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- https://math.libretexts.org/Courses/Coastline_College/Math_C285%3A_Linear_Algebra_and_Diffrential_Equations_(Tran)/03%3A_Determinants/3.E%3A_ExercisesThis inverse is \[\begin{aligned}\frac{1}{-13}\left[\begin{array}{rrr}\left|\begin{array}{cc}2&1 \\ 1&0\end{array}\right| & -\left|\begin{array}{cc}0&1\\3&0\end{array}\right| &\left|\begin{array}{cc}0...This inverse is \[\begin{aligned}\frac{1}{-13}\left[\begin{array}{rrr}\left|\begin{array}{cc}2&1 \\ 1&0\end{array}\right| & -\left|\begin{array}{cc}0&1\\3&0\end{array}\right| &\left|\begin{array}{cc}0&2\\3&1\end{array}\right| \\ -\left|\begin{array}{cc}2&3\\1&0\end{array}\right| &\left|\begin{array}{cc}1&3\\3&0\end{array}\right| &-\left|\begin{array}{cc}1&2\\3&1\end{array}\right| \\ \left|\begin{array}{cc}2&3\\2&1\end{array}\right|&-\left|\begin{array}{cc}1&3\\0&1\end{array}\right|&\left|\begin…
- https://math.libretexts.org/Courses/Mission_College/MAT_04C_Linear_Algebra_(Kravets)/03%3A_Determinants/3.E%3A_ExercisesThis inverse is \[\begin{aligned}\frac{1}{-13}\left[\begin{array}{rrr}\left|\begin{array}{cc}2&1 \\ 1&0\end{array}\right| & -\left|\begin{array}{cc}0&1\\3&0\end{array}\right| &\left|\begin{array}{cc}0...This inverse is \[\begin{aligned}\frac{1}{-13}\left[\begin{array}{rrr}\left|\begin{array}{cc}2&1 \\ 1&0\end{array}\right| & -\left|\begin{array}{cc}0&1\\3&0\end{array}\right| &\left|\begin{array}{cc}0&2\\3&1\end{array}\right| \\ -\left|\begin{array}{cc}2&3\\1&0\end{array}\right| &\left|\begin{array}{cc}1&3\\3&0\end{array}\right| &-\left|\begin{array}{cc}1&2\\3&1\end{array}\right| \\ \left|\begin{array}{cc}2&3\\2&1\end{array}\right|&-\left|\begin{array}{cc}1&3\\0&1\end{array}\right|&\left|\begin…
- https://math.libretexts.org/Courses/Lake_Tahoe_Community_College/A_First_Course_in_Linear_Algebra_(Kuttler)/03%3A_Determinants/3.E%3A_ExercisesThis inverse is \[\begin{aligned}\frac{1}{-13}\left[\begin{array}{rrr}\left|\begin{array}{cc}2&1 \\ 1&0\end{array}\right| & -\left|\begin{array}{cc}0&1\\3&0\end{array}\right| &\left|\begin{array}{cc}0...This inverse is \[\begin{aligned}\frac{1}{-13}\left[\begin{array}{rrr}\left|\begin{array}{cc}2&1 \\ 1&0\end{array}\right| & -\left|\begin{array}{cc}0&1\\3&0\end{array}\right| &\left|\begin{array}{cc}0&2\\3&1\end{array}\right| \\ -\left|\begin{array}{cc}2&3\\1&0\end{array}\right| &\left|\begin{array}{cc}1&3\\3&0\end{array}\right| &-\left|\begin{array}{cc}1&2\\3&1\end{array}\right| \\ \left|\begin{array}{cc}2&3\\2&1\end{array}\right|&-\left|\begin{array}{cc}1&3\\0&1\end{array}\right|&\left|\begin…
- https://math.libretexts.org/Courses/Reedley_College/Differential_Equations_and_Linear_Algebra_(Zook)/03%3A_Determinants/3.05%3A_ExercisesThis inverse is \[\begin{aligned}\frac{1}{-13}\left[\begin{array}{rrr}\left|\begin{array}{cc}2&1 \\ 1&0\end{array}\right| & -\left|\begin{array}{cc}0&1\\3&0\end{array}\right| &\left|\begin{array}{cc}0...This inverse is \[\begin{aligned}\frac{1}{-13}\left[\begin{array}{rrr}\left|\begin{array}{cc}2&1 \\ 1&0\end{array}\right| & -\left|\begin{array}{cc}0&1\\3&0\end{array}\right| &\left|\begin{array}{cc}0&2\\3&1\end{array}\right| \\ -\left|\begin{array}{cc}2&3\\1&0\end{array}\right| &\left|\begin{array}{cc}1&3\\3&0\end{array}\right| &-\left|\begin{array}{cc}1&2\\3&1\end{array}\right| \\ \left|\begin{array}{cc}2&3\\2&1\end{array}\right|&-\left|\begin{array}{cc}1&3\\0&1\end{array}\right|&\left|\begin…
