Matrices and Determinants Practice Questions — JAMB UTME — Page 2

Question 21Question →

Given the matrix $P = \begin{pmatrix} x & 1 & 2 \\ 3 & 0 & x \\ 1 & 2 & 1 \end{pmatrix}$ , for what positive value of $x$ is the determinant of $P$ equal to $3$ ?

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Answer: 2

Answer

The positive value of

x

is

2

.

Expanding the determinant of

P

along the first row yields

\det(P) = -2x^2 + x + 9

. Setting

\det(P) = 3

leads to

-2x^2 + x + 9 = 3

, which rearranges to

2x^2 - x - 6 = 0

. Factoring gives

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

, leading to roots

x = -1.5

and

x = 2

. The positive value is

2

.

Step-by-Step Solution

1

Calculate the determinant of matrix

P

in terms of

x

\det(P) = -2x^2 + x + 9

Expanding along the first row gives

x(0 \cdot 1 - 2x) - 1(3 \cdot 1 - x \cdot 1) + 2(3 \cdot 2 - 0 \cdot 1) = -2x^2 + x + 9

.

2

Set up the quadratic equation using

\det(P) = 3

2x^2 - x - 6 = 0

Subtracting

3

from both sides gives

-2x^2 + x + 6 = 0

, which simplifies to

2x^2 - x - 6 = 0

.

3

Solve the quadratic equation for

x

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

Factoring

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

yields

x = -\frac{3}{2}

and

x = 2

.

4

Select the positive solution as required by the problem statement

x = 2

The problem specifically requests the positive real value of

x

.

Key Concept

3x3 Determinant Evaluation and Quadratic Solving

Question 22Question →

Given the matrix $A = \begin{pmatrix} 1 & 2 & 0 \\ 3 & 4 & 1 \\ 2 & 0 & k \end{pmatrix}$ , if $\det(A) = 10$ , what is the value of $k$ ?

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Answer: -3

Answer

The value of

k

is

-3

.

Expanding the determinant along the first row gives

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

. Setting

-2k + 4 = 10

leads directly to

-2k = 6

, giving

k = -3

.

Step-by-Step Solution

1

Expand the determinant of matrix

A

along the first row.

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

Using cofactor expansion along the top row to find the expression for the determinant.

2

Set the calculated determinant equal to the given value and solve for

k

.

-2k + 4 = 10 \implies -2k = 6 \implies k = -3

Equating the determinant algebraic expression to 10.

Key Concept

Determinant of a 3x3 Matrix

Question 23Question →

Given the matrix $A = \begin{pmatrix} 3 & 2 \\ 1 & 4 \end{pmatrix}$ , if $A^2 - 7A + kI = \mathbf{0}$ , where $I$ is the $2 \times 2$ identity matrix and $\mathbf{0}$ is the $2 \times 2$ zero matrix, what is the value of $k$ ?

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Answer: 10

Answer

The value of

k

is 10.

By matrix multiplication and algebraic evaluation,

A^2 - 7A = -10I

. Substituting into

A^2 - 7A + kI = \mathbf{0}

yields

-10I + kI = \mathbf{0}

, which gives

k = 10

. Alternatively, by the Cayley-Hamilton Theorem, any

2 \times 2

matrix

A

satisfies

A^2 - \text{tr}(A)A + \det(A)I = \mathbf{0}

. Here

\text{tr}(A) = 3 + 4 = 7

and

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

, directly giving

k = \det(A) = 10

.

Step-by-Step Solution

1

Calculate the matrix product

A^2

A^2 = \begin{pmatrix} (3\cdot 3 + 2\cdot 1) & (3\cdot 2 + 2\cdot 4) \\ (1\cdot 3 + 4\cdot 1) & (1\cdot 2 + 4\cdot 4) \end{pmatrix} = \begin{pmatrix} 11 & 14 \\ 7 & 18 \end{pmatrix}

Squaring matrix

A

involves multiplying rows of

A

by columns of

A

.

2

Perform scalar multiplication for

7A

7A = \begin{pmatrix} 21 & 14 \\ 7 & 28 \end{pmatrix}

Each entry of matrix

A

is multiplied by the scalar 7.

3

Subtract

7A

from

A^2

A^2 - 7A = \begin{pmatrix} 11 - 21 & 14 - 14 \\ 7 - 7 & 18 - 28 \end{pmatrix} = \begin{pmatrix} -10 & 0 \\ 0 & -10 \end{pmatrix} = -10I

Subtracting corresponding entries yields a scalar multiple of the identity matrix.

4

Solve for the unknown scalar

k

-10I + kI = \mathbf{0} \implies k = 10

Setting

(k - 10)I = \mathbf{0}

implies

k - 10 = 0

, so

k = 10

.

