About Determinant from Triangular Form:
When a matrix is in upper triangular form, we have a special property that tells us the determinant can be found as the product of the diagonal entries. There are a few things to keep in mind as we use this method. If we swap two rows, the sign of the determinant changes. This is true each time we swap rows, so we must keep track of the sign. Additionally, if we replace a row by the sum of that row and a nonzero constant multiple of some other row, there is no change to our determinant.
Test Objectives
- Demonstrate the ability to place a matrix in upper triangular form
- Demonstrate the ability to find the determinant of a matrix
#1:
Instructions: find the determinant.
$$a)\hspace{.2em}$$ $$\left[ \begin{array}{cc}2 & 6\\ 9 & 1\end{array}\right]$$
$$b)\hspace{.2em}$$ $$\left[ \begin{array}{cc}3 & 14\\ 0 & 0\end{array}\right]$$
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#2:
Instructions: find the determinant.
$$a)\hspace{.2em}$$ $$\left[ \begin{array}{ccc}1 & -1 & 0\\ 0 & 6 & 3 \\-5 & -3 & 0\end{array}\right]$$
$$b)\hspace{.2em}$$ $$\left[ \begin{array}{ccc}1 & -7 & 5\\ 0 & 0 & 2 \\11 & 1 & 0\end{array}\right]$$
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#3:
Instructions: find the determinant.
$$a)\hspace{.2em}$$ $$\left[ \begin{array}{ccc}-1 & 3 & -2\\ -4 & 1 & 0 \\5 & -6 & -1\end{array}\right]$$
$$b)\hspace{.2em}$$ $$\left[ \begin{array}{ccc}-1 & -1 & 0\\ 2 & -4 & 0 \\0 & 1 & 20\end{array}\right]$$
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#4:
Instructions: find the determinant.
$$a)\hspace{.2em}$$ $$\left[ \begin{array}{cccc}-1 & -5 & 7 & 9\\ 0 & 1 & -1 & 3 \\3 & 0 & 2 & 0 \\ 1 & 4 & 3 & 5\end{array}\right]$$
$$b)\hspace{.2em}$$ $$\left[ \begin{array}{cccc}-2 & 1 & 1 & 0\\ 1 & 4 & 2 & -7 \\0 & 0 & -6 & 8 \\ -3 & 0 & 5 & 0\end{array}\right]$$
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#5:
Instructions: find the determinant.
$$a)\hspace{.2em}$$ $$\left[ \begin{array}{cccc}11 & -2 & 5 & 9\\ 0 & 1 & -1 & 3 \\-3 & 0 & 5 & 8 \\ 1 & -2 & 0 & 0\end{array}\right]$$
$$b)\hspace{.2em}$$ $$\left[ \begin{array}{cccc}4 & -2 & -1 & -3\\ 1 & 8 & 1 & -2 \\0 & 7 & 0 & 1 \\ 3 & -2 & 1 & 2\end{array}\right]$$
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Written Solutions:
#1:
Solutions:
$$a)\hspace{.2em}-52$$
$$b)\hspace{.2em}0$$
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#2:
Solutions:
$$a)\hspace{.2em}24$$
$$b)\hspace{.2em}-156$$
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#3:
Solutions:
$$a)\hspace{.2em}-49$$
$$b)\hspace{.2em}120$$
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#4:
Solutions:
$$a)\hspace{.2em}512$$
$$b)\hspace{.2em}186$$
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#5:
Solutions:
$$a)\hspace{.2em}599$$
$$b)\hspace{.2em}-272$$
