Matrices and Determinants Practice Questions — JAMB UTME

Question 1Question →

Given the $3 \times 3$ matrix $A = \begin{pmatrix} x & 2 & 1 \\ 3 & 1 & 2 \\ 1 & x & 0 \end{pmatrix}$ , find the positive value of $x$ for which $\det(A) = -2$ .

Show answer & explanation

Answer: 2.5

Answer

The positive value of

x

is 2.5.

Expanding the determinant of matrix

A

along the third row gives

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

. Setting this equal to

-2

yields

2x^2 - 3x - 5 = 0

. Factoring gives

(2x - 5)(x + 1) = 0

, yielding solutions

x = 2.5

and

x = -1

. Taking the positive value gives

x = 2.5

.

Step-by-Step Solution

1

Calculate the determinant of matrix

A

in terms of

x

\det(A) = -2x^2 + 3x + 3

Expanding along the third row simplifies computation because of the zero entry.

2

Set the determinant expression equal to

-2

and rearrange terms

2x^2 - 3x - 5 = 0

Setting

-2x^2 + 3x + 3 = -2

forms a standard quadratic equation.

3

Factorize the quadratic equation to find the roots

x = 2.5

or

x = -1

Factoring

(2x - 5)(x + 1) = 0

yields two real solutions.

4

Filter for the positive value requested in the stem

x = 2.5

The question specifically requires the positive value of

x

.

Key Concept

Determinant of a 3x3 Matrix and Quadratic Equation Solving

Estimated Time:2m 30s

Question 2Question →

If the matrix $A = \begin{pmatrix} 3 & 2 \\ 7 & 5 \end{pmatrix}$ , which matrix represents the inverse $A^{-1}$ ?

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

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

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

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

Show answer & explanation

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

Answer

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

The correct inverse matrix is computed by evaluating the determinant

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

and constructing the adjugate matrix by swapping the diagonal elements

3

and

5

while changing the signs of

2

and

7

, resulting in

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

.

Step-by-Step Solution

1

Calculate the determinant of matrix

A

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

The inverse requires dividing the adjugate matrix by the determinant of

A

.

2

Find the adjugate of matrix

A

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

For a 2x2 matrix

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

, swap the main diagonal elements (

a

and

d

) and change the signs of the off-diagonal elements (

b

and

c

).

3

Compute

A^{-1} = \frac{1}{\det(A)} \text{adj}(A)

A^{-1} = \frac{1}{1} \begin{pmatrix} 5 & -2 \\ -7 & 3 \end{pmatrix} = \begin{pmatrix} 5 & -2 \\ -7 & 3 \end{pmatrix}

Multiply the adjugate matrix by the reciprocal of the determinant.

Key Concept

2x2 Matrix Inversion

Estimated Time:1m 0s

Question 3Question →

Let $A = \begin{pmatrix} 2 & 1 \\ -1 & 3 \end{pmatrix}$ and $B = \begin{pmatrix} x & 2 \\ 1 & y \end{pmatrix}$ be $2 \times 2$ matrices with integer entries $x$ and $y$ . If the matrix product $AB$ is singular and $\det(A + B) = 0$ , what is the value of $x + y$ ?

-3

3

-5

7

Show answer & explanation

Answer: $-3$

Answer

The value of

x + y

is

-3

.

Since matrix

AB

is singular,

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

. Evaluating

\det(A) = 7 \neq 0

, we find

\det(B) = xy - 2 = 0

, giving

xy = 2

. Computing

A+B = \begin{pmatrix} 2+x & 3 \\ 0 & 3+y \end{pmatrix}

, its determinant is

(2+x)(3+y) = 6 + 3x + 2y + xy = 0

. Substituting

xy = 2

yields

3x + 2y = -8

. The integer pair satisfying both

xy = 2

and

3x + 2y = -8

is

x = -2

and

y = -1

. Therefore,

x + y = -3

.

Step-by-Step Solution

1

Calculate the determinant of matrix

A

.

\det(A) = (2)(3) - (1)(-1) = 6 + 1 = 7

.

Knowing

\det(A)

helps simplify the condition

\det(AB) = 0

using determinant properties.

2

Use the singularity of

AB

to find a relationship between

x

and

y

.

\det(AB) = \det(A)\det(B) = 7(xy - 2) = 0 \implies xy = 2

.

