Matrices and Determinants Practice Questions — JAMB UTME

Question 1Question →

If the matrix $P = \begin{pmatrix} k & 3 & 1 \\ 2 & 1 & k \\ 4 & 2 & 0 \end{pmatrix}$ is singular, find the non-zero value of $k$ .

Show answer & explanation

Answer: 6

Answer

The non-zero value of

k

is 6.

For matrix P to be singular, its determinant must be 0. Expanding along row 3 yields 4(3k - 1) - 2(k^2 - 2) = 12k - 4 - 2k^2 + 4 = -2k^2 + 12k = 0. Factoring gives -2k(k - 6) = 0, which yields k = 0 or k = 6. The non-zero value is 6.

Step-by-Step Solution

1

Set the determinant of matrix P to 0

\det(P) = 0

A matrix is singular if and only if its determinant equals zero.

2

Evaluate the 3x3 determinant by expanding along the third row

4 \cdot (3k - 1) - 2 \cdot (k^2 - 2) + 0 = 0

Expanding along the third row takes advantage of the zero entry to simplify computation.

3

Expand and combine like terms

-2k^2 + 12k = 0

12k - 4 - 2k^2 + 4 reduces to -2k^2 + 12k.

4

Factor out common factors and solve for k

-2k(k - 6) = 0 \implies k = 0 \text{ or } k = 6

Applying the zero-product property.

5

Select the required root

k = 6

The question specifies the non-zero value of k.

Key Concept

Determinant of a 3x3 matrix and singular matrix condition

Estimated Time:2m 0s

Question 2Question →

If $A = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} 5 & 1 \\ 0 & -2 \end{pmatrix}$ , what is the product matrix $AB$ ?

\begin{pmatrix} 10 & -4 \\ 5 & -7 \end{pmatrix}

\begin{pmatrix} 10 & 3 \\ 0 & -8 \end{pmatrix}

\begin{pmatrix} 11 & 19 \\ -2 & -8 \end{pmatrix}

\begin{pmatrix} 10 & 8 \\ 5 & -7 \end{pmatrix}

Show answer & explanation

Answer: $\begin{pmatrix} 10 & -4 \\ 5 & -7 \end{pmatrix}$

Answer

The matrix product

AB

is

\begin{pmatrix} 10 & -4 \\ 5 & -7 \end{pmatrix}

.

Matrix multiplication is evaluated using the row-by-column rule. Evaluating each entry: top-left is

2(5)+3(0)=10

, top-right is

2(1)+3(-2)=-4

, bottom-left is

1(5)+4(0)=5

, and bottom-right is

1(1)+4(-2)=-7

. Thus, the product matrix is

\begin{pmatrix} 10 & -4 \\ 5 & -7 \end{pmatrix}

.

Step-by-Step Solution

1

Calculate the top-left element (Row 1 of A

\cdot

Column 1 of B)

2(5) + 3(0) = 10 + 0 = 10

Matrix multiplication takes the dot product of the corresponding row of the first matrix and column of the second matrix.

2

Calculate the top-right element (Row 1 of A

\cdot

Column 2 of B)

2(1) + 3(-2) = 2 - 6 = -4

Combine products of elements in row 1 of A and column 2 of B.

3

Calculate the bottom-left element (Row 2 of A

\cdot

Column 1 of B)

1(5) + 4(0) = 5 + 0 = 5

Combine products of elements in row 2 of A and column 1 of B.

4

Calculate the bottom-right element (Row 2 of A

\cdot

Column 2 of B)

1(1) + 4(-2) = 1 - 8 = -7

Combine products of elements in row 2 of A and column 2 of B.

Key Concept

Matrix Multiplication (Row-by-Column Rule)

Question 3Question →

What is the determinant of the $2 \times 2$ matrix $P = \begin{pmatrix} 5 & 2 \\ 3 & 4 \end{pmatrix}$ ?

Show answer & explanation

Answer: 14

Answer

The determinant of matrix

P

is

14

.

