x₁ + x₂) …..(ii)

From (i) and (ii), we get T[(x₁, y₁) + (x₂, y₂)] = T(x₁, y₁) + T(x₂, y₂)

Now, T[a(x₁, y₁)] = T(ax₁, ay₂) = (ax₁ + ay₁, ax₁) = a(x₁ + y₁, x₁) = aT(x₁, y₁).

∴ T is a linear transformation.

Question 9:

Show that the given subset of vectors of R³ forms a basis for R³.

{(1, 2, 1), (2, 1, 0), (1, –1, 2)}

Solution:

S = {(1, 2, 1), (2, 1, 0), (1, –1, 2)}

We know that any set of n linearly independent vectors forms the basis of n-dimensional vector space.

Now, dim R³ = 3, we just need to prove that vectors in S are linearly independent.

Let

We reduce this matrix to row echelon form to check the rank of A.

Applying R₂ → R₂ + (–2)R₁ and R₃ → R₃ + ( –1)R₁, we get

Applying R₂ → ( –⅓)R₂ and R₃ → R₃ + 2R₂, we get

Clearly, rank of A = 3 = number of vectors.

Thus, the given vectors are linearly independent.

⇒ S forms the basis of R³.

Question 10:

Given a linear transformation T on V₃(R) defined by T(a, b, c) = (2b + c, a – 4b, 3a) corresponding to the basis B = {(1, 0, 0), (0, 1, 0), (0, 0, 1)}. Find the matrix representation of T.

Solution:

Now, T(1, 0, 0) = (2 × 0 + 0, 1 – 4 × 0, 3 × 1) = (0, 1, 3)

= 0(1, 0, 0) + 1(0, 1, 0) + 3(0, 0, 1)

T(0, 1, 0) = (2 × 1 + 0, 0 – 4 × 1, 3 × 0) = (2, –4, 0)

= 2(1, 0, 0) –4(0, 1, 0) + 0(0, 0, 1)

And T(0, 0, 1) = (2 × 0 + 1, 0 – 4 × 0, 3 × 0) = (1, 0, 0)

= 1(1, 0, 0) + 0(0, 1, 0) + 0(0, 0, 1)

Then, the matrix representation of T with respect to the basis B is

Practice Problems on Linear Algebra

1. Show that the following matrix is diagonalizable:

2. Show that the transformation T: V₃(R) → V₂(R) defined by T(a, b, c) = (b, c) ∀ a, b, c ∈ R is a linear transformation.

3. Show that the given subset of vectors of R³ forms a basis for V₃(R).

{(1, 0, –1), (1, 2, 1), (0, –3, 2)}.

4. Given a linear transformation T on V₃(R) defined by T(a, b, c) = (2b + c, a – 4b, 3a) corresponding to the basis B = {(1, 1, 1), (1, 1, 0), (1, 0, 0)}. Find the matrix representation of T.

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