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

3

Form the matrix

A + B

and compute its determinant.

A + B = \begin{pmatrix} 2+x & 3 \\ 0 & 3+y \end{pmatrix} \implies \det(A+B) = (2+x)(3+y) - (3)(0) = (2+x)(3+y) = 6 + 3x + 2y + xy

.

Setting this determinant to zero gives a second equation involving

x

and

y

.

4

Substitute

xy = 2

into

\det(A+B) = 0

and solve for integer values of

x

and

y

.

6 + 3x + 2y + 2 = 0 \implies 3x + 2y = -8

. Testing integer pairs

(x,y)

for

xy = 2

gives

x = -2

and

y = -1

.

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

is satisfied only by

x = -2, y = -1

among integer pairs.

5

Compute

x + y

.

x + y = -2 + (-1) = -3

.

This answers the question directly.

Key Concept

Determinants of matrix products and sums, matrix singularity, and integer solutions to matrix equations.

Question 4Question →

If the matrix $P = \begin{pmatrix} x & 3 \\ 8 & 6 \end{pmatrix}$ is a singular matrix, what is the value of $x$ ?

-4

4

5

16

Show answer & explanation

Answer: 4

Answer

The value of

x

is

4

.

For a

2 \times 2

matrix to be singular, its determinant must be zero. For matrix

P

, the determinant is

(x \cdot 6) - (3 \cdot 8) = 6x - 24

. Setting

6x - 24 = 0

gives

6x = 24

, which simplifies to

x = 4

.

Step-by-Step Solution

1

State the condition for a matrix to be singular.

A matrix is singular if its determinant is equal to zero:

\det(P) = 0

.

By definition, a singular matrix has a determinant equal to zero.

2

Calculate the determinant of matrix

P

.

\det(P) = (x \times 6) - (3 \times 8) = 6x - 24.

The determinant of a

2 \times 2

matrix

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

is calculated using the formula

ad - bc

.

3

Solve the linear equation

6x - 24 = 0

for

x

.

6x = 24 \implies x = 4.

Adding

24

to both sides and dividing by

6

isolates

x

.

Key Concept

Singular matrix condition and

2 \times 2

determinant calculation

Question 5Question →

Given the matrices $A = \begin{pmatrix} k & 2 \\ 3 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & -1 \\ 2 & 0 \end{pmatrix}$ , what is the smaller positive integer value of $k$ for which $\det(A^2 + B) = 4$ ?

5

6

4

8

Show answer & explanation

Answer: 5

Answer

The smaller positive integer value of

k

is 5.

Evaluating

A^2

via matrix multiplication gives

\begin{pmatrix} k^2 + 6 & 2k + 2 \\ 3k + 3 & 7 \end{pmatrix}

. Adding

B

gives

A^2 + B = \begin{pmatrix} k^2 + 7 & 2k + 1 \\ 3k + 5 & 7 \end{pmatrix}

. Computing

\det(A^2 + B) = (k^2 + 7)(7) - (2k + 1)(3k + 5) = k^2 - 13k + 44

. Setting this to 4 results in

k^2 - 13k + 40 = 0

, which yields roots

k = 5

and

k = 8

. The smaller value is 5.

Step-by-Step Solution

1

Compute matrix

A^2

using matrix multiplication (row by column).

A^2 = \begin{pmatrix} k & 2 \\ 3 & 1 \end{pmatrix} \begin{pmatrix} k & 2 \\ 3 & 1 \end{pmatrix} = \begin{pmatrix} k^2 + 6 & 2k + 2 \\ 3k + 3 & 7 \end{pmatrix}

Matrix multiplication requires taking the dot product of rows of the first matrix with columns of the second matrix.

2

Add matrix

B

to

A^2

.

A^2 + B = \begin{pmatrix} k^2 + 6 & 2k + 2 \\ 3k + 3 & 7 \end{pmatrix} + \begin{pmatrix} 1 & -1 \\ 2 & 0 \end{pmatrix} = \begin{pmatrix} k^2 + 7 & 2k + 1 \\ 3k + 5 & 7 \end{pmatrix}

Matrix addition is performed by adding corresponding entries.

3

Calculate the determinant of

A^2 + B

and set it equal to 4.