For any

2 \times 2

matrix

\begin{pmatrix} a & b \\ c & d \end{pmatrix}

, the determinant is given by

ad - bc

. Substituting

a=5

,

b=2

,

c=3

, and

d=4

yields

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

.

Step-by-Step Solution

1

Apply the determinant formula for a

2 \times 2

matrix

\begin{pmatrix} a & b \\ c & d \end{pmatrix}

, which is

\det(P) = ad - bc

.

\det(P) = (5)(4) - (2)(3)

The determinant of a

2 \times 2

matrix is defined as the product of the main diagonal elements minus the product of the off-diagonal elements.

2

Evaluate the arithmetic expression.

20 - 6 = 14

Perform multiplication followed by subtraction to get the final determinant value.

Key Concept

Determinant of a 2x2 Matrix

Estimated Time:45s

Question 4Question →

Given the matrices $A = \begin{pmatrix} 4 & x \\ -1 & 3 \end{pmatrix}$ and $B = \begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix}$ , if the determinant of the product matrix $AB$ is $95$ , what is the value of $x$ ?

Show answer & explanation

Answer: 7

Answer

The value of

x

is

7

.

Using the property

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

, we find

\det(B) = (2)(4) - (1)(3) = 5

and

\det(A) = (4)(3) - (x)(-1) = 12 + x

. Setting

5(12 + x) = 95

yields

60 + 5x = 95

, which solves to

x = 7

.

Step-by-Step Solution

1

Calculate the determinant of matrix

B

\det(B) = (2)(4) - (1)(3) = 5

The determinant of a

2 \times 2

matrix

\begin{pmatrix} a & b \\ c & d \end{pmatrix}

is

ad - bc

.

2

Express the determinant of matrix

A

in terms of

x

\det(A) = (4)(3) - (x)(-1) = 12 + x

Applying the determinant formula to matrix

A

gives

12 - (-x) = 12 + x

.

3

Apply the determinant product property

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

5(12 + x) = 95

The determinant of the product of two square matrices is equal to the product of their individual determinants.

4

Solve the linear equation for

x

60 + 5x = 95 \implies 5x = 35 \implies x = 7

Subtract

60

from both sides and divide by

5

to isolate

x

.

Key Concept

Determinant of Matrix Product Property

Question 5Question →

Given the matrices $A = \begin{pmatrix} 2 & 1 \\ 4 & 3 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & -1 \\ 2 & 5 \end{pmatrix}$ , find the determinant of the matrix $C = 2A - B$ .

Show answer & explanation

Answer: -15

Answer

The determinant of the matrix

C = 2A - B

is

-15

.

Scalar multiplication gives

2A = \begin{pmatrix} 4 & 2 \\ 8 & 6 \end{pmatrix}

. Subtracting

B

yields

C = \begin{pmatrix} 3 & 3 \\ 6 & 1 \end{pmatrix}

. Evaluating the determinant gives

\det(C) = (3)(1) - (3)(6) = 3 - 18 = -15

.

Step-by-Step Solution

1

Multiply matrix

A

by scalar

2

2A = \begin{pmatrix} 4 & 2 \\ 8 & 6 \end{pmatrix}

Scalar multiplication requires multiplying each entry of matrix

A

by

2

.

2

Subtract matrix

B

from matrix

2A

entry-wise to find matrix

C

C = \begin{pmatrix} 4 - 1 & 2 - (-1) \\ 8 - 2 & 6 - 5 \end{pmatrix} = \begin{pmatrix} 3 & 3 \\ 6 & 1 \end{pmatrix}

Subtract corresponding entries of matrix

B

from

2A

.

3

Calculate the determinant of matrix

C

\det(C) = (3)(1) - (3)(6) = 3 - 18 = -15

The determinant of a

2 \times 2

matrix

\begin{pmatrix} a & b \\ c & d \end{pmatrix}

is

ad - bc

.

