Question 8.9.1: Find |A| if A =[2 3 0 4, 0 5 -1 6, 1 0 -2 3, -3 2 0 -5]....
\text { Find }|A| \text { if } A=\left[\begin{array}{rrrr}2 & 3 & 0 & 4 \\0 & 5 & -1 & 6 \\1 & 0 & -2 & 3 \\-3 & 2 & 0 & -5\end{array}\right]
A = [[2, 3, 0, 4], [0, 5, -1, 6], [1, 0, -2, 3], [-3, 2, 0, -5]]
Step 1:
We plan to use property 3 of the theorem on row and column transformations of a determinant to introduce three zeros in some row or column. This will make it easier to calculate the determinant.
Step 2:
We choose to work with an element of the matrix that equals 1, since this enables us to avoid the use of fractions. If 1 is not an element of the original matrix, it is always possible to introduce the number 1 by using property 2 or 3 of the theorem.
Step 3:
In this example, 1 appears in row 3. We proceed by applying the transformation -2 R_3 + R_1 to row 1 and 3 R_3 + R_4 to row 4. These transformations are allowed according to property 3 of the theorem.
Step 4:
After applying the transformations, we have a new matrix. We can then calculate the determinant of this new matrix.
Step 5:
To calculate the determinant, we expand by the first column. This means we multiply each element in the first column by its cofactor and add them up with appropriate signs.
Step 6:
After expanding, we have a new matrix. We can then apply the transformation 5 C_2 + C_1 to column 1 and 6 C_2 + C_3 to column 3. These transformations are allowed according to property 3 of the theorem.
Step 7:
After applying the transformations, we have a new matrix. We can then calculate the determinant of this new matrix.
Step 8:
To calculate the determinant, we expand by the second row. This means we multiply each element in the second row by its cofactor and add them up with appropriate signs.
Step 9:
After expanding, we have a 2x2 matrix. We can calculate the determinant of this matrix using the definition of the determinant.
Step 10:
Using the definition of the determinant, we calculate the determinant of the 2x2 matrix to be 120.
Final Answer
We plan to use property 3 of the theorem on row and column transformations of a determinant to introduce three zeros in some row or column. It is convenient to work with an element of the matrix that equals 1 , since this enables us to avoid the use of fractions. If 1 is not an element of the original matrix, it is always possible to introduce the number 1 by using property 2 or 3 of the theorem. In this example, 1 appears in row 3 , and we proceed as follows, with the reason for each equality stated at the right.
\left|\begin{array}{rrrr}2 & 3 & 0 & 4 \\0 & 5 & -1 & 6 \\1 & 0 & -2 & 3 \\-3 & 2 & 0 &-5\end{array}\right|=\left|\begin{array}{rrrr}0 & 3 & 4 & -2 \\0 & 5 & -1 & 6 \\1 & 0 & -2 & 3 \\0 & 2 & -6 & 4\end{array}\right| \quad \begin{aligned}& -2 R_3+R_1 \rightarrow R_1 \\\\\\& 3 R_3+R_4\rightarrow R_4\end{aligned}
=(1) \cdot(-1)^{3+1}\left|\begin{array}{rrr}3 & 4 & -2 \\5 & -1 & 6 \\2 & -6 & 4\end{array}\right| \begin{aligned}\text { expand by the }\\\text {first column}\end{aligned}
=\left|\begin{array}{rrr}23 & 4 & 22 \\0 & -1 & 0 \\-28 & -6 & -32\end{array}\right| \quad \begin{aligned}& 5 C_2+C_1 \rightarrow C_1 \\\\& \quad 6 C_2+C_3 \rightarrow C_3\end{aligned}
\begin{aligned}&\begin{array}{ll}=(-1) \cdot(-1)^{2+2}\left|\begin{array}{rr}23 & 22 \\-28 & -32\end{array}\right| & \begin{array}{l}\text { expand by the } \\\text { second row }\end{array} \\\begin{aligned}=(-1)[(23)(-32)-(-28)(22)]\\=120\end{aligned} & & \begin{array}{l}\text { definition of } \\\text { determinant }\end{array} \\\end{array}\\\end{aligned}
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