What is the determinant of the matrix (073107300)\begin{pmatrix} 0 & 7 & 3 \\ 1 & 0 & 7 \\ 3 & 0 & 0 \end{pmatrix} ? Simplify your answer. (determinant, matrix, linear algebra, cofactor expansion, square matrix)

Answer: The determinant of the matrix is 147147.

Step-by-step solution

Step 1: Choose a row or column for cofactor expansion

To find the determinant of a 3×33 \times 3 matrix, we can use the cofactor expansion method. This involves choosing a row or a column and then calculating the sum of the products of each element with its corresponding cofactor. It's generally easiest to choose a row or column that contains the most zeros, as this simplifies calculations. In this matrix, the third row and the second column both have two zeros.

Step 2: Expand along the third row

Let's expand along the third row. The elements are a31=3a_{31}=3, a32=0a_{32}=0, and a33=0a_{33}=0. The cofactor for an element aija_{ij} is (−1)i+j(-1)^{i+j} times the determinant of the submatrix obtained by removing the ii -th row and jj -th column. Since the second and third elements of the third row are zero, their contributions to the determinant will be zero, simplifying the calculation significantly.

Step 3: Calculate the determinant of the 2×22 \times 2 submatrix

Now we need to calculate the determinant of the 2×22 \times 2 submatrix: ∣7307∣\begin{vmatrix} 7 & 3 \\ 0 & 7 \end{vmatrix}. The determinant of a 2×22 \times 2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix} is given by ad−bcad - bc.

Step 4: Substitute and simplify

Substitute the determinant of the 2×22 \times 2 submatrix back into the expansion formula. The terms involving zeros will vanish, leaving only the product of 33, 11 (from (−1)3+1(-1)^{3+1}), and the determinant of the submatrix.

Step 5: Final calculation

Perform the final multiplication to get the determinant of the matrix.

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