Key Concept

Matrix Operations and Determinants

Question 6Question →

If matrix $A$ has dimension $2 \times 3$ and matrix $B$ has dimension $3 \times 4$ , what is the dimension of the matrix product $AB$ ?

3 \times 3

2 \times 4

4 \times 2

2 \times 3

Show answer & explanation

Answer: $2 \times 4$

Answer

The dimension of the matrix product

AB

is

2 \times 4

.

For two matrices to be multiplied, the number of columns in the first matrix must equal the number of rows in the second matrix. When a matrix of size

m \times n

is multiplied by a matrix of size

n \times p

, the resulting product matrix has size

m \times p

. For

A_{2 \times 3}

and

B_{3 \times 4}

, the outer dimensions give

2 \times 4

.

Step-by-Step Solution

1

Identify the dimensions of the given matrices

A

and

B

Matrix

A

is of order

2 \times 3

(

m = 2, n = 3

) and Matrix

B

is of order

3 \times 4

(

n = 3, p = 4

).

Matrix multiplication requirement requires the number of columns of the first matrix to match the number of rows of the second matrix.

2

Apply the matrix multiplication dimension rule

The product matrix

AB

has

m

rows and

p

columns, resulting in dimension

2 \times 4

.

When multiplying an

m \times n

matrix by an

n \times p

matrix, the resulting matrix has dimensions

m \times p

.

Key Concept

Matrix Multiplication Dimension Compatibility and Resulting Order

Estimated Time:45s

Question 7Question →

If the determinant of the matrix $M = \begin{pmatrix} x & 3 \\ 2 & 5 \end{pmatrix}$ is $14$ , find the value of $x$ .

Show answer & explanation

Answer: 4

Answer

The value of

x

is

4

.

For matrix

M = \begin{pmatrix} x & 3 \\ 2 & 5 \end{pmatrix}

, the determinant is calculated as

(x)(5) - (3)(2) = 5x - 6

. Equating this to

14

gives

5x - 6 = 14

, which simplifies to

5x = 20

, yielding

x = 4

.

Step-by-Step Solution

1

Apply the

2 \times 2

determinant formula

\det = ad - bc

\det(M) = (x \times 5) - (3 \times 2) = 5x - 6

The determinant of a

2 \times 2

matrix is the product of the main diagonal minus the product of the anti-diagonal.

2

Set the determinant equal to the given value

14

5x - 6 = 14

The problem states that the determinant is equal to 14.

3

Solve the linear equation for

x

5x = 20 \implies x = 4

Adding 6 to both sides gives

5x = 20

, and dividing by 5 yields

x = 4

.

Key Concept

Determinant of a 2x2 Matrix

Question 8Question →

What is the determinant of the matrix $P = \begin{pmatrix} 5 & 3 \\ 2 & 7 \end{pmatrix}$ ?

Show answer & explanation

Answer: 29

Answer

29

For a

2 \times 2

matrix

\begin{pmatrix} a & b \\ c & d \end{pmatrix}

, the determinant is given by

ad - bc

. Substituting the values

a = 5

,

b = 3

,

c = 2

, and

d = 7

yields

(5 \times 7) - (3 \times 2) = 35 - 6 = 29

.

Step-by-Step Solution

1

Identify the values of

a, b, c,

and

d

from the given

2 \times 2

matrix

P = \begin{pmatrix} 5 & 3 \\ 2 & 7 \end{pmatrix}

.

a = 5, b = 3, c = 2, d = 7

To set up the values for the

2 \times 2

determinant formula.

2

Multiply the elements of the main diagonal and the secondary diagonal.

Main diagonal product:

5 \times 7 = 35

; Secondary diagonal product:

3 \times 2 = 6

.

The determinant of a

2 \times 2

matrix is the difference between the main diagonal product and the secondary diagonal product.

3

Subtract the product of the secondary diagonal from the product of the main diagonal.

Determinant =

35 - 6 = 29

.

Completes the formula

\det(P) = ad - bc

.

