Question

Evaluate det A where: \(\mathrm{A}=\begin{array}{rrr}10 & 1 & 5 \\ 3 & -6 & 9 \\ 12 & 6 & 1\end{array}\)

Short answer

The determinant of matrix A is calculated using the first-row expansion to obtain sub-matrices and their determinants: det(A) = a_11 * det(A_11) - a_12 * det(A_12) + a_13 * det(A_13) = (10)(-60) - (1)(-105) + (5)(90) = -600 + 105 + 450 = -45. Thus, det(A) = -45.

Step-by-step solution

Set up the first-row expansion

For a 3x3 matrix, the determinant can be calculated using the first-row expansion as follows: det(A) = a_11 * det(A_11) - a_12 * det(A_12) + a_13 * det(A_13), where a_ij denotes the (i, j)-th element of A, and A_ij is the sub-matrix obtained by deleting the i-th row and j-th column of A. In the given matrix: \(A = \begin{bmatrix} 10 & 1 & 5 \\ 3 & -6 & 9 \\ 12 & 6 & 1 \end{bmatrix}\), the first-row elements are: a_11 = 10, a_12 = 1, and a_13 = 5. We need to find the sub-matrices A_11, A_12, and A_13.

Calculate sub-matrices

To obtain the sub-matrices A_11, A_12, and A_13, we delete the first row and corresponding columns of A: \(A_{11} = \begin{bmatrix} -6 & 9 \\ 6 & 1 \end{bmatrix}\), \(A_{12} = \begin{bmatrix} 3 & 9 \\ 12 & 1 \end{bmatrix}\), \(A_{13} = \begin{bmatrix} 3 & -6 \\ 12 & 6 \end{bmatrix}\). Now, we will find the determinants of these sub-matrices: det(A_11) = |\(A_{11}\)| = (-6)(1) - (9)(6) = -6 - 54 = -60, det(A_12) = |\(A_{12}\)| = (3)(1) - (9)(12) = 3 - 108 = -105, det(A_13) = |\(A_{13}\)| = (3)(6) - (-6)(12) = 18 + 72 = 90.

Combine the results

Finally, we calculate the determinant of A using the first-row expansion: det(A) = a_11 * det(A_11) - a_12 * det(A_12) + a_13 * det(A_13) = (10)(-60) - (1)(-105) + (5)(90) = -600 + 105 + 450 = -45. So, the determinant of matrix A is -45.

Key concepts

First-Row Expansion

When evaluating the determinant of a 3x3 matrix, first-row expansion is a widely used method. The process is straightforward yet crucial to understand. Starting off, the matrix is symbolically written out to highlight the first row from which the expansion will be conducted.

For our matrix \(A\), the first-row elements are \(a_{11} = 10\), \(a_{12} = 1\), and \(a_{13} = 5\). The first row is pivotal because each element here is multiplied by the determinant of its respective sub-matrix, which is obtained by removing the first row and the column of the current element.

To account for the alternating signs in the determinant calculation, we apply a pattern of plus and minus signs starting with a plus for \(a_{11}\) and alternating thereafter, resulting in the formula: \(\text{det}(A) = a_{11} \cdot \text{det}(A_{11}) - a_{12} \cdot \text{det}(A_{12}) + a_{13} \cdot \text{det}(A_{13})\). This pattern is a critical aspect of matrix algebra known as the cofactor expansion.

Sub-Matrix Calculation

The sub-matrix calculation is essential for simplifying a larger matrix into smaller, more manageable pieces. Once the elements for the first-row expansion are identified, we proceed to find the respective 2x2 sub-matrices for each.

To find \(A_{11}\), \(A_{12}\), and \(A_{13}\), we must remove the entirety of the row and column for each first-row element. For instance, \(A_{11}\) is created by excluding the first row and first column of the original matrix, leaving us with \[\begin{bmatrix} -6 & 9 \ 6 & 1 \end{bmatrix}\]. The same process is applied to obtain \(A_{12}\) and \(A_{13}\).

The determinants of these sub-matrices are then calculated using the simple formula for a 2x2 matrix: \(\text{det}(B) = b_{11} \cdot b_{22} - b_{12} \cdot b_{21}\), where \(B\) is any 2x2 sub-matrix and \(b_{ij}\) are the elements of \(B\). Calculating these determinants is a stepping stone towards finding the determinant of the original matrix.

3x3 Matrix Determinant

The determinant of a 3x3 matrix condenses its properties into a single value, fundamental in various areas of linear algebra such as solving systems of linear equations and understanding matrix invertibility.

Once the determinants of the sub-matrices are obtained, as we demonstrated with the calculations of \(\text{det}(A_{11})\), \(\text{det}(A_{12})\), and \(\text{det}(A_{13})\), these values are incorporated back into the expansion formula to find the determinant of the original 3x3 matrix. By multiplying each first-row element by the corresponding sub-matrix determinant and applying the correct sign, we are able to sum these values to find the final determinant.

Applying the formula for our matrix \(A\), we arrived at \(\text{det}(A) = -45\). This single number, the determinant, is packed with information—it can signal if matrix \(A\) is invertible (a non-zero determinant) or if there are linear dependencies among its rows (a zero determinant). Understanding this process deeply is essential for grasping broader concepts in linear algebra.