Matrix Determinant : Practice Problems with Solutions

Matrix determinant are a fundamental concept in linear algebra, linking square matrices to scalar values that encode geometric and algebraic information. This collection of exercises covers every essential topic: computing determinants of 2×2 and 3×3 matrices, applying cofactor expansion along any row or column, exploiting row reduction and elementary row operations, using the key properties of determinants (multilinearity, alternating sign, multiplicativity), testing matrix invertibility, applying Cramer’s rule to linear systems, and extending to 4×4 and n×n determinants. Problems progress from easy direct calculations to challenging theoretical proofs, making this page suitable for high school students, undergraduates, and anyone preparing for exams in linear algebra or advanced mathematics.

Contents7 sections

Determinants of 2×2 Matrices

The determinant of a 2×2 matrix is the simplest case and the foundation for all higher-order computations. For a matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ , the determinant is defined as $\det(A) = ad - bc$ . Mastering this formula, including cases with fractions, negative entries, and parameters, prepares you for cofactor expansion and more advanced topics.

Problem 1: Direct Computation of a 2×2 Determinant

Easy

Consider the matrix $A = \begin{pmatrix} 5 & 3 \\ 2 & 4 \end{pmatrix}$ .

Compute $\det(A)$ using the standard formula.
State whether $A$ is invertible and justify your answer.

Hint

Recall that for a 2×2 matrix the determinant equals the product of the main diagonal entries minus the product of the off-diagonal entries. A matrix is invertible if and only if its determinant is nonzero.

View Solution

Solution to question 1:

\det(A) = (5)(4) - (3)(2) = 20 - 6 = 14

Solution to question 2: Since

\det(A) = 14 \neq 0

, the matrix

A

is invertible. A square matrix is invertible if and only if its determinant is different from zero.

Problem 2: Determinant with Negative and Fractional Entries

Easy

Let $B = \begin{pmatrix} -3 & \tfrac{1}{2} \\ 6 & -4 \end{pmatrix}$ .

Calculate $\det(B)$ .
Find $\det(2B)$ without recomputing the full product matrix, using only $\det(B)$ .

Hint

Apply the formula directly. For question 2, use the scalar multiplication property:

\det(kA) = k^n \det(A)

for an

n \times n

matrix.

View Solution

Solution to question 1:

\det(B) = (-3)(-4) - \left(\tfrac{1}{2}\right)(6) = 12 - 3 = 9

Solution to question 2: Since

B

is a

2 \times 2

matrix,

\det(2B) = 2^2 \det(B) = 4 \times 9 = 36

Problem 3: Finding a Parameter from a Given Determinant

Medium

The matrix $C = \begin{pmatrix} x+1 & 3 \\ 2 & x-2 \end{pmatrix}$ has determinant equal to $-4$ .

Set up and solve the equation $\det(C) = -4$ for $x$ .
For each value of $x$ found, determine whether $C$ is invertible.

Hint

Expand the determinant as a polynomial in

x

, then solve the resulting quadratic equation. Check the discriminant carefully.

View Solution

Solution to question 1:

\det(C) = (x+1)(x-2) - (3)(2) = x^2 - x - 2 - 6 = x^2 - x - 8

Setting

x^2 - x - 8 = -4

x^2 - x - 4 = 0 \implies x = \frac{1 \pm \sqrt{17}}{2}

Solution to question 2: For both values,

\det(C) = -4 \neq 0

, so

C

is invertible in both cases.

Determinants of 3×3 Matrices: Cofactor Expansion and Sarrus’ Rule

Computing the determinant of a 3×3 matrix requires expanding along a row or column using cofactors, or applying Sarrus’ rule as a mnemonic. Choosing the row or column with the most zeros significantly reduces calculation effort. This section develops fluency with both techniques across a range of entry types.

Problem 4: Sarrus’ Rule on a Standard 3×3 Matrix

Easy

Compute the determinant of the matrix

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

Use Sarrus’ rule to evaluate $\det(D)$ .
Verify your result by cofactor expansion along the first row.