\det(A^2 + B) = (k^2 + 7)(7) - (2k + 1)(3k + 5) = 7k^2 + 49 - (6k^2 + 13k + 5) = k^2 - 13k + 44 = 4

The determinant of a

2 \times 2

matrix

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

is

ad - bc

.

4

Solve the quadratic equation for

k

.

k^2 - 13k + 40 = 0 \implies (k - 5)(k - 8) = 0 \implies k = 5 \text{ or } k = 8

Factoring the quadratic equation gives the two possible integer roots.

Key Concept

Matrix multiplication, matrix addition, determinant of a 2x2 matrix, and quadratic equation solving.

Question 6Question →

Two $2 \times 2$ matrices are given as $A = \begin{pmatrix} 2 & -1 \\ 4 & 3 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 0 \\ 2 & k \end{pmatrix}$ . If the determinant of the product matrix $AB$ is equal to $30$ , what is the value of $k$ ?

Show answer & explanation

Answer: 3

Answer

The value of k is 3.

The determinant of a product of square matrices equals the product of their individual determinants:

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

. Evaluating

\det(A)

yields

(2)(3) - (-1)(4) = 10

, and

\det(B)

yields

(1)(k) - (0)(2) = k

. Substituting these into

\det(AB) = 30

gives

10k = 30

, leading directly to

k = 3

.

Step-by-Step Solution

1

Evaluate the determinant of matrix A

\det(A) = 10

Using the

2 \times 2

determinant formula

\det\begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad - bc

, we calculate

\det(A) = (2)(3) - (-1)(4) = 6 + 4 = 10

.

2

Evaluate the determinant of matrix B

\det(B) = k

Calculating the determinant of matrix

B

yields

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

.

3

Use the product property of determinants to solve for k

k = 3

Since

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

, we have

10k = 30

. Dividing both sides by 10 yields

k = 3

.

Key Concept

Determinant of Matrix Product

Estimated Time:1m 30s

Question 7Question →

Given the matrices $A = \begin{pmatrix} 2 & 3 \\ 1 & m \end{pmatrix}$ and $B = \begin{pmatrix} 4 \\ -2 \end{pmatrix}$ , if the matrix product $AB = \begin{pmatrix} 2 \\ 6 \end{pmatrix}$ , what is the value of $m$ ?

-1

1

-3

-5

Show answer & explanation

Answer: $-1$

Answer

The value of

m

is

-1

.

Multiplying the

2 \times 2

matrix by the

2 \times 1

column vector yields a

2 \times 1

result where the second row entry is

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

. Setting this equal to the second row element

6

gives

4 - 2m = 6

, which simplifies to

m = -1

.

Step-by-Step Solution

1

Compute the matrix product

AB

AB = \begin{pmatrix} 2(4) + 3(-2) \\ 1(4) + m(-2) \end{pmatrix} = \begin{pmatrix} 8 - 6 \\ 4 - 2m \end{pmatrix} = \begin{pmatrix} 2 \\ 4 - 2m \end{pmatrix}

Matrix multiplication requires taking the dot product of each row of the first matrix with the column vector.

2

Equate the resulting vector to the given vector

\begin{pmatrix} 2 \\ 6 \end{pmatrix}

4 - 2m = 6

Corresponding elements of equal matrices must be equal.

3

Solve the linear equation for

m

-2m = 6 - 4 \Rightarrow -2m = 2 \Rightarrow m = -1

Isolating the variable

m

by subtracting

4

from both sides and dividing by

-2

.

Key Concept

Matrix Multiplication and Equality of Matrices

Question 8Question →

Given that the determinant of the $3 \times 3$ matrix $M = \begin{pmatrix} 3 & 1 & 0 \\ 2 & x & -1 \\ 0 & 4 & 2 \end{pmatrix}$ is equal to $20$ , calculate the value of $x$ .

Show answer & explanation

Answer: 2

Answer

The value of

x

is

2

.

Expanding the matrix

M

along its top row yields

\det(M) = 3(2x + 4) - 1(4) = 6x + 8

. Setting this expression equal to

20

gives

6x + 8 = 20

, which simplifies to

6x = 12

, yielding

x = 2

.

Step-by-Step Solution

1

Expand the 3x3 matrix along the first row

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

Cofactor expansion along a row containing a zero simplifies the computation of a 3x3 determinant.

2

Simplify the algebraic expression for the determinant

\det(M) = 6x + 8

Distribute the coefficients and combine like constant terms.

