Find the determinants in Exercises 5–10 by row reduction to echelon form

# Find the determinants in Exercises 5–10 by row reduction to echelon form

H
240 points

Find the determinants in Exercises 5–10 by row reduction to echelon form.

\begin{vmatrix} 1 & 3 & 2 & -4 \\ 0 & 1 & 2 & -5 \\ 2 & 7 & 6 & -3 \\ -3 & -10 & -7 & 2 \end{vmatrix}

Find the
hansel

8.7k points

Keep in mind the following when solving this type of problem

If A is a square matrix then

• A multiple of one row added to another row does not change the determinant of the matrix
• If two rows are swapped with each other then the new matrix B is the negative of matrix A, or det B = -det A
• If one row of A is multiplied by a constant k, then for the new matrix det B = k det A

In this case I will first do,

$$R_3=R_3-2R_1$$ $$R_4=R_4+3R_1$$

and then

$$R_3=R_3-R_2$$ $$R_4=R_4+R_2$$

to arrive at

$$\begin{vmatrix} 1 & 3 & 2 & -4 \\ 0 & 1 & 2 & -5 \\ 0 & 0 & 0 & 10 \\ 0 & 0 & 1 & -15 \end{vmatrix}$$

now swap $R_3$ and $R_4$ which means the we take the negative of the determinant

$$-\begin{vmatrix} 1 & 3 & 2 & -4 \\ 0 & 1 & 2 & -5 \\ 0 & 0 & 1 & -15 \\ 0 & 0 & 0 & 10 \end{vmatrix}$$

this is a triangular matrix so the determinant is the product of the diagonals,

$$=(-1)(1 \cdot 1 \cdot 1 \cdot 10)=-10$$

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