Question

In Exercises 53–54, evaluate each determinant. ﻿∣∣31−23∣∣7015∣∣3007∣∣9−635∣∣\begin{vmatrix}
\begin{vmatrix} 3 & 1 \ -2 & 3 \end{vmatrix} & \begin{vmatrix} 7 & 0 \ 1 & 5 \end{vmatrix} \
\begin{vmatrix} 3 & 0 \ 0 & 7 \end{vmatrix} & \begin{vmatrix} 9 & -6 \ 3 & 5 \end{vmatrix}
\end{vmatrix}​​3−2​13​​​30​07​​​​71​05​​​93​−65​​​​

Accepted Answer

Step 1: Recognize that the problem asks to evaluate the determinant of the product of two matrices. The determinant of a product of matrices equals the product of their determinants. So, we use the property: $$\det(AB) = \det(A) 	imes \det(B). $$Step 2: Identify the two matrices from the problem. Let matrix $$A = \begin{bmatrix} 3 & 1 \ -2 & 3 \end{bmatrix}$$ and matrix $$B = \begin{bmatrix} 7 & 0 \ 1 & 5 \end{bmatrix}. $$Step 3: Calculate the determinant of matrix $$A$$ using the formula for a 2x2 matrix: $$\det(A) = a d - b c$$, where $$A = \begin{bmatrix} a & b \ c & d \end{bmatrix}. So$$, $$\det(A) = (3)(3) - (1)(-2). $$Step 4: Calculate the determinant of matrix $$B$$ similarly: $$\det(B) = (7)(5) - (0)(1). $$Step 5: Multiply the two determinants found in steps 3 and 4 to get the determinant of the product matrix: $$\det(AB) = \det(A) 	imes \det(B).$$

In Exercises 53–54, evaluate each determinant. $\begin{vmatrix}\[\begin{vmatrix}\) 3 & 1 \\ -2 & 3 \(\end{vmatrix}\) & \(\begin{vmatrix}\) 7 & 0 \\ 1 & 5 \(\end{vmatrix}\) \(\begin{vmatrix}\) 3 & 0 \\ 0 & 7 \(\end{vmatrix}\) & \(\begin{vmatrix}\) 9 & -6 \\ 3 & 5 \(\end{vmatrix}\]\end{vmatrix}$

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Key Concepts

Determinant of a 2x2 Matrix

Properties of Determinants

Matrix Notation and Evaluation