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

To solve this problem, we need to understand the structure of the given 4x4 matrix \( M \) and the conditions described in the problem statement.

### Step 1: Understanding Matrix \( M \)

The matrix \( M \) is:

\[
M = \begin{bmatrix}
5 & 4 & -2 & 3 \\
5 & 7 & -1 & 8 \\
5 & 7 & 6 & 10 \\
5 & 7 & 1 & 9
\end{bmatrix}
\]

### 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.

Learn how to solve a determinant problem involving a 4x4 matrix M, where the determinant is expressed as a multiple of a 3x3 submatrix. Follow a detailed step-by-step solution to find the values of a, b, c, and d. Suitable for advanced high school students.

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