Math Problem Statement
◗✉❛❧ ❞❛6 ❛✜2♠❛AM❡6 6❡❣✉✐♥9❡6 E ✈❡2❞❛❞❡✐2❛ ❛❝❡2❝❛ ❞❛ ♠❛92✐③ A = 1 2 a 3 0 −1 2a 0 0 0 3a 1 0 0 0 1 ❄ det A 9❡♠ ✈❛❧♦2❡6 ❞✐❢❡2❡♥9❡6 ♣❛2❛ ❞✐❢❡2❡♥9❡6 a ∈ R✳ A E ✉♠❛ ♠❛92✐③ ✐♥✈❡29N✈❡❧ ♣❛2❛ 3✉❛❧3✉❡2 a ∈ R✳ det A = − det 1 2 a 0 −1 2a 0 0 3a det A = − det 3 a 2 1 0 2a −1 0 1 3a 0 0 1 0 0 0
Solution
It looks like you're dealing with the determinant of a 4×4 matrix and its properties. Let's go step by step to determine .
Given:
\begin{bmatrix} 1 & 2 & a & 3 \\ 0 & -1 & 2a & 0 \\ 0 & 0 & 3a & 1 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$ ### Step 1: Expanding the Determinant Since the last column has only one nonzero entry at $$A_{4,4} = 1$$, we expand along the fourth column: $$\det(A) = 1 \times \det \begin{bmatrix} 1 & 2 & a \\ 0 & -1 & 2a \\ 0 & 0 & 3a \end{bmatrix}$$ Since the last row of this 3×3 matrix has only one nonzero entry at $$(3,3)$$, we expand along the third column: $$\det(A) = 1 \times (3a) \times \begin{vmatrix} 1 & 2 \\ 0 & -1 \end{vmatrix}$$ ### Step 2: Computing the 2×2 Determinant $$\begin{vmatrix} 1 & 2 \\ 0 & -1 \end{vmatrix} = (1)(-1) - (2)(0) = -1$$ ### Step 3: Final Calculation $$\det(A) = 1 \times (3a) \times (-1) = -3a$$ Thus, $$\det(A) = -3a$$ If $$a \neq 0$$, the matrix is invertible. If $$a = 0$$, then $$\det(A) = 0$$, meaning $$A$$ is singular. --- ### Additional Questions for You: 1. Would you like a deeper explanation of determinant expansion? 2. Do you need verification for the determinant using a different method? 3. Should we check the inverse of $$A$$ if it exists? 4. Would you like an interpretation of how the determinant affects matrix properties? 5. Do you need help with determinant properties in general? #### **Tip**: If a matrix has a row or column with mostly zeros, expand along that row or column to simplify calculations.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Determinants
Matrix Expansion
Properties of Determinants
Formulas
det(A) = 1 × det(matrix)
Theorems
Determinant Expansion Theorem
Properties of Determinants (det(A) = -det(A))
Suitable Grade Level
Grades 11-12