3

Solve the linear equation for x

x = 2

Subtract 8 from 20 to get 12, then divide by 6.

Key Concept

Determinant of a 3x3 Matrix via Cofactor Expansion

Question 9Question →

Given that the matrix $A = \begin{pmatrix} x & 2 \\ 5 & x - 3 \end{pmatrix}$ is a singular matrix, what is the positive value of $x$ ?

2

3

5

10

Show answer & explanation

Answer: $5$

Answer

The positive value of

x

is

5

.

For a matrix to be singular, its determinant must be zero. Calculating the determinant of

A

yields

\det(A) = x(x - 3) - (2)(5) = x^2 - 3x - 10

. Setting this to zero and factoring gives

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

, leading to

x = 5

or

x = -2

. The positive solution is

5

.

Step-by-Step Solution

1

Set the determinant of the matrix equal to zero.

\det(A) = (x)(x - 3) - (2)(5) = 0

By definition, a matrix is singular if and only if its determinant is zero.

2

Expand and simplify the algebraic equation.

x^2 - 3x - 10 = 0

Expanding

x(x - 3)

gives

x^2 - 3x

, and subtracting

10

forms a standard quadratic equation.

3

Factor the quadratic equation to solve for

x

.

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

The factors of

-10

that sum to

-3

are

-5

and

+2

.

4

Select the positive value requested by the question.

x = 5

The question specifically asks for the positive value of

x

.

Key Concept

Singular Matrices and Determinants

Question 10Question →

If the matrix $M = \begin{pmatrix} 2x & 4 \\ 3 & 6 \end{pmatrix}$ is a singular matrix, what is the value of $x$ ?

1

-1

2

0

Show answer & explanation

Answer: $1$

Answer

The value of

x

is

1

.

A matrix is singular when its determinant equals zero. Evaluating the determinant of

\begin{pmatrix} 2x & 4 \\ 3 & 6 \end{pmatrix}

gives

(2x)(6) - (4)(3) = 12x - 12

. Setting

12x - 12 = 0

leads directly to

x = 1

.

Step-by-Step Solution

1

Recall the condition for a matrix to be singular.

A matrix is singular if and only if its determinant is equal to zero, so

\det(M) = 0

.

Singular matrices have no inverse because their determinant is zero.

2

Calculate the determinant of matrix

M = \begin{pmatrix} 2x & 4 \\ 3 & 6 \end{pmatrix}

.

\det(M) = (2x)(6) - (4)(3) = 12x - 12.

The determinant of a

2 \times 2

matrix

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

is

ad - bc

.

3

Set the determinant equal to zero and solve for

x

.

12x - 12 = 0 \implies 12x = 12 \implies x = 1

.

Solving the linear equation yields the required parameter value.

Key Concept

Condition for Singular Matrix and 2x2 Determinant Evaluation

Estimated Time:45s

Question 11Question →

For the matrix $M = \begin{pmatrix} k & 3 & 1 \\ 2 & k & 0 \\ 1 & 5 & 2 \end{pmatrix}$ , the determinant of $M$ is equal to $4$ . What is the positive value of $k$ ?

Show answer & explanation

Answer: 2

Answer

The positive value of

k

is

2

.

Expanding the determinant along the second row yields

\det(M) = -2(6-5) + k(2k-1) = 2k^2 - k - 2

. Setting this equal to

4

yields

2k^2 - k - 6 = 0

, which factors into

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

. The positive value is

2

.

Step-by-Step Solution

1

Evaluate the determinant of

M

in terms of

k

using row 2 cofactor expansion

\det(M) = 2k^2 - k - 2

Expanding along the second row gives

-2(6 - 5) + k(2k - 1) - 0 = -2 + 2k^2 - k

.

2

Set the determinant equal to the given value 4 and rearrange into standard quadratic form

2k^2 - k - 6 = 0

Subtracting 4 from both sides gives

2k^2 - k - 6 = 0

.

3

Solve the quadratic equation by factorization

k = -1.5 \text{ or } k = 2

Factoring

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

gives roots

k = -1.5

and

k = 2

.

4

Select the positive root

k = 2

The question specifically asks for the positive value of

k

.

Key Concept

Evaluating a 3x3 matrix determinant using cofactor expansion and solving the resulting quadratic equation for an unknown parameter.