- https://math.libretexts.org/Courses/Cosumnes_River_College/Math_420%3A_Differential_Equations_(Breitenbach)/11%3A_Appendices/04%3A_Determinants_and_Cramer's_Rule_for_n_x_n_Matrices/11.4.1%3A_Determinants_and_Cramer's_Rule_for_n_x_n_Matrices_(Exercises)Find the following determinant by expanding along the first row and second column. \[\left|\begin{array}{ccc}1&2&1\\2&1&3\\2&1&1\end{array}\right|\nonumber\] Find the following determinant by expandin...Find the following determinant by expanding along the first row and second column. \[\left|\begin{array}{ccc}1&2&1\\2&1&3\\2&1&1\end{array}\right|\nonumber\] Find the following determinant by expanding along the first column and third row. \[\left|\begin{array}{ccc}1&2&1\\1&0&1\\2&1&1\end{array}\right|\nonumber\] Find the following determinant by expanding along the second row and first column. \[\left|\begin{array}{ccc}1&2&1\\2&1&3\\2&1&1\end{array}\right|\nonumber\]
- https://math.libretexts.org/Courses/SUNY_Schenectady_County_Community_College/A_First_Journey_Through_Linear_Algebra/03%3A_Determinants/3.E%3A_ExercisesThis inverse is \[\begin{aligned}\frac{1}{-13}\left[\begin{array}{rrr}\left|\begin{array}{cc}2&1 \\ 1&0\end{array}\right| & -\left|\begin{array}{cc}0&1\\3&0\end{array}\right| &\left|\begin{array}{cc}0...This inverse is \[\begin{aligned}\frac{1}{-13}\left[\begin{array}{rrr}\left|\begin{array}{cc}2&1 \\ 1&0\end{array}\right| & -\left|\begin{array}{cc}0&1\\3&0\end{array}\right| &\left|\begin{array}{cc}0&2\\3&1\end{array}\right| \\ -\left|\begin{array}{cc}2&3\\1&0\end{array}\right| &\left|\begin{array}{cc}1&3\\3&0\end{array}\right| &-\left|\begin{array}{cc}1&2\\3&1\end{array}\right| \\ \left|\begin{array}{cc}2&3\\2&1\end{array}\right|&-\left|\begin{array}{cc}1&3\\0&1\end{array}\right|&\left|\begin…
- https://math.libretexts.org/Courses/Cosumnes_River_College/Math_420%3A_Differential_Equations_(Breitenbach)/11%3A_Appendices/03%3A_Determinants_and_Cramer's_Rule_for_2_X_2_Matrices/11.3.1%3A_Determinants_and_Cramer's_Rule_for_2_X_2_Matrices_(Exercises)In Exercises 1-3 find the determinant of the given matrix. 1. \(\left[\begin{array}{cc}1&-3\\3&2\end{array}\right]\) 2. \(\left[\begin{array}{cc}4&3\\5&-2\end{array}\right]\) 3. \(\left[\begin{array}{...In Exercises 1-3 find the determinant of the given matrix. 1. \(\left[\begin{array}{cc}1&-3\\3&2\end{array}\right]\) 2. \(\left[\begin{array}{cc}4&3\\5&-2\end{array}\right]\) 3. \(\left[\begin{array}{cc}4&3\\6&2\end{array}\right]\) In Exercises 4-6 use Cramer's Rule to find the solution to the given systems. 4. \(\begin{aligned}x+2y&=1 \\ 2x-y&=2\end{aligned}\) 5. \(\begin{aligned}5x-8y&=-3 \\ 7x-y&=6\end{aligned}\) 6. \(\begin{aligned}3x+4y&=23 \\ 4x-3y&=14\end{aligned}\)
