Two 2 \times 2 matrices are given as A = \begin{pmatrix} 2 & -1 \\ 4 & 3 \end{pmatrix} and… — JAMB UTME

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 $k$ ?

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

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

Evaluate the determinant of matrix B

\det(B) = k

Calculating the determinant of matrix

B

yields

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

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

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