Let A, B, C be three 4 x 4 matrices such that det \ A = 5, det \ B = -3, and det \ C = 1/2. Then the det 2AB^(-1)C^3B^T is

Modern Math

Matrices & Determinants

Hard

Let $A, B, C$ be three $4 \times 4$ matrices such that $det \ A = 5, det \ B = -3$ , and $det \ C = \frac{1}{2}$ . Then the $det$ $2AB^{-1}C^3B^T$ is

Entered answer:

✅ Correct Answer: 10

Properties you need:

$|kA| = k^n|A|$ where $n$ is the order of the matrix

$|AB| = |A| \cdot |B|$

$|A^{-1}| = \dfrac{1}{|A|}$

$|A^T| = |A|$

$|A^m| = |A|^m$

Given: $|A| = 5, \quad |B| = -3, \quad |C| = \dfrac{1}{2}$

Find: $|2AB^{-1}C^3B^T|$

Split using $|XY| = |X| \cdot |Y|$ :

$|2AB^{-1}C^3B^T| = |2A| \cdot |B^{-1}| \cdot |C^3| \cdot |B^T|$

Apply each property:

$|2A| = 2^4 \cdot |A| = 16 \times 5 = 80$

Since these are $4 \times 4$ matrices, the scalar $2$ gets raised to the power $4$ , not $1$ . This is a very common mistake — be careful!

$|B^{-1}| = \dfrac{1}{|B|} = \dfrac{1}{-3}$

$|C^3| = |C|^3 = \left(\dfrac{1}{2}\right)^3 = \dfrac{1}{8}$

$|B^T| = |B| = -3$

Multiply everything together:

$= 80 \times \dfrac{1}{-3} \times \dfrac{1}{8} \times (-3)$

Notice that $\dfrac{1}{-3}$ and $(-3)$ cancel each other out — this happens because $B^{-1}$ and $B^T$ are both present, and their determinants always cancel.

$= 80 \times \dfrac{1}{8}$

$= 10$

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Let $A, B, C$ be three $4 \times 4$ matrices such that $det \ A = 5, det \ B = -3$ , and $det \ C = \frac{1}{2}$ . Then the $det$ $2AB^{-1}C^3B^T$ is