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Example text

Matrices ⎞ 6 13 + a ⎠ , −4 where a and b are arbitrary real numbers. 1. Find the pairs (a, b), for which the matrix equation AX = B has precisely one solution. 2. Find the complete solution of the matrix equation for (a, b) = (−3, 4). 1. The total matrix is equivalent to ⎛ ⎞ 1 2 −2 2 6 ∼ 4 a b 13 + a ⎠ (A | B) = ⎝ 1 R2 := R2 − R1 0 −2 1 −2 ⎞−4 ⎛ 1 2 −2 2 6 ∼ ⎝ 0 2 a+2 b−2 7+a ⎠ R3 := R2 + R3 1 −2 −4 ⎞ ⎛ 0 −2 1 2 −2 2 6 ⎝ 0 2 a+2 b−2 7+a ⎠ 0 0 a+3 b−4 3+a It follows that the matrix of coefficients is of rank 3 for a = −3, hence there exists precisely one solution, when a = −3 and b ∈ R.

By transposing the equation is transferred into the standard form AT XT = 0 with the unknown matrix XT to the right of A. The matrix of coefficients AT is equivalent to ⎛ ⎞ 1 1 2 ∼ AT = ⎝ 2 3 5 ⎠ R2 := R2 − 2R1 R3 := R3 − R1 1 ⎞ 5 6 ⎛ 1 1 2 ∼ ⎝ 0 1 1 ⎠ R3 := R3 − 4R2 ⎛ 0 4 4 ⎞ R1 := R1 − R2 1 0 1 ⎝ 0 1 1 ⎠. 0 0 0 Hence, every column y, for which AT y is the zero column must therefore be of the form s(1, 1, −1)T , where s is a parameter, thus ⎛ ⎞ s t u t u ⎠, XT = ⎝ s s, t, u ∈ R, −s −t −u and we get by transposing ⎛ ⎞ s s −s X = ⎝ t t −t ⎠ , u u −y 2.

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