Given the matrices A = \begin{pmatrix} 2 & x \\ -1 & 3 \end{pmatrix} and B = \begin{pmatri… — JAMB UTME

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 $x$ ?

4Answer
B
-4
C
14
D
-1

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

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

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.

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)

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

Answer

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Common Mistakes

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