Let A be a 3 \times 3 square matrix such that \det(A) > 0. If the matrix satisfies the pro… — JAMB UTME

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 $\det(A)$ ?

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

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)}

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)

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

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