Question 12Question →

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

-3

10

-1

1

Show answer & explanation

Answer: $-3$

Answer

The value of

x + y

is

-3

.

Evaluating the matrix products

AB

and

BA

using standard matrix multiplication rules yields

AB = \begin{pmatrix} 2+2x & 8+xy \\ 5 & -4+3y \end{pmatrix}

and

BA = \begin{pmatrix} -2 & x+12 \\ 4-y & 2x+3y \end{pmatrix}

. Equating entry-by-entry gives

2+2x = -2 \implies x = -2

and

5 = 4-y \implies y = -1

. Adding these together gives

x + y = -3

.

Step-by-Step Solution

1

Compute the product matrix

AB

AB = \begin{pmatrix} 2(1) + x(2) & 2(4) + x(y) \\ -1(1) + 3(2) & -1(4) + 3(y) \end{pmatrix} = \begin{pmatrix} 2 + 2x & 8 + xy \\ 5 & -4 + 3y \end{pmatrix}

Matrix multiplication requires taking the dot product of rows from the first matrix and columns from the second matrix.

2

Compute the product matrix

BA

BA = \begin{pmatrix} 1(2) + 4(-1) & 1(x) + 4(3) \\ 2(2) + y(-1) & 2(x) + y(3) \end{pmatrix} = \begin{pmatrix} -2 & x + 12 \\ 4 - y & 2x + 3y \end{pmatrix}

Evaluate

BA

by multiplying rows of

B

by columns of

A

.

3

Equate corresponding entries of

AB

and

BA

since

AB = BA

From row 1, col 1:

2 + 2x = -2 \implies 2x = -4 \implies x = -2

.
From row 2, col 1:

5 = 4 - y \implies y = -1

.

Two matrices are equal if and only if all corresponding entries are equal.

4

Verify consistency on remaining entries and calculate

x + y

Row 1, col 2 check:

8 + (-2)(-1) = 10

and

-2 + 12 = 10

.
Row 2, col 2 check:

-4 + 3(-1) = -7

and

2(-2) + 3(-1) = -7

.
Sum:

x + y = -2 + (-1) = -3

.

Verifying consistency ensures the system of equations has a valid unique solution.

Key Concept

Matrix Multiplication Commutativity and Matrix Equality

Estimated Time:2m 30s

Question 13Question →

If the matrix $A = \begin{pmatrix} 1 & 0 & 2 \\ 2 & k & 1 \\ 3 & 1 & 4 \end{pmatrix}$ is singular, what is the value of $k$ ?

\frac{3}{2}

-\frac{3}{2}

\frac{7}{2}

-\frac{1}{2}

Show answer & explanation

Answer: $\frac{3}{2}$

Answer

\frac{3}{2}

A matrix is singular if its determinant equals zero. Expanding the determinant of matrix

A

along its first row gives

1(4k - 1) - 0 + 2(2 - 3k) = -2k + 3

. Setting

-2k + 3 = 0

yields

k = \frac{3}{2}

.

Step-by-Step Solution

1

Apply the condition for a singular matrix

A matrix is singular when its determinant equals zero, so

\det(A) = 0

.

By definition, square matrices with zero determinant are singular.

2

Expand

\det(A)

along the first row

\det(A) = 1(4k - 1) - 0(8 - 3) + 2(2 - 3k) = 4k - 1 + 4 - 6k = -2k + 3

.

Laplace expansion along the first row simplifies computation due to the zero entry.

3

Solve for

k

-2k + 3 = 0 \implies 2k = 3 \implies k = \frac{3}{2}

.

Isolating the variable gives the required value of

k

.

Key Concept

Singular Matrix Condition and 3x3 Determinant Expansion

Question 14Question →

Find the product matrix $AB$ given that $A = \begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix}$ and $B = \begin{pmatrix} 4 & 5 \\ 1 & 2 \end{pmatrix}$ .

\begin{pmatrix} 6 & 9 \\ 3 & 6 \end{pmatrix}

\begin{pmatrix} 4 & 10 \\ 0 & 6 \end{pmatrix}

\begin{pmatrix} 4 & 23 \\ 1 & 8 \end{pmatrix}

\begin{pmatrix} 5 & 7 \\ 1 & 5 \end{pmatrix}

Show answer & explanation

Answer: $\begin{pmatrix} 6 & 9 \\ 3 & 6 \end{pmatrix}$

Answer

\begin{pmatrix} 6 & 9 \\ 3 & 6 \end{pmatrix}

The matrix product

AB

is found by performing row-by-column multiplication. Computing row 1 gives

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

and

(1)(5)+(2)(2) = 9

. Computing row 2 gives

(0)(4)+(3)(1) = 3

and

(0)(5)+(3)(2) = 6

. Thus, the resulting matrix is

\begin{pmatrix} 6 & 9 \\ 3 & 6 \end{pmatrix}

.

