New version page

UW AMATH 352 - Lecture 3: Matrices and Linear Equations

Pages: 32
Documents in this Course

26 pages

19 pages

This preview shows page 1-2-15-16-31-32 out of 32 pages.

View Full Document
Do you want full access? Go Premium and unlock all 32 pages.
Do you want full access? Go Premium and unlock all 32 pages.
Do you want full access? Go Premium and unlock all 32 pages.
Do you want full access? Go Premium and unlock all 32 pages.
Do you want full access? Go Premium and unlock all 32 pages.
Do you want full access? Go Premium and unlock all 32 pages.

Unformatted text preview:

Lecture 3: Matrices and Linear EquationsAMath 352Fri., Apr. 21 / 17Linear EquationsSolve for x and y:x + 2y = 33x − y = 2.Use eq. 1 to eliminate x in eq. 2. Subtract 3 times eq. 1 from eq. 2:3x − y − 3(x + 2y) = 2 − 3 · 3 ⇒ −7y = −7 ⇒ y = 1.Now substitute y = 1 into eq. 1 and solve for x:x + 2 · 1 = 3 ⇒ x = 1.Now check:1 + 2 · 1?= 3 yes3 · 1 − 1?= 2 yes2 / 17Linear EquationsSolve for x and y:x + 2y = 33x − y = 2.Use eq. 1 to eliminate x in eq. 2. Subtract 3 times eq. 1 from eq. 2:3x − y − 3(x + 2y) = 2 − 3 · 3 ⇒ −7y = −7 ⇒ y = 1.Now substitute y = 1 into eq. 1 and solve for x:x + 2 · 1 = 3 ⇒ x = 1.Now check:1 + 2 · 1?= 3 yes3 · 1 − 1?= 2 yes2 / 17Linear EquationsSolve for x and y:x + 2y = 33x − y = 2.Use eq. 1 to eliminate x in eq. 2. Subtract 3 times eq. 1 from eq. 2:3x − y − 3(x + 2y) = 2 − 3 · 3 ⇒ −7y = −7 ⇒ y = 1.Now substitute y = 1 into eq. 1 and solve for x:x + 2 · 1 = 3 ⇒ x = 1.Now check:1 + 2 · 1?= 3 yes3 · 1 − 1?= 2 yes2 / 17Linear EquationsSolve for x and y:x + 2y = 33x − y = 2.Use eq. 1 to eliminate x in eq. 2. Subtract 3 times eq. 1 from eq. 2:3x − y − 3(x + 2y) = 2 − 3 · 3 ⇒ −7y = −7 ⇒ y = 1.Now substitute y = 1 into eq. 1 and solve for x:x + 2 · 1 = 3 ⇒ x = 1.Now check:1 + 2 · 1?= 3 yes3 · 1 − 1?= 2 yes2 / 17Linear Equations, Cont.This technique can be extended to any number of linear equationsin the same number of unknowns. Solve for x, y, and z:x + 2y + 3z = 42x + 3y + 4z = 5−x − 6y + z = 2.Use eq. 1 to eliminate x in eqs. 2 and 3. Subtract 2 times the first eq.from the second; add the first eq. to the third:2x + 3y + 4z − 2(x + 2y + 3z) = 5 − 2 · 4 ⇒ −y − 2z = −3.−x − 6y + z + (x + 2y + 3z) = 2 + 4 ⇒ −4y + 4z = 6.3 / 17Linear Equations, Cont.This technique can be extended to any number of linear equationsin the same number of unknowns. Solve for x, y, and z:x + 2y + 3z = 42x + 3y + 4z = 5−x − 6y + z = 2.Use eq. 1 to eliminate x in eqs. 2 and 3. Subtract 2 times the first eq.from the second; add the first eq. to the third:2x + 3y + 4z − 2(x + 2y + 3z) = 5 − 2 · 4 ⇒ −y − 2z = −3.