Key Concept

Determinant of a 2x2 Matrix

Question 9Question →

Given that the matrix $A = \begin{pmatrix} 1 & 2 & k \\ 0 & k & 3 \\ 2 & -1 & 1 \end{pmatrix}$ is singular, what is the positive value of $k$ ?

Show answer & explanation

Answer: 3

Answer

The positive value of

k

is

3

.

A square matrix is singular when its determinant is zero. Expanding

\det(A)

along the first column yields

1 \cdot (k + 3) + 2 \cdot (6 - k^2) = -2k^2 + k + 15

. Equating this to zero gives the quadratic equation

2k^2 - k - 15 = 0

, which factors as

(2k + 5)(k - 3) = 0

. The roots are

k = -2.5

and

k = 3

. The positive value is

3

.

Step-by-Step Solution

1

Calculate the determinant of matrix A

\det(A) = -2k^2 + k + 15

Expanding along the first column simplifies the calculation because of the zero entry.

2

Set the determinant to zero for singularity

2k^2 - k - 15 = 0

A matrix is singular if and only if its determinant equals zero.

3

Solve the quadratic equation for k

k = 3 or k = -2.5

Factoring

2k^2 - k - 15 = (2k + 5)(k - 3) = 0

yields two roots.

4

Choose the positive solution

k = 3

The question explicitly requires the positive value of

k

.

Key Concept

Singular matrix definition and 3x3 determinant evaluation

Question 10Question →

Given that $x$ is a positive real number and the determinant of the matrix $A = \begin{pmatrix} x & 1 & 0 \\ 2 & x & 3 \\ 0 & 1 & 4 \end{pmatrix}$ is $2$ , find the value of $x$ .

Show answer & explanation

Answer: 2

Answer

The positive value of

x

is

2

.

Expanding the determinant along the first row yields

\det(A) = x(4x - 3) - 1(8 - 0) = 4x^2 - 3x - 8

. Equating this to

2

gives

4x^2 - 3x - 10 = 0

. Factoring the quadratic gives

(4x + 5)(x - 2) = 0

, which yields

x = 2

or

x = -1.25

. Because

x

must be positive, the correct value is

2

.

Step-by-Step Solution

1

Expand the

3 \times 3

determinant along the first row

\det(A) = x(4x - 3) - 1(8 - 0) + 0 = 4x^2 - 3x - 8

Expanding along the first row leverages the zero entry to simplify calculation of the determinant.

2

Equate the determinant expression to the given determinant value

4x^2 - 3x - 10 = 0

Setting the calculated determinant equal to

2

creates a quadratic equation in terms of

x

.

3

Factor the quadratic equation to find the candidate values for

x

(4x + 5)(x - 2) = 0 \implies x = 2 \text{ or } x = -1.25

Factoring determines all algebraic solutions that satisfy the determinant equation.

4

Apply the positivity constraint given in the problem statement

x = 2

The question restricts

x

to positive real numbers, discarding the negative root.

Key Concept

Determinant of a 3x3 matrix and quadratic equations

Estimated Time:2m 0s

Question 11Question →

What is the determinant of the matrix $A = \begin{pmatrix} 6 & -3 \\ 2 & 1 \end{pmatrix}$ ?

12

0

-12

-36

Show answer & explanation

Answer: 12

Answer

12

For any

2 \times 2

matrix

\begin{pmatrix} a & b \\ c & d \end{pmatrix}

, the determinant is given by

ad - bc

. Substituting

a = 6

,

b = -3

,

c = 2

, and

d = 1

yields

(6)(1) - (-3)(2) = 6 - (-6) = 12

.

Step-by-Step Solution

1

Identify the elements of the 2x2 matrix

A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}

a = 6

,

b = -3

,

c = 2

,

d = 1

Standard entries for determinant formula

\det(A) = ad - bc

.

2

Compute the product of the main diagonal elements (

ad

) and off-diagonal elements (

bc

)

ad = (6)(1) = 6

and

bc = (-3)(2) = -6

These form the components of the 2x2 determinant formula.