Step-by-Step Solution

1

Set up row-by-column matrix multiplication formula for

2 \times 2

matrices.

The entry at row

i

, column

j

of product

AB

is the dot product of row

i

of

A

and column

j

of

B

.

Matrix multiplication requires combining rows of the left matrix with columns of the right matrix.

2

Calculate row 1 entries of the product matrix.

First element:

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

. Second element:

(1)(5) + (2)(2) = 5 + 4 = 9

.

Multiply elements of row 1 of

A

by corresponding elements of columns 1 and 2 of

B

and sum them.

3

Calculate row 2 entries of the product matrix.

Third element:

(0)(4) + (3)(1) = 0 + 3 = 3

. Fourth element:

(0)(5) + (3)(2) = 0 + 6 = 6

.

Multiply elements of row 2 of

A

by corresponding elements of columns 1 and 2 of

B

and sum them.

4

Assemble the product matrix.

AB = \begin{pmatrix} 6 & 9 \\ 3 & 6 \end{pmatrix}

.

Combine all computed row and column entries into a single

2 \times 2

matrix.

Key Concept

Matrix Multiplication (

2 \times 2

Matrix Product)

Question 15Question →

Given the matrices $A = \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & x \\ -2 & 5 \end{pmatrix}$ , if the determinant of the product matrix $AB$ is equal to $121$ , find the value of $x$ .

Show answer & explanation

Answer: 3

Answer

The value of

x

is

3

.

By applying the property

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

, we compute

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

and

\det(B) = (1)(5) - (x)(-2) = 5 + 2x

. Substituting into the equation gives

11(5 + 2x) = 121

, which yields

5 + 2x = 11

and leads to

x = 3

.

Step-by-Step Solution

1

Find the determinant of matrix

A

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

The determinant of a

2 \times 2

matrix

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

is calculated as

ad - bc

.

2

Express the determinant of matrix

B

in terms of

x

\det(B) = (1)(5) - (x)(-2) = 5 + 2x

Apply the

2 \times 2

determinant formula to matrix

B

.

3

Apply the determinant product rule

\det(AB) = \det(A) \cdot \det(B) = 11(5 + 2x) = 121

For any square matrices

A

and

B

,

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

.

4

Solve the resulting linear equation for

x

x = 3

Dividing

121

by

11

gives

5 + 2x = 11

, which simplifies to

2x = 6

and

x = 3

.

Key Concept

Determinant of a Matrix Product

Estimated Time:1m 30s

Question 16Question →

Given the matrices $A = \begin{pmatrix} 2 & x \\ -1 & 3 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 0 \\ 4 & 2 \end{pmatrix}$ , if $\det(A^{-1} B^T) = \frac{1}{5}$ , what is the value of $x$ ?

4

-4

14

-1

Show answer & explanation

Answer: 4

Answer

The value of

x

is 4.

Using the properties

\det(B^T) = \det(B) = 2

and

\det(A^{-1}) = \frac{1}{\det(A)} = \frac{1}{6+x}

, the product rule gives

\det(A^{-1}B^T) = \frac{2}{6+x}

. Equating this to

\frac{1}{5}

gives

6+x = 10

, so

x = 4

.

Step-by-Step Solution

1

Calculate the determinants of matrices

A

and

B

.

\det(A) = (2)(3) - (-1)(x) = 6 + x

and

\det(B) = (1)(2) - (0)(4) = 2$.

The determinant of a 2x2 matrix is computed using

ad - bc

.

2

Apply determinant rules for inverse and transpose matrices.

\det(B^T) = \det(B) = 2

and

\det(A^{-1}) = \frac{1}{\det(A)} = \frac{1}{6+x}$.

The determinant of a transpose equals the original determinant, and the determinant of an inverse is its reciprocal.

3

Express

\det(A^{-1} B^T)

using the product rule of determinants.