Key Concept

Matrix Polynomial Equations and Cayley-Hamilton Theorem

Question 24Question →

Let $A$ be a $3 \times 3$ square matrix such that $\det(A) > 0$ . If the matrix satisfies the property $\det(3A^{-1}) = \det(A^T A)$ , what is the value of $\det(A)$ ?

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Answer: 3

Answer

The value of

\det(A)

is 3.

Using fundamental determinant identities for a

3 \times 3

matrix (

n = 3

):

\det(3A^{-1}) = 3^3 \det(A^{-1}) = \frac{27}{\det(A)}

and

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

. Equating them yields

\frac{27}{\det(A)} = (\det(A))^2 \implies (\det(A))^3 = 27

, which yields

\det(A) = 3

.

Step-by-Step Solution

1

Express

\det(3A^{-1})

in terms of

\det(A)

using determinant scaling and inverse properties.

\det(3A^{-1}) = 3^3 \det(A^{-1}) = \frac{27}{\det(A)}.

For an

n \times n

matrix

M

, scaling by constant

k

gives

\det(kM) = k^n \det(M)

, and

\det(M^{-1}) = \frac{1}{\det(M)}

.

2

Express

\det(A^T A)

in terms of

\det(A)

using transpose and multiplication properties.

\det(A^T A) = \det(A^T)\det(A) = (\det(A))^2.

The determinant of a product is the product of determinants, and

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

.

3

Set the two simplified expressions equal to each other and solve for

\det(A)

.

\frac{27}{\det(A)} = (\det(A))^2 \implies (\det(A))^3 = 27 \implies \det(A) = 3.

Taking the cube root of both sides gives the unique real value since

\det(A) > 0

.

Key Concept

Properties of Determinants (Scalar Multiplication, Transpose, Inverse, and Matrix Products)

Estimated Time:1m 30s

Question 25Question →

Given the matrices $A = \begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & -2 \\ 0 & 3 \end{pmatrix}$ , if $C = AB - 2I$ , where $I$ is the $2 \times 2$ identity matrix, what is the determinant of matrix $C$ ?

3

-3

0

15

Show answer & explanation

Answer: $3$

Answer

The determinant of matrix C is 3.

Multiplying matrices A and B using standard row-by-column multiplication yields \begin{pmatrix} 2 & -1 \\ 3 & 6 \end{pmatrix}. Subtracting the scaled identity matrix \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} produces matrix C = \begin{pmatrix} 0 & -1 \\ 3 & 4 \end{pmatrix}. Taking the determinant ad - bc gives (0)(4) - (-1)(3) = 3.

Step-by-Step Solution

1

Compute the matrix product AB.

AB = \begin{pmatrix} (2)(1) + (1)(0) & (2)(-2) + (1)(3) \\ (3)(1) + (4)(0) & (3)(-2) + (4)(3) \end{pmatrix} = \begin{pmatrix} 2 & -1 \\ 3 & 6 \end{pmatrix}

Matrix multiplication uses dot products of rows of the first matrix and columns of the second matrix.

2

Subtract 2I from AB to find matrix C.

C = \begin{pmatrix} 2 & -1 \\ 3 & 6 \end{pmatrix} - \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ 3 & 4 \end{pmatrix}

The identity matrix I multiplied by scalar 2 has 2 on its main diagonal and 0 elsewhere.

3

Calculate the determinant of matrix C.

\det(C) = (0)(4) - (-1)(3) = 0 + 3 = 3

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

Key Concept

Matrix multiplication, matrix arithmetic operations, and evaluation of 2x2 determinants.

Estimated Time:1m 30s

Question 26Question →

Given the matrices $A = \begin{pmatrix} 3 & y \\ 2 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 4 \\ x & 2 \end{pmatrix}$ , if $AB = \begin{pmatrix} 9 & 16 \\ 5 & 10 \end{pmatrix}$ , what is the value of $x + y$ ?

5

6

7

9

Show answer & explanation

Answer: $5$

Answer

The value of

x + y

is

5

.

Multiplying matrix

A

by matrix

B

using standard row-by-column multiplication yields

AB = \begin{pmatrix} 3 + xy & 12 + 2y \\ 2 + x & 10 \end{pmatrix}

. Setting this equal to

\begin{pmatrix} 9 & 16 \\ 5 & 10 \end{pmatrix}

gives

2 + x = 5 \implies x = 3

and

12 + 2y = 16 \implies y = 2

. Thus,

x + y = 3 + 2 = 5

.

Step-by-Step Solution

1

Compute the product matrix

AB

using matrix multiplication rules.

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

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

2

Equate the elements of

AB

with the given matrix

\begin{pmatrix} 9 & 16 \\ 5 & 10 \end{pmatrix}

to solve for

x

and

y

.