3

Subtract

bc

from

ad

\det(A) = 6 - (-6) = 6 + 6 = 12

Subtracting a negative quantity is equivalent to addition.

Key Concept

Determinant of a 2x2 Matrix

Estimated Time:45s

Question 12Question →

Given the $3 \times 3$ matrix $M = \begin{pmatrix} k & 2 & 1 \\ 3 & k-1 & 2 \\ 1 & 4 & 1 \end{pmatrix}$ , where $k > 5$ . If $\det(M^2) = 100$ and $\det(M) < 0$ , what is the value of $k$ ?

Show answer & explanation

Answer: 7

Answer

7

Using the matrix determinant power identity

\det(M^2) = (\det(M))^2 = 100

and the condition

\det(M) < 0

, we find

\det(M) = -10

. Expanding

\det(M)

along the first row yields

\det(M) = k(k-9) - 2(1) + 1(13-k) = k^2 - 10k + 11

. Setting this equal to

-10

gives the quadratic equation

k^2 - 10k + 21 = 0

, which factors as

(k-3)(k-7) = 0

. Given

k > 5

, the unique solution is

k = 7

.

Step-by-Step Solution

1

Apply determinant properties for matrix powers

\det(M) = -10

Because

\det(M^2) = (\det(M))^2 = 100

and it is given that

\det(M) < 0

, taking the negative square root gives

\det(M) = -10

.

2

Evaluate the determinant of matrix

M

using first-row expansion

\det(M) = k^2 - 10k + 11

Expanding along row 1 gives

k((k-1)(1) - 8) - 2(3(1) - 2(1)) + 1(3(4) - (k-1)(1)) = k^2 - 9k - 2 + 13 - k = k^2 - 10k + 11

.

3

Set up and simplify the quadratic equation for

k

k^2 - 10k + 21 = 0

Equating

k^2 - 10k + 11

to

-10

yields

k^2 - 10k + 21 = 0

.

4

Solve the quadratic equation and enforce the inequality condition

k > 5

k = 7

Factoring

(k-3)(k-7) = 0

gives

k = 3

or

k = 7

. Applying the restriction

k > 5

selects

k = 7

.

Key Concept

Determinant Properties of Matrix Powers and 3x3 Matrix Expansion

Question 13Question →

Given the matrices $P = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}$ and $Q = \begin{pmatrix} 5 & 0 \\ -2 & 1 \end{pmatrix}$ , what is the product matrix $PQ$ ?

\begin{pmatrix} 10 & 0 \\ -2 & 4 \end{pmatrix}

\begin{pmatrix} 4 & 3 \\ -3 & 4 \end{pmatrix}

\begin{pmatrix} 10 & 15 \\ -3 & -2 \end{pmatrix}

\begin{pmatrix} 16 & 3 \\ 13 & 4 \end{pmatrix}

Show answer & explanation

Answer: $\begin{pmatrix} 4 & 3 \\ -3 & 4 \end{pmatrix}$

Answer

The product matrix

PQ

is

\begin{pmatrix} 4 & 3 \\ -3 & 4 \end{pmatrix}

.

To find

PQ

, multiply each row of matrix

P

by each column of matrix

Q

. The entry in row 1, column 1 is

2(5) + 3(-2) = 4

. The entry in row 1, column 2 is

2(0) + 3(1) = 3

. The entry in row 2, column 1 is

1(5) + 4(-2) = -3

. The entry in row 2, column 2 is

1(0) + 4(1) = 4

. Putting these together yields

\begin{pmatrix} 4 & 3 \\ -3 & 4 \end{pmatrix}

.

Step-by-Step Solution

1

Compute the entries of the top row of

PQ

using the first row of

P

and both columns of

Q

.

First entry:

(2)(5) + (3)(-2) = 10 - 6 = 4

. Second entry:

(2)(0) + (3)(1) = 0 + 3 = 3

.

Matrix entry

(1,1)

is the dot product of Row 1 of

P

and Column 1 of

Q

; entry

(1,2)

is the dot product of Row 1 of

P

and Column 2 of

Q

.