\det(A^{-1} B^T) = \det(A^{-1}) \cdot \det(B^T) = \frac{1}{6+x} \cdot 2 = \frac{2}{6+x}$.

For square matrices,

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

.

4

Equate the expression to

\frac{1}{5}

and solve for

x

.

\frac{2}{6+x} = \frac{1}{5} \implies 10 = 6 + x \implies x = 4$.

Cross-multiplying and simplifying gives

x = 4

.

Key Concept

Properties of Determinants (Product, Inverse, and Transpose)

Question 17Question →

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

\begin{pmatrix} 3 & -1 \\ 2 & 16 \end{pmatrix}

\begin{pmatrix} 3 & -2 \\ 0 & 20 \end{pmatrix}

\begin{pmatrix} -1 & -7 \\ 10 & 20 \end{pmatrix}

\begin{pmatrix} 3 & 11 \\ 2 & 16 \end{pmatrix}

Show answer & explanation

Answer: $\begin{pmatrix} 3 & -1 \\ 2 & 16 \end{pmatrix}$

Answer

The product matrix

PQ

is

\begin{pmatrix} 3 & -1 \\ 2 & 16 \end{pmatrix}

.

The matrix containing elements 3, -1 in the top row and 2, 16 in the bottom row correctly applies the row-by-column multiplication rule for matrices

P

and

Q

.

Step-by-Step Solution

1

Set up the row-by-column matrix multiplication

PQ

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

Matrix multiplication requires computing dot products of rows of the first matrix with columns of the second matrix.

2

Calculate the first row elements of the product matrix

Row 1, Column 1:

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

; Row 1, Column 2:

3(-2) + 1(5) = -6 + 5 = -1

Multiply elements of Row 1 of

P

by corresponding elements of Columns 1 and 2 of

Q

and sum them.

3

Calculate the second row elements of the product matrix

Row 2, Column 1:

2(1) + 4(0) = 2 + 0 = 2

; Row 2, Column 2:

2(-2) + 4(5) = -4 + 20 = 16

Multiply elements of Row 2 of

P

by corresponding elements of Columns 1 and 2 of

Q

and sum them.

4

Assemble the resulting

2 \times 2

matrix

PQ = \begin{pmatrix} 3 & -1 \\ 2 & 16 \end{pmatrix}

Combine all computed entry values into matrix form.

Key Concept

Matrix Multiplication (Row-by-Column Rule)

Estimated Time:1m 30s

Question 18Question →

Given the matrices $A = \begin{pmatrix} x & 1 \\ 4 & 2 \end{pmatrix}$ and $B = \begin{pmatrix} 2 & 3 \\ 1 & x \end{pmatrix}$ , what is the sum of all real values of $x$ for which the matrix $C = AB - B^T$ is singular?

\frac{5}{2}

\frac{13}{2}

1

-\frac{5}{2}

Show answer & explanation

Answer: $\frac{5}{2}$

Answer

The sum of all real values of

x

is

\frac{5}{2}

.

Evaluating

AB

yields

\begin{pmatrix} 2x + 1 & 4x \\ 10 & 2x + 12 \end{pmatrix}

and subtracting

B^T = \begin{pmatrix} 2 & 1 \\ 3 & x \end{pmatrix}

gives matrix

C = \begin{pmatrix} 2x - 1 & 4x - 1 \\ 7 & x + 12 \end{pmatrix}

. Setting

\det(C) = (2x - 1)(x + 12) - 7(4x - 1) = 0

results in the quadratic equation

2x^2 - 5x - 5 = 0

. Since the discriminant

\Delta = 65 > 0

, both roots are real, and their sum is given by

-\frac{b}{a} = \frac{5}{2}

.

Step-by-Step Solution

1

Compute the matrix product

AB

AB = \begin{pmatrix} x & 1 \\ 4 & 2 \end{pmatrix} \begin{pmatrix} 2 & 3 \\ 1 & x \end{pmatrix} = \begin{pmatrix} 2x + 1 & 4x \\ 10 & 2x + 12 \end{pmatrix}

Matrix multiplication requires taking the dot product of rows of the first matrix with columns of the second matrix.

2

Find the transpose of matrix

B

, denoted

B^T

B^T = \begin{pmatrix} 2 & 1 \\ 3 & x \end{pmatrix}

The transpose of a matrix is formed by swapping its rows and columns.

