11.4.1: Determinants and Cramer's Rule for n x n Matrices (Exercises)
- Page ID
- 108898
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1. Let \(A=\left[\begin{array}{ccc}1&2&4\\0&1&3\\-2&5&1\end{array}\right]\). Find the following.
- \(minor(A)_{11}\)
- \(minor(A)_{21}\)
- \(minor(A)_{32}\)
- \(C_{11}\)
- \(C_{21}\)
- \(C_{32}\)
2. Find the determinants of the following matrices.
- \(\left[\begin{array}{ccc}1&2&3\\3&2&2\\0&9&8\end{array}\right]\)
- \(\left[\begin{array}{ccc}4&3&2\\1&7&8\\3&-9&3\end{array}\right]\)
- \(\left[\begin{array}{cccc}1&2&3&2\\1&3&2&3\\4&1&5&0\\1&2&1&2\end{array}\right]\)
- Answer
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- The answer is \(31\).
- The answer is \(375\).
- The answer is \(-2\).
3. Find the following determinant by expanding along the first row and second column. \[\left|\begin{array}{ccc}1&2&1\\2&1&3\\2&1&1\end{array}\right|\nonumber\]
- Answer
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\[\left|\begin{array}{ccc}1&2&1\\2&1&3\\2&1&1\end{array}\right|=6\nonumber\]
4. Find the following determinant by expanding along the first column and third row. \[\left|\begin{array}{ccc}1&2&1\\1&0&1\\2&1&1\end{array}\right|\nonumber\]
- Answer
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\[\left|\begin{array}{ccc}1&2&1\\1&0&1\\2&1&1\end{array}\right|=2\nonumber\]
5. Find the following determinant by expanding along the second row and first column. \[\left|\begin{array}{ccc}1&2&1\\2&1&3\\2&1&1\end{array}\right|\nonumber\]
- Answer
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\[\left|\begin{array}{ccc}1&2&1\\2&1&3\\2&1&1\end{array}\right|=6\nonumber\]
6. Compute the determinant by cofactor expansion. Pick the easiest row or column to use. \[\left|\begin{array}{cccc}1&0&0&1\\2&1&1&0\\0&0&0&2\\2&1&3&1\end{array}\right|\nonumber\]
- Answer
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\[\left|\begin{array}{cccc}1&0&0&1\\2&1&1&0\\0&0&0&2\\2&1&3&1\end{array}\right|=-4\nonumber\]
7. Find the determinant of the following matrices.
- \(A=\left[\begin{array}{cc}1&-34\\0&2\end{array}\right]\)
- \(A=\left[\begin{array}{ccc}4&3&14\\0&-2&0\\0&0&5\end{array}\right]\)
- \(A=\left[\begin{array}{cccc}2&3&15&0\\0&4&1&7\\0&0&-3&5\\0&0&0&1\end{array}\right]\)
8. Use Cramer’s rule to find the solution to \[\begin{array}{c}x_1+2x_2+x_3=1 \\ 2x_1-x_2-x_3=2 \\ x_1+x_3=1\end{array}\nonumber\]
- Answer
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Solution is: \([x_1 = 1, x_2 = 0,x_3 = 0]\). For example, \[y=\frac{\left|\begin{array}{ccc}1&1&1\\2&2&-1\\1&1&1\end{array}\right|}{\left|\begin{array}{ccc}1&2&1\\2&-1&-1\\1&0&1\end{array}\right|}=0\nonumber\]
9. Use Cramer’s rule to find the solution to \[\begin{array}{c}x_1+x_2-2x_3=14 \\ 2x_1-x_2+x_3=0 \\ 6x_1+3x_2+4x_3=1\end{array}\nonumber\]
10. Use Cramer’s rule to find the solution to \[\begin{array}{c}2x_1+x_2+x_3=4 \\ 10x_1-2x_2+2x_3=-1 \\ 6x_1-2x_2+4x_3=8\end{array}\nonumber\]