2

Compute the entries of the bottom row of

PQ

using the second row of

P

and both columns of

Q

.

Third entry:

(1)(5) + (4)(-2) = 5 - 8 = -3

. Fourth entry:

(1)(0) + (4)(1) = 0 + 4 = 4

.

Matrix entry

(2,1)

is the dot product of Row 2 of

P

and Column 1 of

Q

; entry

(2,2)

is the dot product of Row 2 of

P

and Column 2 of

Q

.

3

Assemble the computed entries into a

2 \times 2

matrix.

PQ = \begin{pmatrix} 4 & 3 \\ -3 & 4 \end{pmatrix}

.

Combining row results gives the final product matrix.

Key Concept

Matrix Multiplication (Row-by-Column Dot Product)

Estimated Time:1m 15s

Question 14Question →

Given the matrices $A = \begin{pmatrix} x & 3 \\ 1 & 2 \end{pmatrix}$ and $B = \begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix}$ , what is the value of $x$ such that the determinant of the product matrix $AB$ is $25$ ?

4

2

-1

11.5

Show answer & explanation

Answer: 4

Answer

The value of

x

is 4.

The determinant of a product of matrices equals the product of their determinants,

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

. Since

\det(B) = 5

and

\det(A) = 2x - 3

, setting

5(2x - 3) = 25

yields

2x - 3 = 5

, giving

x = 4

.

Step-by-Step Solution

1

Calculate the determinant of matrix B

\det(B) = (2)(4) - (1)(3) = 8 - 3 = 5

The determinant of a 2x2 matrix

\begin{pmatrix} a & b \\ c & d \end{pmatrix}

is given by

ad - bc

.

2

Express the determinant of matrix A in terms of x

\det(A) = (x)(2) - (3)(1) = 2x - 3

Apply the 2x2 determinant formula to matrix A.

3

Use the determinant product rule det(AB) = det(A) * det(B)

\det(AB) = (2x - 3) \cdot 5 = 10x - 15

The determinant of the product of two square matrices equals the product of their individual determinants.

4

Set det(AB) equal to 25 and solve for x

10x - 15 = 25 \implies 10x = 40 \implies x = 4

Equate the determinant expression to the given value 25 to isolate x.

Key Concept

Determinant of a Matrix Product

Question 15Question →

Given two matrices $A = \begin{pmatrix} x & 2 \\ 2 & 3 \end{pmatrix}$ and $B = \begin{pmatrix} x & 1 \\ -1 & 1 \end{pmatrix}$ , where $x > 0$ . If the determinant of the product matrix $AB$ is $20$ , find the value of $x$ .

Show answer & explanation

Answer: 3

Answer

The positive value of

x

is 3.

Using the property

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

, we compute

\det(A) = 3x - 4

and

\det(B) = x + 1

. Equating their product to 20 gives

(3x - 4)(x + 1) = 20

, which simplifies to the quadratic

3x^2 - x - 24 = 0

. Factorizing yields

(3x + 8)(x - 3) = 0

. Since

x

must be a positive number, the correct value is

x = 3

.

Step-by-Step Solution

1

Calculate the determinant of matrix A

\det(A) = 3x - 4

The determinant of a 2x2 matrix \begin{pmatrix} a & b \\ c & d \end{pmatrix} is ad - bc.

2

Calculate the determinant of matrix B

\det(B) = x + 1

Applying ad - bc gives (x)(1) - (1)(-1) = x + 1.

3

Use the determinant product property det(AB) = det(A) * det(B)

(3x - 4)(x + 1) = 20

The determinant of the product of two square matrices equals the product of their individual determinants.

4

Form and solve the quadratic equation

3x^2 - x - 24 = 0, which factorizes into (3x + 8)(x - 3) = 0

Expanding (3x - 4)(x + 1) gives 3x^2 - x - 4. Subtracting 20 yields 3x^2 - x - 24 = 0.