Hint

Sarrus’ rule: copy the first two columns to the right, sum the three downward diagonal products, subtract the three upward diagonal products. For cofactor expansion along row 1, recall

\det(D) = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13}

where

C_{ij} = (-1)^{i+j}M_{ij}

View Solution

Solution to question 1 (Sarrus):

\det(D) = (2)(3)(1) + (1)(2)(4) + (-1)(0)(-1) - (4)(3)(-1) - (-1)(2)(2) - (1)(0)(1)

= 6 + 8 + 0 - (-12) - (-4) - 0 = 6 + 8 + 12 + 4 = 30

Solution to question 2 (cofactor expansion):

\det(D) = 2\det\begin{pmatrix}3&2\\-1&1\end{pmatrix} - 1\det\begin{pmatrix}0&2\\4&1\end{pmatrix} + (-1)\det\begin{pmatrix}0&3\\4&-1\end{pmatrix}

= 2(3+2) - 1(0-8) + (-1)(0-12) = 2(5) - 1(-8) + (-1)(-12) = 10 + 8 + 12 = 30 \checkmark

Problem 5: Strategic Cofactor Expansion Along the Best Row or Column

Medium

Evaluate the determinant of

E = \begin{pmatrix} 3 & 0 & 0 \\ -2 & 5 & 0 \\ 7 & -4 & 6 \end{pmatrix}

Identify the most efficient row or column for cofactor expansion and justify your choice.
Compute $\det(E)$ using that expansion and state a general rule suggested by this result.

Hint

Notice the structure of the matrix. A matrix of this form is called lower triangular. Consider expanding along the first row, or recall the determinant rule for triangular matrices.

View Solution

Solution to question 1: The first row contains two zeros, so cofactor expansion along row 1 involves only one nonzero term, making the computation fastest. Solution to question 2:

\det(E) = 3 \cdot \det\begin{pmatrix}5&0\\-4&6\end{pmatrix} - 0 + 0 = 3(30 - 0) = 90

General rule: The determinant of any triangular matrix (upper or lower) equals the product of its main diagonal entries:

\det(E) = 3 \times 5 \times 6 = 90

Problem 6: 3×3 Determinant with a Symbolic Parameter

Medium

Let

F = \begin{pmatrix} 1 & 2 & t \\ 3 & 1 & -1 \\ 0 & t & 2 \end{pmatrix}

Express $\det(F)$ as a polynomial in $t$ .
Find all values of $t$ for which $F$ is singular (not invertible).

Hint

Expand along the first row. Collect all powers of

t

and set the resulting polynomial to zero. Factor if possible, or use the quadratic formula.

View Solution

Solution to question 1:

\det(F) = 1\cdot\det\begin{pmatrix}1&-1\\t&2\end{pmatrix} - 2\cdot\det\begin{pmatrix}3&-1\\0&2\end{pmatrix} + t\cdot\det\begin{pmatrix}3&1\\0&t\end{pmatrix}

= (2+t) - 2(6) + t(3t) = 2 + t - 12 + 3t^2 = 3t^2 + t - 10

Solution to question 2:

3t^2 + t - 10 = 0 \implies t = \frac{-1 \pm \sqrt{1+120}}{6} = \frac{-1 \pm 11}{6}

t = \frac{10}{6} = \frac{5}{3} \quad \text{or} \quad t = \frac{-12}{6} = -2

The matrix

F

is singular for

t = \dfrac{5}{3}

and

t = -2

Properties of Determinants and Row Operations

Rather than expanding every determinant from scratch, the properties of determinants allow powerful shortcuts. These include: swapping two rows changes the sign, multiplying a row by a scalar multiplies the determinant by that scalar, adding a multiple of one row to another leaves the determinant unchanged, and the determinant of a product equals the product of determinants. These exercises train you to apply these rules fluently.

Problem 7: Identifying the Effect of Row Operations

Easy

Suppose $\det(A) = 6$ for a $3\times 3$ matrix $A$ . For each operation below, state the determinant of the resulting matrix without full computation.

Swap row 1 and row 3 of $A$ to obtain matrix $A_1$ .
Multiply row 2 of $A$ by $-5$ to obtain matrix $A_2$ .
Replace row 3 of $A$ by row 3 $+ 7 \cdot$ row 1 to obtain matrix $A_3$ .

Hint

Each part corresponds to one elementary row operation. Recall: (i) row swap multiplies determinant by

-1

; (ii) scalar multiplication of a row multiplies the determinant by that scalar; (iii) adding a multiple of one row to another does not change the determinant.