3

Compute matrix

C = AB - B^T

C = \begin{pmatrix} 2x + 1 & 4x \\ 10 & 2x + 12 \end{pmatrix} - \begin{pmatrix} 2 & 1 \\ 3 & x \end{pmatrix} = \begin{pmatrix} 2x - 1 & 4x - 1 \\ 7 & x + 12 \end{pmatrix}

Perform element-wise subtraction of matrix

B^T

from matrix

AB

.

4

Set the determinant of matrix

C

to zero for singularity

\det(C) = (2x - 1)(x + 12) - 7(4x - 1) = 2x^2 + 23x - 12 - (28x - 7) = 2x^2 - 5x - 5 = 0

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

5

Find the sum of the real roots of the quadratic equation

2x^2 - 5x - 5 = 0

Discriminant

\Delta = (-5)^2 - 4(2)(-5) = 65 > 0

. By Vieta's formula,

x_1 + x_2 = -\frac{b}{a} = -\frac{-5}{2} = \frac{5}{2}

.

Since the discriminant is positive, two distinct real roots exist, and their sum is given by

-\frac{b}{a}

.

Key Concept

Singular matrix condition, matrix multiplication, transpose operations, and Vieta's formulas.

Question 19Question →

Given the $2 \times 2$ matrix $M = \begin{pmatrix} 7 & -2 \\ 4 & 3 \end{pmatrix}$ , what is the value of $\det(M)$ ?

Show answer & explanation

Answer: 29

Answer

The determinant of matrix

M

is 29.

The determinant of a

2 \times 2

matrix

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

is calculated as

ad - bc

. For the matrix

M = \begin{pmatrix} 7 & -2 \\ 4 & 3 \end{pmatrix}

, we compute

(7)(3) - (-2)(4) = 21 - (-8) = 29

.

Step-by-Step Solution

1

Identify the entries

a, b, c, d

from the matrix

M = \begin{pmatrix} 7 & -2 \\ 4 & 3 \end{pmatrix}

.

a = 7

,

b = -2

,

c = 4

, and

d = 3

.

These entries correspond to the standard

2 \times 2

matrix representation

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

.

2

Compute the determinant using the formula

\det(M) = ad - bc

.

\det(M) = (7)(3) - (-2)(4) = 21 + 8 = 29

.

Multiplying the main diagonal entries and subtracting the product of the off-diagonal entries yields the determinant.

Key Concept

Determinant of a 2x2 Matrix

Question 20Question →

Given the matrices $A = \begin{pmatrix} x & 2 & 1 \\ 3 & 0 & 2 \\ 1 & 4 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 2 & -1 \\ 5 & 3 \end{pmatrix}$ , find the value of $x$ if $\det(A) + \det(B) = 5$ .

Show answer & explanation

Answer: 2

Answer

The value of x is 2.

Evaluating \det(B) gives 11, and cofactor expansion of \det(A) along the top row yields 10 - 8x. Summing these determinants gives (10 - 8x) + 11 = 5, which simplifies to 8x = 16, resulting in x = 2.

Step-by-Step Solution

1

Calculate the determinant of the 2x2 matrix B

\det(B) = 11

For a 2x2 matrix \begin{pmatrix} a & b \\ c & d \end{pmatrix}, the determinant is ad - bc. Thus, \det(B) = (2)(3) - (-1)(5) = 6 + 5 = 11.

2

Calculate the determinant of the 3x3 matrix A in terms of x

\det(A) = 10 - 8x

Expanding along the first row: \det(A) = x\begin{vmatrix} 0 & 2 \\ 4 & 1 \end{vmatrix} - 2\begin{vmatrix} 3 & 2 \\ 1 & 1 \end{vmatrix} + 1\begin{vmatrix} 3 & 0 \\ 1 & 4 \end{vmatrix} = x(0 - 8) - 2(3 - 2) + 1(12 - 0) = -8x - 2 + 12 = 10 - 8x.

3

Set up and solve the linear equation \det(A) + \det(B) = 5

x = 2

Substituting the computed determinants gives (10 - 8x) + 11 = 5, which simplifies to 21 - 8x = 5. Rearranging yields 8x = 16, so x = 2.

Key Concept

Determinants of 2x2 and 3x3 matrices and linear equation solving

Matrices and Determinants

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept