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Additional info for Linear algebra c-1 - Linear equations, matrices and determinants
Sample text
Find the complete solution of the system of equations ⎛ ⎞ ⎛ ⎞ x1 0 A ⎝ x2 ⎠ = ⎝ 0 ⎠ . x3 0 2. Find the complete solution of the matrix equation AX = A2 . (One may benefit from the fact that the matrix A is a particular solution of this matrix equation). 1. The matrix of coefficients is equivalent to ⎛ ⎞ ⎞ ∼ 1 0 1 0 1 1 R := R3 ⎝ 0 1 1 ⎠, A = ⎝ 1 −1 0 ⎠ 1 R2 := R3 − R2 0 0 0 1 0 1 R3 := R1 + R2 − R3 ⎛ corresponding to the system of equations x1 + x3 = 0 and x2 + x3 = 0, thus x1 = x2 = −x3 . Choosing x3 = −s as parameter, the complete solution is ⎫ ⎧⎛ ⎞ s ⎬ ⎨ ⎝ s ⎠ s∈R .
Since the three columns are independent, the complete solution is given by ⎛ ⎞ s t u t u ⎠, Y=⎝ s s, t, u ∈ R, −s −t −u and hence ⎛ s+1 X=I+Y =⎝ s −s ⎞ t u t+1 u ⎠, −t 1 − u s, t, u ∈ R. 16 is precisely of this type. com 48 Linear Algebra Examples c-1 2. 18 Find the solution of each of the following two matrix equations AX = B n˚ ar and XA = B, ⎛ ⎞ 2 1 0 A=⎝ 1 1 2 ⎠ 1 1 1 ⎛ og 1 2 B=⎝ 4 2 1 1 ⎞ 3 −1 ⎠ . 2 1. First consider the equation AX = B. The total matrix is equivalent to ⎞ ∼ 1 2 3 2 1 0 R := R1 − R2 ⎝ 1 1 2 4 2 −1 ⎠ 1 R2 := R2 + R3 − R1 1 1 2 1 1 1 R3 := ⎞ R2 − R3 ⎛ −3 0 4 ∼ 1 0 −2 ⎝ 0 +1 +3 +4 +1 −2 ⎠ R1 := R1 + 2R3 3 1 −3 R2 := R2 − 3R3 0 1 ⎞ ⎛ 0 3 0 2 1 0 0 ⎝ 0 1 0 −5 −2 7 ⎠, 3 1 −3 0 0 1 ⎛ hence the (unique) solution is given by ⎛ ⎞ 3 0 2 7 ⎠.
Since YA = B is equivalent to AT YT = BT , where Y is a (2 × 2) matrix, the total matrix (AT | BT ) is equivalent to ⎞ ⎛ 1 3 ∼ 1 2 1 −1 ⎠ R2 := R1 + R2 (AT | BT ) = ⎝ −1 −1 3 5 R3 := R3 − R1 1 3 ⎞ ⎛ 1 3 ∼ 1 2 ⎝ 0 1 2 2 ⎠ R1 := R1 − 2R2 2 2 R 0 1 ⎞3 := R2 − R3 ⎛ −3 −1 1 0 ⎝ 0 1 2 2 ⎠ 0 0 0 0 which is of rank 2. It follows that YT = −3 −1 2 2 , thus −3 2 −1 2 Y= . 3 There are infinitely many solutions of 1, (and only one to 2). Furthermore, X is a (3 × 3) matrix, while Y is a (2 × 2) matrix. 21 Given the matrices ⎛ ⎞ 1 2 3 5 ⎠ og A=⎝ 2 a+3 −1 a − 3 a + 2 ⎛ ⎞ 1 −1 2 1 4 ⎠, B=⎝ 4 1 4 a+4 a ∈ R.