L-A First Course in Linear Algebra

Chapter D: Determinants

Section DM

Page 353

C24

Doing the computations by hand, find the determinant of the matrix below.

\[\mathbf{A} = \left[\begin{array}{rrr}-2 & 3 & -2\\-4 & -2 & 1\\2 & 4 & 2\end{array}\right] \]

(a) Solution using R
##      [,1] [,2] [,3]
## [1,]   -2    3   -2
## [2,]   -4   -2    1
## [3,]    2    4    2
## [1] 70
(b) Solution using D33M

\[ det(A) = |A| = \left|\begin{array}{rrr}-2 & 3 & -2\\-4 & -2 & 1 \\2 & 4 & 2\end{array}\right| \] \[ = (-2)\left|\begin{array}{rrr}-2 & 1\\4 & 2\end{array}\right| - (3)\left|\begin{array}{rrr}-4 & 1\\2 & 2\end{array}\right| + (-2)\left|\begin{array}{rrr}-4 & -2\\2 & 4\end{array}\right|\] \[ = (-2)((-2*2) - (1*4)) - (3)((-4*2) - (1*2)) + (-2)((-4*4) - (-2*2)) \] \[ = (-2)((-4) - (4)) - (3)((-8) - (2)) + (-2)((-16) - (-4)) \] \[ = (-2)(-8) - (3)(-10) + (-2)(-12) \] \[ = (16) - (-30) + (24) \] \[ = 70 \]

(c) Solution using Upper Triangular Matrix trhough Gauss Elimination

\[\mathbf{A} = \left[\begin{array}{rrr}-2 & 3 & -2\\-4 & -2 & 1\\2 & 4 & 2\end{array}\right] \]

Step1:Gaussian Elimination to Create Upper Triangular Matrix
  • R2 <- R2-2R1
  • R3 <- R3+R1

\[\mathbf{A} = \left[\begin{array}{rrr}-2 & 3 & -2\\-4-2(-2) & -2-2(3) & 1-2(-2)\\2+(-2) & 4+(3) & 2+(-2)\end{array}\right] = \left[\begin{array}{rrr}-2 & 3 & -2\\0 & -8 & 5\\0 & 7 & 0\end{array}\right] \]

  • R3 <- R3+(7/8)R2

\[\mathbf{A} = \left[\begin{array}{rrr}-2 & 3 & -2\\0 & -8 & 5\\(0)+(7/8)*(0) & 7+(7/8)(-8) & (0)+(7/8)(5)\end{array}\right] = \left[\begin{array}{rrr}-2 & 3 & -2\\0 & -8 & 5\\0 & 0 & 35/8\end{array}\right] \]

Step2:Determinant of Upper Triangular Matrix

\[ det(A) = |A| = \left|\begin{array}{rrr}-2 & 3 & -2\\-4 & -2 & 1 \\2 & 4 & 2\end{array}\right| = \left|\begin{array}{rrr}-2 & 3 & -2\\0 & -8 & 5\\0 & 0 & 35/8\end{array}\right| = (-2)*(-8)*(35/8) = 70 \]