5

Determine the positive solution for x

x = 3

Solving the factors gives x = -8/3 or x = 3. Since x must be positive (x > 0), x = 3.

Key Concept

Determinant of Matrix Product and 2x2 Determinants

Estimated Time:2m 0s

Question 16Question →

Given the matrices $A = \begin{pmatrix} 4 & 2 \\ 1 & 3 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & -1 \\ 2 & k \end{pmatrix}$ , find the value of $k$ such that the determinant of the product matrix $AB$ is equal to 50.

Show answer & explanation

Answer: 3

Answer

The value of

k

is 3.

By applying the determinant product rule

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

, we find

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

. Given

\det(AB) = 50

, it follows that

\det(B) = 50 / 10 = 5

. Since

\det(B) = (1)(k) - (-1)(2) = k + 2

, setting

k + 2 = 5

gives

k = 3

.

Step-by-Step Solution

1

Calculate the determinant of matrix A

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

The determinant of a 2x2 matrix \begin{pmatrix} a & b \\ c & d \end{pmatrix} is ad - bc.

2

Apply the product property of determinants

\det(B) = \frac{\det(AB)}{\det(A)} = \frac{50}{10} = 5

For any two square matrices of the same dimension, \det(AB) = \det(A) \cdot \det(B).

3

Express the determinant of matrix B in terms of k and solve

\det(B) = (1)(k) - (-1)(2) = k + 2 = 5 \implies k = 3

Equating the calculated determinant formula for B to its numerical value of 5.

Key Concept

Determinant Product Property and 2x2 Matrix Determinant

Estimated Time:1m 30s

Question 17Question →

Which of the following is the inverse of the matrix $M = \begin{pmatrix} 5 & 2 \\ 7 & 3 \end{pmatrix}$ ?

\begin{pmatrix} 3 & -2 \\ -7 & 5 \end{pmatrix}

\begin{pmatrix} 3 & 2 \\ 7 & 5 \end{pmatrix}

\begin{pmatrix} -5 & 2 \\ 7 & -3 \end{pmatrix}

\begin{pmatrix} 5 & 7 \\ 2 & 3 \end{pmatrix}

Show answer & explanation

Answer: $\begin{pmatrix} 3 & -2 \\ -7 & 5 \end{pmatrix}$

Answer

\begin{pmatrix} 3 & -2 \\ -7 & 5 \end{pmatrix}

For any non-singular 2x2 matrix

\begin{pmatrix} a & b \\ c & d \end{pmatrix}

, the inverse is

\frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}

. Here,

ad - bc = (5)(3) - (2)(7) = 1

, so the inverse matrix is

\begin{pmatrix} 3 & -2 \\ -7 & 5 \end{pmatrix}

.

Step-by-Step Solution

1

Calculate the determinant of matrix M

\det(M) = (5)(3) - (2)(7) = 15 - 14 = 1

For a 2x2 matrix

M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}

, the determinant is given by

\det(M) = ad - bc

.

2

Find the adjugate matrix of M

\text{adj}(M) = \begin{pmatrix} 3 & -2 \\ -7 & 5 \end{pmatrix}

The adjugate of a 2x2 matrix is obtained by swapping the main diagonal elements (

a

and

d

) and negating the off-diagonal elements (

b

and

c

).

3

Apply the matrix inverse formula

M^{-1} = \frac{1}{\det(M)} \text{adj}(M) = \frac{1}{1}\begin{pmatrix} 3 & -2 \\ -7 & 5 \end{pmatrix} = \begin{pmatrix} 3 & -2 \\ -7 & 5 \end{pmatrix}

Multiplying the adjugate matrix by the reciprocal of the determinant gives the exact inverse matrix.

Key Concept

Inverse of a 2x2 Matrix

Question 18Question →

Given the matrix $A = \begin{pmatrix} 3 & 1 \\ 4 & 2 \end{pmatrix}$ , what is the determinant of the matrix $3A$ ?