−x − 6y + z + (x + 2y + 3z) = 2 + 4 ⇒ −4y + 4z = 6.3 / 17Linear Equations, Cont.x + 2y + 3z = 4−y − 2z = −3−4y + 4z = 6.Subtract 4 times the second eq. from the third to find:−4y + 4z − 4(−y − 2z) = 6 − 4 · (−3) ⇒ 12z = 18 ⇒ z =32.Substitute this into the second eq. to find:−y − 2 ·32= −3 ⇒ y = 0.Substitute for y and z in the first eq. to find:x + 2 · 0 + 3 ·32= 4 ⇒ x = −12.4 / 17Linear Equations, Cont.x + 2y + 3z = 4−y − 2z = −3−4y + 4z = 6.Subtract 4 times the second eq. from the third to find:−4y + 4z − 4(−y − 2z) = 6 − 4 · (−3) ⇒ 12z = 18 ⇒ z =32.Substitute this into the second eq. to find:−y − 2 ·32= −3 ⇒ y = 0.Substitute for y and z in the first eq. to find:x + 2 · 0 + 3 ·32= 4 ⇒ x = −12.4 / 17Linear Equations, Cont.x + 2y + 3z = 4−y − 2z = −3−4y + 4z = 6.Subtract 4 times the second eq. from the third to find:−4y + 4z − 4(−y − 2z) = 6 − 4 · (−3) ⇒ 12z = 18 ⇒ z =32.Substitute this into the second eq. to find:−y − 2 ·32= −3 ⇒ y = 0.Substitute for y and z in the first eq. to find:x + 2 · 0 + 3 ·32= 4 ⇒ x = −12.4 / 17Linear Equations, Cont.x + 2y + 3z = 42x + 3y + 4z = 5−x − 6y + z = 2.Check: x = −12, y = 0, z =32.−12+92?= 4 yes2 · (−12) + 4 ·32?= 5 yes12+32?= 2 yes.5 / 17Matrix NotationWe don’t have to write x, y, and z so many times!x + 2y + 3z = 42x + 3y + 4z = 5−x − 6y + z = 2.can be written as1 2 32 3 4−1 −6 1xyz=452.The 3 by 3 array of numbers on the left is a matrix, and theproduct of this matrix with the vector [x, y, z]Tof unknowns isjust the left-hand side of the above equations:1 2 32 3 4−1 −6 1xyz=x + 2y + 3z2x + 3y + 4z−x − 6y + z.6 / 17Solving Equations Using Matrix NotationAppend the right-hand side vector to the matrix:1 2 3 | 42 3 4 | 5−1 −6 1 | 2.Eliminate x from the second and third eqs. by subtracting 2 timesthe first eq. from the second and adding the first eq. to the third:1 2 3 | 40 −1 −2 | −30 −4 4 | 6.Eliminate y from the third eq. by subtracting 4 times the secondeq.:1 2 3 | 40 −1 −2 | −30 0 12 | 18.7 / 17Solving Equations Using Matrix NotationAppend the right-hand side vector to the matrix:1 2 3 | 42 3 4 | 5−1 −6 1 | 2.Eliminate x from the second and third eqs. by subtracting 2 timesthe first eq. from the second and adding the first eq. to the third:1 2 3 | 40 −1 −2 | −30 −4 4 | 6.Eliminate y from the third eq. by subtracting 4 times the secondeq.:1 2 3 | 40 −1 −2 | −30 0 12 | 18.7 / 17Solving Equations Using Matrix NotationAppend the right-hand side vector to the matrix:1 2 3 | 42 3 4 | 5−1 −6 1 | 2.Eliminate x from the second and third eqs. by subtracting 2 timesthe first eq. from the second and adding the first eq. to the third:1 2 3 | 40 −1 −2 | −30 −4 4 | 6.Eliminate y from the third eq. by subtracting 4 times the secondeq.:1 2 3 | 40 −1 −2 | …

View Full Document