Compute the determinant by cofactor expansion. At each​ step, choose a row or column that involves the least amount of computation.
StartAbsoluteValue Start 4 By 4 Matrix 1st Row 1st Column 9 2nd Column 0 3rd Column 0 4st Column 5 2nd Row 1st Column 1 2nd Column 7 3rd Column 2 4st Column negative 3 3rd Row 1st Column 3 2nd Column 0 3rd Column 0 4st Column 0 4st Row 1st Column 2 2nd Column 3 3rd Column 1 4st Column 1 EndMatrix EndAbsoluteValue
 


To compute the determinant of the given \( 4 \times 4 \) matrix using cofactor expansion, we will choose a row or column that minimizes computation, which means selecting one with the most zeros.

The matrix is:

\[
\begin{vmatrix}
9 & 0 & 0 & 5 \\
1 & 7 & 2 & -3 \\
3 & 0 & 0 & 0 \\
2 & 3 & 1 & 1 \\
\end{vmatrix}
\]

We can expand along the third row since it contains the most zeros.

### Cofactor expansion along the third row:

\[
\text{det}(A) = 3 \cdot C_{31} + 0 \cdot C_{32} + 0 \cdot C_{33} + 0 \cdot C_{34}
\]

Only the first term remains (because of the zeros in the third row):

\[
\text{det}(A) = 3 \cdot C_{31}
\]

Where \( C_{31} \) is the cofactor of the element at position (3,1). The cofactor is calculated as:

\[
C_{31} = (-1)^{3+1} \cdot \text{det of the } 3 \times 3 \text{ submatrix formed by deleting row 3 and column 1}
\]

The submatrix formed by deleting row 3 and column 1 is:

\[
\begin{pmatrix}
0 & 0 & 5 \\
7 & 2 & -3 \\
3 & 1 & 1
\end{pmatrix}
\]

Now, we calculate the determinant of this \( 3 \times 3 \) submatrix by expanding along the first row (which has two zeros):

\[
\text{det} \begin{pmatrix}
0 & 0 & 5 \\
7 & 2 & -3 \\
3 & 1 & 1
\end{pmatrix} = 0 \cdot C_{11} + 0 \cdot C_{12} + 5 \cdot C_{13}
\]

Only the last term remains, so we calculate \( C_{13} \):

\[
C_{13} = (-1)^{1+3} \cdot \text{det of the } 2 \times 2 \text{ submatrix} \begin{pmatrix} 7 & 2 \\ 3 & 1 \end{pmatrix}
\]

The determinant of the \( 2 \times 2 \) submatrix is:

\[
\text{det} \begin{pmatrix} 7 & 2 \\ 3 & 1 \end{pmatrix} = (7 \cdot 1) - (2 \cdot 3) = 7 - 6 = 1
\]

So, \( C_{13} = (-1)^{1+3} \cdot 1 = 1 \).

Therefore, the determinant of the \( 3 \times 3 \) submatrix is:

\[
\text{det} \begin{pmatrix}
0 & 0 & 5 \\
7 & 2 & -3 \\
3 & 1 & 1
\end{pmatrix} = 5 \cdot 1 = 5
\]

### Final determinant calculation:

\[
\text{det}(A) = 3 \cdot 5 = 15
\]

Thus, the determinant of the matrix is \( \boxed{15} \).

---

Would you like more details on any step or have any questions?

Here are 5 related questions you could explore:
1. What is cofactor expansion in a general sense for any matrix size?
2. How is a cofactor defined mathematically?
3. How does choosing a row or column with the most zeros simplify calculations?
4. What other methods exist for finding the determinant of a \( 4 \times 4 \) matrix?
5. What are the applications of determinants in linear algebra?

**Tip:** Always look for rows or columns with zeros when calculating determinants using cofactor expansion to minimize computation.

Compute the determinant by cofactor expansion. At each step, choose a row or column that involves the least amount of computation. StartAbsoluteValue Start 4 By 4 Matrix 1st Row 1st Column 9 2nd Column 0 3rd Column 0 4st Column 5 2nd Row 1st Column 1 2nd Column 7 3rd Column 2 4st Column negative 3 3rd Row 1st Column 3 2nd Column 0 3rd Column 0 4st Column 0 4st Row 1st Column 2 2nd Column 3 3rd Column 1 4st Column 1 EndMatrix EndAbsoluteValue

Learn how to compute the determinant of a 4x4 matrix using cofactor expansion, minimizing computation by selecting a row with zeros. This example demonstrates key linear algebra concepts such as cofactor expansion and determinant properties. Suitable for high school students (Grades 10-12).

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