Question

Evaluate each determinant in Exercises 49–52. ﻿∣−2−3351−400122−32011∣\begin{vmatrix}
-2 & -3 & 3 & 5 \
1 & -4 & 0 & 0 \
1 & 2 & 2 & -3 \
2 & 0 & 1 & 1
\end{vmatrix}​−2112​−3−420​3021​50−31​​

Accepted Answer

Step 1: Write down the determinant of the 4x4 matrix as given:

$$\left| \begin{array}{cccc} -2 & -3 & 3 & 5 \ 1 & -4 & 0 & 0 \ 1 & 2 & 2 & -3 \ 2 & 0 & 1 & 1 \end{array} ight|$$ Step 2: Choose a row or column to expand the determinant. It is often easiest to pick a row or column with zeros or smaller numbers to simplify calculations. For this matrix, the second row has two zeros, so expanding along the second row is a good choice. Step 3: Use the cofactor expansion formula along the chosen row. For each element $$a_{ij} in $$the row, calculate the minor determinant by removing the $$ith $$row and $$jth $$column, then multiply by $$(-1)^{i+j}$$ and the element $$a_{ij}. $$Sum all these products to get the determinant:

$$	ext{det}(A) = \sum_{j=1}^4 (-1)^{2+j} a_{2j} M_{2j}$$

where $$M_{2j} is $$the minor determinant after removing row 2 and column $$j. $$Step 4: Calculate each minor determinant $$M_{2j}$$, which are 3x3 determinants. For each 3x3 matrix, you can use the rule of Sarrus or cofactor expansion again to find their values. Step 5: Substitute the values of the minors back into the cofactor expansion formula, multiply by the corresponding elements and signs, then sum all terms to find the value of the original 4x4 determinant.

Evaluate each determinant in Exercises 49–52. $\begin{vmatrix}\)-2 & -3 & 3 & 5 \\1 & -4 & 0 & 0 \\1 & 2 & 2 & -3 \\2 & 0 & 1 & 1\(\end{vmatrix}$

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Key Concepts

Determinant of a Matrix

Expansion by Minors (Cofactor Expansion)

Properties of Determinants

Evaluate each determinant in Exercises 49–52. ∣−2−3351−400122−32011∣\(\begin{vmatrix}\)-2 & -3 & 3 & 5 \\1 & -4 & 0 & 0 \\1 & 2 & 2 & -3 \\2 & 0 & 1 & 1\(\end{vmatrix}\)−2112−3−420302150−31