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

Answered by Aaron 2 years ago

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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Asked: 2 years ago