Math Problem Statement
Let M be the matrix [[[5],[4],[-2],[3]],[[5],[7],[-1],[8]],[[5],[7],[6],[10]],[[5],[7],[1],[9]]]
The determinant of M can be expressed as the constant 5 times the [315] determinant of the single 3 x 3 matrix [[[3],[1],[5]],[[3],[a],[b]],[[3],[c],[d]]] then a+b+c+d=
a) 28
b) 30
c) 32
d) 34
e) NOTA
Solve it sep by step
Solution
To solve this problem, we need to understand the structure of the given 4x4 matrix and the conditions described in the problem statement.
Step 1: Understanding Matrix
The matrix is:
5 & 4 & -2 & 3 \\ 5 & 7 & -1 & 8 \\ 5 & 7 & 6 & 10 \\ 5 & 7 & 1 & 9 \end{bmatrix}$$ ### Step 2: Recognize the Common Factor and Reduced Matrix The problem states that the determinant of $$M$$ can be expressed as $$5 \times$$ the determinant of a 3x3 matrix. This suggests that by factoring out the common elements in the first column of $$M$$, we might simplify the problem. Looking at $$M$$, notice that the first column consists entirely of the number 5: $$M = \begin{bmatrix} 5 & 4 & -2 & 3 \\ 5 & 7 & -1 & 8 \\ 5 & 7 & 6 & 10 \\ 5 & 7 & 1 & 9 \end{bmatrix}$$ We can factor out a 5 from each element of the first column, effectively factoring out a 5 from the entire matrix. Let's perform this operation: $$M = 5 \times \begin{bmatrix} 1 & \frac{4}{5} & -\frac{2}{5} & \frac{3}{5} \\ 1 & \frac{7}{5} & -\frac{1}{5} & \frac{8}{5} \\ 1 & \frac{7}{5} & \frac{6}{5} & 2 \\ 1 & \frac{7}{5} & \frac{1}{5} & \frac{9}{5} \end{bmatrix}$$ The determinant of $$M$$ will then be: $$\text{det}(M) = 5 \times \text{det}\left(\begin{bmatrix} 1 & \frac{4}{5} & -\frac{2}{5} & \frac{3}{5} \\ 1 & \frac{7}{5} & -\frac{1}{5} & \frac{8}{5} \\ 1 & \frac{7}{5} & \frac{6}{5} & 2 \\ 1 & \frac{7}{5} & \frac{1}{5} & \frac{9}{5} \end{bmatrix}\right)$$ ### Step 3: Simplify the Matrix Notice that all rows in the new matrix start with 1. Subtracting the first row from each subsequent row, we obtain a simpler matrix: $$\text{Subtracting Row 1 from Rows 2, 3, and 4:}$$ $$\begin{bmatrix} 1 & \frac{4}{5} & -\frac{2}{5} & \frac{3}{5} \\ 0 & \frac{3}{5} & \frac{1}{5} & \frac{5}{5} \\ 0 & \frac{3}{5} & \frac{8}{5} & \frac{7}{5} \\ 0 & \frac{3}{5} & \frac{3}{5} & \frac{6}{5} \end{bmatrix}$$ Removing the first column of 1's (since the determinant of the new matrix is the same when the first row is not full of zeros), the matrix becomes a 3x3 matrix: $$\text{New 3x3 Matrix:} \quad \begin{bmatrix} 3 & 1 & 5 \\ 3 & 8 & 7 \\ 3 & 3 & 6 \end{bmatrix}$$ ### Step 4: Understand the New Matrix Structure We were provided with the format of the matrix: $$\begin{bmatrix} 3 & 1 & 5 \\ 3 & a & b \\ 3 & c & d \end{bmatrix}$$ Equating with our obtained 3x3 matrix: - $$a = 8$$ - $$b = 7$$ - $$c = 3$$ - $$d = 6$$ Thus: $$a + b + c + d = 8 + 7 + 3 + 6 = 24$$ ### Step 5: Final Comparison with Options According to the options provided, $$a + b + c + d$$: $$\text{a) } 28, \quad \text{b) } 30, \quad \text{c) } 32, \quad \text{d) } 34, \quad \text{e) NOTA$$ Since $$a + b + c + d = 24$$ does not match any of the given options, the correct answer is: **e) NOTA (None of the Above)**. ### Conclusion: $$\boxed{\text{The answer is e) NOTA.}}$$ --- Would you like more details on any part of this solution, or have other questions? Here are some questions you might consider next: 1. How to find determinants of larger matrices efficiently? 2. Why does factoring out a column value help in simplifying matrix determinants? 3. How to determine if a matrix has linearly dependent rows or columns? 4. What are some properties of determinants useful in problem-solving? 5. How does cofactor expansion work in finding determinants? **Tip:** When dealing with matrices, always check for common factors or repeating rows/columns that might help simplify the determinant calculation.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Matrices
Determinants
Matrix operations
Formulas
Matrix determinant properties
Theorems
Properties of determinants
Suitable Grade Level
Grades 11-12