From row 2, column 1:

2 + x = 5 \implies x = 3

. From row 1, column 2:

12 + 2y = 16 \implies 2y = 4 \implies y = 2

.

Two matrices are equal if and only if their corresponding elements are equal.

3

Calculate the sum

x + y

.

x + y = 3 + 2 = 5

Substituting the values found for

x

and

y

into the requested expression.

Key Concept

Matrix Multiplication and Equality of Matrices

Alternative Method

Verification can be done by checking row 1, column 1:

3 + xy = 3 + (3)(2) = 9

, which matches the given matrix entry.

Estimated Time:1m 15s

Question 27Question →

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

\begin{pmatrix} 4 & 7 \\ 6 & -3 \end{pmatrix}

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

\begin{pmatrix} 2 & 13 \\ 4 & -1 \end{pmatrix}

\begin{pmatrix} 4 & 9 \\ 6 & -3 \end{pmatrix}

Show answer & explanation

Answer: $\begin{pmatrix} 4 & 7 \\ 6 & -3 \end{pmatrix}$

Answer

The matrix product

AB

is

\begin{pmatrix} 4 & 7 \\ 6 & -3 \end{pmatrix}

.

To find

AB

, each element

c_{ij}

is formed by taking the dot product of row

i

of matrix

A

and column

j

of matrix

B

. Performing these steps gives

c_{11} = 2(1) + 1(2) = 4

,

c_{12} = 2(4) + 1(-1) = 7

,

c_{21} = 0(1) + 3(2) = 6

, and

c_{22} = 0(4) + 3(-1) = -3

, yielding

\begin{pmatrix} 4 & 7 \\ 6 & -3 \end{pmatrix}

.

Step-by-Step Solution

1

Multiply the first row of

A

by the first column of

B

to find element

(1,1)

.

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

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

2

Multiply the first row of

A

by the second column of

B

to find element

(1,2)

.

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

Evaluates the element in row 1, column 2 of the resulting matrix.

3

Multiply the second row of

A

by the first column of

B

to find element

(2,1)

.

0(1) + 3(2) = 0 + 6 = 6

Evaluates the element in row 2, column 1 of the resulting matrix.

4

Multiply the second row of

A

by the second column of

B

to find element

(2,2)

.

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

Evaluates the element in row 2, column 2 of the resulting matrix.

Key Concept

Matrix Multiplication

Estimated Time:1m 0s

Question 28Question →

If the determinant of the matrix $A = \begin{pmatrix} k & 1 & 2 \\ 3 & -1 & 0 \\ 2 & 4 & 1 \end{pmatrix}$ is equal to $17$ , what is the value of $k$ ?

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Answer: 8

Answer

The value of

k

is

8

.

Expanding the matrix determinant along the first row gives

\det(A) = k(-1) - 1(3) + 2(14) = -k + 25

. Setting

-k + 25 = 17

leads to

-k = -8

, so

k = 8

.

Step-by-Step Solution

1

Perform cofactor expansion along the first row of matrix

A

.

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

Expanding along the first row uses the formula \det(A) = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13}.

2

Evaluate the products and simplify the algebraic expression for the determinant.

\det(A) = -k - 3 + 28 = -k + 25

Simplifying each sub-determinant term yields a linear expression in

k

.

3

Equate the expression to

17

and solve for

k

.

-k + 25 = 17 \implies k = 8

Subtracting 25 from both sides gives

-k = -8

, which simplifies to

k = 8

.

Key Concept

Determinant of a 3x3 Matrix

Question 29Question →

A $2 \times 2$ matrix is given by $A = \begin{pmatrix} x + 2 & 3 \\ 4 & x - 2 \end{pmatrix}$ . If $A$ is a singular matrix, what is the positive value of $x$ ?

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Answer: 4

Answer

The positive value of

x

is

4

.

A matrix is singular when its determinant is zero. Expanding

\det(A) = (x+2)(x-2) - 12 = 0

yields

x^2 - 16 = 0

. Solving

x^2 = 16

gives

x = \pm 4

. The positive value is

4

.

Step-by-Step Solution

1

Apply the condition for a matrix to be singular

\det(A) = 0

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

2

Formulate the determinant equation for matrix

A

(x + 2)(x - 2) - (3)(4) = 0

The determinant of a

2 \times 2

matrix

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

is calculated as

ad - bc

.

3

Expand and simplify the algebraic equation

x^2 - 4 - 12 = 0 \implies x^2 - 16 = 0

Expanding

(x+2)(x-2)

gives

x^2 - 4

, and subtracting

12

yields

x^2 - 16

.

4

Solve for

x

and select the positive root

x^2 = 16 \implies x = 4 \text{ or } x = -4 \implies x = 4

Taking the square root gives two roots, and the problem explicitly specifies the positive value.

Key Concept

Singular Matrices and Determinants

Estimated Time:1m 15s

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