View Solution

Solution to question 1: Swapping two rows multiplies the determinant by

-1

\det(A_1) = -\det(A) = -6

Solution to question 2: Multiplying one row by a scalar

k

multiplies the determinant by

k

\det(A_2) = (-5)\det(A) = -30

Solution to question 3: Adding a multiple of one row to another does not change the determinant:

\det(A_3) = \det(A) = 6

Problem 8: Determinant via Row Reduction to Triangular Form

Medium

Use elementary row operations to compute the determinant of

G = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 5 & 4 \\ 3 & 4 & 9 \end{pmatrix}

Reduce $G$ to upper triangular form, tracking all row operations and their effect on the determinant.
Read off the determinant from the triangular form.

Hint

Use

R_2 \leftarrow R_2 - 2R_1

and

R_3 \leftarrow R_3 - 3R_1

first to clear the first column below the pivot. Since these are of the type “add a multiple of one row to another”, they do not change the determinant.

View Solution

Solution to question 1: Step 1:

R_2 \leftarrow R_2 - 2R_1

R_3 \leftarrow R_3 - 3R_1

(no change to

\det

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

Step 2:

R_3 \leftarrow R_3 + 2R_2

(no change to

\det

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

Solution to question 2: The matrix is now upper triangular; its determinant is the product of diagonal entries:

\det(G) = 1 \times 1 \times (-4) = -4

Problem 9: Determinant of a Product and Scalar Multiple

Medium

Let $A$ and $B$ be $3\times 3$ matrices with $\det(A) = 4$ and $\det(B) = -3$ .

Find $\det(AB)$ and $\det(BA)$ .
Find $\det(3A)$ and $\det(A^{-1}B)$ .

Hint

Use

\det(AB) = \det(A)\det(B)

\det(kA) = k^n\det(A)

for

n\times n

, and

\det(A^{-1}) = 1/\det(A)

View Solution

Solution to question 1:

\det(AB) = \det(A)\det(B) = (4)(-3) = -12

\det(BA) = \det(B)\det(A) = (-3)(4) = -12

(The determinant of a product is commutative even though matrix multiplication is not.) Solution to question 2:

\det(3A) = 3^3 \det(A) = 27 \times 4 = 108

\det(A^{-1}B) = \det(A^{-1})\det(B) = \frac{1}{4}\times(-3) = -\frac{3}{4}

Invertibility, Singularity, and the Determinant Criterion

One of the most important applications of the determinant is deciding whether a matrix is invertible (nonsingular) or singular. A square matrix is invertible if and only if its determinant is nonzero. This section presents problems that connect the determinant to the existence of an inverse, linear independence of rows or columns, and the solvability of linear systems.

Problem 10: Testing Invertibility and Finding the Inverse via the Adjugate

Medium

Consider the matrix $H = \begin{pmatrix} 2 & 1 & 0 \\ -1 & 3 & 2 \\ 0 & 1 & 4 \end{pmatrix}$ .

Compute $\det(H)$ and confirm that $H$ is invertible.
Find $H^{-1}$ using the formula $H^{-1} = \dfrac{1}{\det(H)}\,\mathrm{adj}(H)$ .

Hint

Expand along the first row to find

\det(H)

. For the adjugate, compute all nine cofactors

C_{ij} = (-1)^{i+j}M_{ij}

and transpose the resulting cofactor matrix.

View Solution

Solution to question 1:

\det(H) = 2\det\begin{pmatrix}3&2\\1&4\end{pmatrix} - 1\det\begin{pmatrix}-1&2\\0&4\end{pmatrix} + 0 = 2(10) - 1(-4) = 20 + 4 = 24

Since

\det(H) = 24 \neq 0

, the matrix

H

is invertible. Solution to question 2: The cofactor matrix is:

C = \begin{pmatrix} 10 & 4 & -1 \\ -4 & 8 & -2 \\ 2 & -4 & 7 \end{pmatrix}

The adjugate is the transpose of the cofactor matrix:

\mathrm{adj}(H) = C^T = \begin{pmatrix} 10 & -4 & 2 \\ 4 & 8 & -4 \\ -1 & -2 & 7 \end{pmatrix}

Therefore:

H^{-1} = \frac{1}{24}\begin{pmatrix} 10 & -4 & 2 \\ 4 & 8 & -4 \\ -1 & -2 & 7 \end{pmatrix}

Problem 11: Singularity and Linear Dependence

Medium

Determine whether the vectors $\mathbf{v}_1 = (1,\, 2,\, -1)$ , $\mathbf{v}_2 = (3,\, 0,\, 4)$ , and $\mathbf{v}_3 = (5,\, 4,\, 2)$ are linearly independent.

Form the matrix whose rows are $\mathbf{v}_1,\, \mathbf{v}_2,\, \mathbf{v}_3$ and compute its determinant.
Interpret the result in terms of linear independence and explain the connection to the determinant.

Hint

Three vectors in

\mathbb{R}^3

are linearly dependent if and only if the determinant of the matrix formed by these vectors (as rows or columns) is zero.

View Solution

Solution to question 1:

\det\begin{pmatrix}1&2&-1\\3&0&4\\5&4&2\end{pmatrix}

Expanding along row 1:

= 1(0-16) - 2(6-20) + (-1)(12-0) = -16 - 2(-14) - 12 = -16 + 28 - 12 = 0

Solution to question 2: Since

\det = 0

, the three vectors are linearly dependent. Indeed, one can verify that

\mathbf{v}_3 = 2\mathbf{v}_1 + \mathbf{v}_2

. A zero determinant means the rows (or columns) of the matrix are linearly dependent, and equivalently the matrix is singular.

Cramer’s Rule: Solving Linear Systems with Determinants

Cramer’s rule expresses each variable in a square linear system $A\mathbf{x} = \mathbf{b}$ as a ratio of determinants: $x_i = \det(A_i)/\det(A)$ , where $A_i$ is $A$ with its $i$ -th column replaced by $\mathbf{b}$ . While not the most computationally efficient method for large systems, Cramer’s rule provides theoretical insight and is practical for $2\times 2$ and $3\times 3$ systems.

Problem 12: Cramer’s Rule for a 2×2 System

Easy

Solve the linear system using Cramer’s rule:

\begin{cases} 3x + 2y = 8 \\ x - 4y = -6 \end{cases}

Find $\det(A)$ and confirm the system has a unique solution.
Apply Cramer’s rule to find $x$ and $y$ .

Hint

Form the coefficient matrix

A

and compute its determinant. Then replace the first column with the right-hand side vector

\mathbf{b}

to get

A_1

, and the second column for

A_2

View Solution

Solution to question 1:

A = \begin{pmatrix}3&2\\1&-4\end{pmatrix}, \quad \det(A) = (3)(-4)-(2)(1) = -12-2 = -14 \neq 0

The system has a unique solution. Solution to question 2:

A_1 = \begin{pmatrix}8&2\\-6&-4\end{pmatrix}, \quad \det(A_1) = (8)(-4)-(2)(-6) = -32+12 = -20

A_2 = \begin{pmatrix}3&8\\1&-6\end{pmatrix}, \quad \det(A_2) = (3)(-6)-(8)(1) = -18-8 = -26

x = \frac{\det(A_1)}{\det(A)} = \frac{-20}{-14} = \frac{10}{7}, \qquad y = \frac{\det(A_2)}{\det(A)} = \frac{-26}{-14} = \frac{13}{7}

Problem 13: Cramer’s Rule for a 3×3 System

Medium

Use Cramer’s rule to solve:

\begin{cases} x + y + z = 6 \\ 2x - y + z = 3 \\ x + 2y - z = 2 \end{cases}

Compute $\det(A)$ .
Find $x$ , $y$ , and $z$ using Cramer’s rule.

Hint

Replace the first, second, and third columns of

A

with the vector

\mathbf{b} = (6, 3, 2)^T

to form

A_1

A_2

A_3

respectively. Use cofactor expansion efficiently by choosing rows or columns with small entries.

View Solution

Solution to question 1:

A = \begin{pmatrix}1&1&1\\2&-1&1\\1&2&-1\end{pmatrix}

\det(A) = 1(1-2) - 1(-2-1) + 1(4+1) = -1+3+5 = 7

Solution to question 2:

\det(A_1) = \begin{vmatrix}6&1&1\\3&-1&1\\2&2&-1\end{vmatrix} = 6(1-2) - 1(-3-2) + 1(6+2) = -6+5+8 = 7

x = \frac{7}{7} = 1

\det(A_2) = \begin{vmatrix}1&6&1\\2&3&1\\1&2&-1\end{vmatrix} = 1(-3-2) - 6(-2-1) + 1(4-3) = -5+18+1 = 14

y = \frac{14}{7} = 2

\det(A_3) = \begin{vmatrix}1&1&6\\2&-1&3\\1&2&2\end{vmatrix} = 1(-2-6) - 1(4-3) + 6(4+1) = -8-1+30 = 21

z = \frac{21}{7} = 3

The solution is

x = 1,\; y = 2,\; z = 3

Determinants of 4×4 Matrices and Block Structures

Extending determinant calculations to $4\times 4$ matrices demands efficient strategies: cofactor expansion combined with careful choice of the expansion row or column, or exploiting block matrix structure when the matrix decomposes into smaller blocks. These problems build the algebraic stamina required for higher linear algebra courses and competitive mathematics.

Problem 14: Determinant of a 4×4 Matrix by Cofactor Expansion

Medium

Compute the determinant of

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

Identify the most efficient column for cofactor expansion and state why.
Evaluate $\det(K)$ completely.

Hint

Column 2 contains two zeros, making it ideal for expansion. Only entries

k_{22} = -1

and

k_{42} = 2

are nonzero in that column. Each will produce a

3\times 3

minor to evaluate.

View Solution

Solution to question 1: Column 2 has entries

(0, -1, 0, 2)

, so only two terms are nonzero in the expansion, minimising work. Solution to question 2: Expanding along column 2:

\det(K) = 0 \cdot C_{12} + (-1)\cdot C_{22} + 0\cdot C_{32} + 2\cdot C_{42}

C_{22} = (+1)\det\begin{pmatrix}2&1&3\\4&3&1\\0&0&-3\end{pmatrix}

Expanding the

3\times 3

minor along its third row:

= (-3)\det\begin{pmatrix}2&1\\4&3\end{pmatrix} = (-3)(6-4) = -6

C_{22} = -6

C_{42} = -1\cdot\det\begin{pmatrix}2&1&3\\0&0&2\\4&3&1\end{pmatrix}

Expanding the

3\times 3

minor along row 2:

= -(2)\det\begin{pmatrix}2&1\\4&3\end{pmatrix} \cdot (-1)^{2+3} = -2(2) \cdot (-1) = 4

Wait, let us expand row 2 properly: only entry

a_{23}=2

is nonzero:

\det = 0 - 0 + 2\cdot(-1)^{2+3}\det\begin{pmatrix}2&1\\4&3\end{pmatrix} = 2(-1)(2) = -4

C_{42} = (-1)^{4+2}(-4) = (+1)(-4) = -4

.Therefore:

\det(K) = (-1)(-6) + 2(-4) = 6 - 8 = -2

Problem 15: Block Triangular Matrix Determinant

Hard

Let

M = \begin{pmatrix} A & C \\ 0 & B \end{pmatrix}

where

A = \begin{pmatrix}3&1\\2&4\end{pmatrix}

B = \begin{pmatrix}1&5\\0&2\end{pmatrix}

, and

C

is a

2\times 2

matrix with arbitrary entries. Here

0

denotes the

2\times 2

zero matrix.

State and prove the block triangular determinant formula $\det(M) = \det(A)\det(B)$ .
Compute $\det(M)$ for this specific $A$ and $B$ .

Hint

For part 1, expand

\det(M)

along the last two rows (which form the block

(0 \mid B)

), or use cofactor expansion along columns 1 and 2 to reduce to

\det(A)

times a

2\times 2

determinant. Note that the result is independent of

C

View Solution

Solution to question 1: Write

M

as a

4\times 4

matrix. Expanding along the first two columns, the only nonzero contributions from rows 3 and 4 come from the

B

-block. More rigorously, one uses the Leibniz formula for determinants: every permutation

\sigma

that picks an entry from the lower-left zero block produces a zero term. Hence:

\det(M) = \det(A)\det(B)

regardless of

C

. This result generalises to any block upper (or lower) triangular square matrix. Solution to question 2:

\det(A) = (3)(4)-(1)(2) = 12-2 = 10

\det(B) = (1)(2)-(5)(0) = 2

\det(M) = 10 \times 2 = 20

Advanced Problems: Theoretical and Proof-Based Determinant Questions

At the highest level of difficulty, determinant problems move beyond calculation into proof and theory: proving that the determinant of a skew-symmetric odd-order matrix is zero, showing that $\det(AB) = \det(A)\det(B)$ for a specific pair, working with Vandermonde determinants, or analysing how determinants interact with rank and linear independence. These exercises are suitable for upper-division undergraduates and graduate preparation.

Problem 16: Skew-Symmetric Matrices and Odd Order

Hard

A square matrix $A$ is called skew-symmetric if $A^T = -A$ .

Prove that if $A$ is an $n\times n$ skew-symmetric matrix and $n$ is odd, then $\det(A) = 0$ .
Give a specific example of a $3\times 3$ skew-symmetric matrix and verify that its determinant is zero.

Hint

Use two determinant properties:

\det(A^T) = \det(A)

and

\det(-A) = (-1)^n\det(A)

. Combine them with the condition

A^T = -A

View Solution

Solution to question 1: Since

A^T = -A

, taking determinants of both sides:

\det(A^T) = \det(-A)

\det(A) = (-1)^n \det(A)

When

n

is odd,

(-1)^n = -1

, so:

\det(A) = -\det(A) \implies 2\det(A) = 0 \implies \det(A) = 0

Solution to question 2: The matrix

A = \begin{pmatrix}0&1&-2\\-1&0&3\\2&-3&0\end{pmatrix}

satisfies

A^T = -A

. Expanding along row 1:

\det(A) = 0(0-(-9)) - 1(0-6) + (-2)(3-0) = 0 + 6 - 6 = 0 \checkmark

Problem 17: Vandermonde Determinant

Hard

The Vandermonde matrix of order 3 is defined as

V = \begin{pmatrix}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2\end{pmatrix}

Compute $\det(V)$ by row-reducing: subtract row 1 from rows 2 and 3, then factor.
State the Vandermonde formula and give the condition under which $V$ is invertible.

Hint

After

R_2 \leftarrow R_2 - R_1

and

R_3 \leftarrow R_3 - R_1

, factor

(b-a)

from row 2 and

(c-a)

from row 3. The remaining

2\times 2

determinant factors further with another row operation.

View Solution

Solution to question 1: Apply

R_2 \leftarrow R_2 - R_1

and

R_3 \leftarrow R_3 - R_1

\det(V) = \begin{vmatrix}1&a&a^2\\0&b-a&b^2-a^2\\0&c-a&c^2-a^2\end{vmatrix}

Factor

(b-a)

from row 2 and

(c-a)

from row 3:

= (b-a)(c-a)\begin{vmatrix}1&a&a^2\\0&1&b+a\\0&1&c+a\end{vmatrix}

Apply

R_3 \leftarrow R_3 - R_2

in the

2\times 2

block:

= (b-a)(c-a)\begin{vmatrix}1&b+a\\0&c-b\end{vmatrix} = (b-a)(c-a)(c-b)

Solution to question 2: The Vandermonde formula is:

\det(V) = (b-a)(c-a)(c-b) = \prod_{1\le i < j \le 3}(x_j - x_i)

where

x_1=a,\, x_2=b,\, x_3=c

. The matrix

V

is invertible if and only if all three values

a

b

c

are distinct.

Problem 18: Determinant, Rank, and the Characterisation of Singular Matrices

Hard

Let $A$ be an $n\times n$ real matrix.

Prove that $\det(A) \neq 0$ if and only if $\mathrm{rank}(A) = n$ .
Let $A$ and $B$ be $n\times n$ matrices and assume $AB$ is invertible. Show that both $A$ and $B$ must be invertible.

Hint

For part 1, use the equivalence between full rank and invertibility, and between invertibility and nonzero determinant. For part 2, use the multiplicativity of the determinant:

\det(AB) = \det(A)\det(B)

. If a product of real numbers is nonzero, what can you say about each factor?

View Solution

Solution to question 1: The following statements are equivalent for an

n\times n

matrix

A

: (i)

A

is invertible; (ii)

\mathrm{rank}(A) = n

; (iii) the rows (equivalently columns) of

A

are linearly independent; (iv)

\det(A) \neq 0

. The chain of implications (i)

\Leftrightarrow

(ii) follows from the rank-nullity theorem (the null space is trivial iff rank is full), and (i)

\Leftrightarrow

(iv) follows from the Leibniz formula: if any row is a linear combination of others, cofactor expansion yields zero. Conversely, row-reducing a full-rank matrix yields an upper triangular matrix with nonzero diagonal, whose determinant is nonzero. Solution to question 2: Since

AB

is invertible,

\det(AB) \neq 0

. By the multiplicativity property:

\det(AB) = \det(A)\det(B) \neq 0

A product of real numbers is nonzero if and only if each factor is nonzero. Therefore

\det(A) \neq 0

and

\det(B) \neq 0

, which means both

A

and

B

are invertible.

\blacksquare