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Lecture 28

# MATH125 Lecture 28: 28

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Department
Mathematics
Course Code
MATH125
Professor
Nikita

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Lecture 28, March 22, 2017. Theorem. Let A be a square matrix. (a) If A has a zero row or column then detA = 0. (b) If a multiple of one row of A is added to another row to produce a matrix B then detB = detA. (c) If two rows of A are interchanged to produce B then detB = ▯detA: (d) If one row of A is multiplied by a scalar c to produce B then detB = c ▯ detA: (e) If A has two identical rows or columns then detA = 0. We will make comments about the proof of this theorem later and now we only note that the above theorem provide us with the power- ful tool of computation of determinants. The strategy of computing detA is to reduce A to an echelon form and then to use the fact that the determinant of a triangular matrix is the product of the diagonal entries. Example. Compute the determinant of 2 1 ▯4 2 3 A = 4 ▯2 8 ▯9 5 : ▯1 7 0 Solution. We ▯rst create zeroes below the ▯rst entry in the ▯rst col- umn. By the ▯bove theorem t▯e d▯terminant does▯not▯change. ▯ ▯ 1 ▯4 2 ▯ ▯ 1 ▯4 2 ▯ ▯ 1 ▯4 2 ▯ ▯ ▯ ▯ ▯ ▯ ▯ detA = ▯ ▯2 8 ▯9 ▯ = ▯ 0 0 ▯5 ▯ = ▯ 0 0 ▯5 ▯: ▯ ▯1 7 0 ▯ ▯ ▯1 7 0 ▯ ▯ 0 3 2 ▯ Next, we interchange rows 2 and 3. After this operation the determi- nant change the sign. ▯ ▯ ▯ 1 ▯4 2 ▯ detA = ▯ ▯ 0 3 2 ▯: ▯ 0 0 ▯5 ▯ Now we see that our matrix is triangular, so to compute its determinant we must take the product of its diagonal entries. Hence we have detA = (▯1) ▯ 1 ▯ 3 ▯ (▯5) = 15: Note that from the main theorem, part (c) it follows that we can factor out a common factor of a matrix while computing its determinant. For instance ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯5k ▯2k 3k ▯ = k ▯▯ 5 ▯2 3 ▯: ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ Example. Compute detA where 2 3 2 ▯8 6 8 6 3 ▯9 5 10 7 A = 4 ▯3 0 1 ▯2 5 : 1 ▯4 0 6 Solution. To simplify computations we want 1 in the position ▯rst row and ▯rst column. We could interchange rows 1 and 4. Instea
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