6

18

30

54

Show answer & explanation

Answer: $18$

Answer

The determinant of the matrix

3A

is

18

.

For any

2 \times 2

matrix, scaling the matrix by a factor

k

scales its determinant by

k^2

. Since

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

, multiplying the matrix by

3

yields

\det(3A) = 3^2 \times 2 = 9 \times 2 = 18

. Alternatively, computing

3A = \begin{pmatrix} 9 & 3 \\ 12 & 6 \end{pmatrix}

directly gives

\det(3A) = (9)(6) - (3)(12) = 54 - 36 = 18

.

Step-by-Step Solution

1

Calculate the determinant of matrix

A

.

\det(A) = (3)(2) - (1)(4) = 6 - 4 = 2.

The determinant of a

2 \times 2

matrix

\begin{pmatrix} a & b \\ c & d \end{pmatrix}

is given by

ad - bc

.

2

Apply the determinant scalar multiplication property for an

n \times n

matrix.

\det(3A) = 3^2 \det(A) = 9 \times 2 = 18.

For any

n \times n

square matrix

A

and scalar

k

, the identity

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

holds. Here,

k=3

and

n=2

.

Key Concept

Scalar Multiplication Property of Determinants

Estimated Time:1m 15s

Question 19Question →

Given the matrix $A = \begin{pmatrix} y & 5 \\ 3 & y+2 \end{pmatrix}$ , if the determinant of $A$ is $9$ and $y > 0$ , calculate the value of $y$ .

Show answer & explanation

Answer: 4

Answer

The value of

y

is 4.

Expanding the determinant of

A

yields

\det(A) = y(y+2) - 15 = y^2 + 2y - 15

. Equating this to 9 gives

y^2 + 2y - 24 = 0

, which factors as

(y+6)(y-4) = 0

. The roots are

y = -6

and

y = 4

. Given that

y > 0

, the required value is 4.

Step-by-Step Solution

1

Find the expression for the determinant of matrix

A

\det(A) = y(y+2) - (5)(3) = y^2 + 2y - 15

The determinant of a

2 \times 2

matrix

\begin{pmatrix} a & b \\ c & d \end{pmatrix}

is given by

ad - bc

.

2

Form and solve the quadratic equation

y^2 + 2y - 15 = 9 \implies y^2 + 2y - 24 = 0 \implies (y + 6)(y - 4) = 0

Set the determinant expression equal to the given determinant value of 9 and rearrange into standard quadratic form.

3

Apply the given domain restriction

y > 0

y = 4

The root

y = -6

is discarded because

y

must be strictly positive.

Key Concept

Determinant of a 2x2 matrix and solving non-linear determinant equations

Question 20Question →

Given the matrices $M = \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix}$ and $N = \begin{pmatrix} 1 & 0 \\ 2 & 3 \end{pmatrix}$ , what is the determinant of the matrix product $MN$ ?

14

15

24

33

Show answer & explanation

Answer: 33

Answer

33

Using the property that the determinant of a product of matrices equals the product of their individual determinants,

\det(M) = (2)(4) - (-1)(3) = 11

and

\det(N) = (1)(3) - (0)(2) = 3

. Multiplying these gives

\det(MN) = 11 \times 3 = 33

.

Step-by-Step Solution

1

Calculate the determinant of matrix

M

\det(M) = (2)(4) - (-1)(3) = 8 + 3 = 11

The determinant of a

2 \times 2

matrix

\begin{pmatrix} a & b \\ c & d \end{pmatrix}

is given by

ad - bc

.

2

Calculate the determinant of matrix

N

\det(N) = (1)(3) - (0)(2) = 3 - 0 = 3

Apply the determinant formula

ad - bc

to matrix

N

.

3

Apply the determinant product property to find

\det(MN)

\det(MN) = \det(M) \times \det(N) = 11 \times 3 = 33

For any two square matrices of the same order,

\det(MN) = \det(M) \cdot \det(N)

.

Key Concept

Determinant Property of Matrix Products

Estimated Time:1m 30s

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