Answer to Question #135939 in Linear Algebra for Florence

Question #135939

Find all solutions of this system.

4x + 2y + 3z = 0

3x − y + 2z = 0

x + 2y − z = 0

Compare the value of the determinant of the coefficient matrix to zero using the nature

of the solution.


1
Expert's answer
2020-10-01T15:13:10-0400

"\\mathbf{To\\;find: Solutions\\;of\\;the\\;system:-}"


4x + 2y + 3z = 0

3x − y + 2z = 0

x + 2y − z = 0


"\\mathbf{The \\;augmented\\;matrix\\;associated\\;with\\;above\\;system\\;is-}"


"\\left[\n\\begin{array}{ccc|c}\n4 & 2 & 3 & 0 \\\\ \n3 & -1 & 2 & 0 \\\\ \n1 & 2 & -1 & 0 \\\\ \n \\end{array}\n\\right]"


"\\mathbf{Row\\;reducing\\;above\\;matrix\\;R_1 \\to R_1-4R_3\\;and\\;R_2\\to R_2-3R_3}"


"\\left[\n\\begin{array}{ccc|c}\n0 & -6 & 7 & 0 \\\\ \n0 & -7 & 5 & 0 \\\\ \n1 & 2 & -1 & 0 \\\\ \n \\end{array}\n\\right]"


"\\mathbf{Again\\;R_1\\to R_1-R_2\\;we\\;get-}"


"\\left[\n\\begin{array}{ccc|c}\n0 & 1 & 2 & 0 \\\\ \n0 & -7 & 5 & 0 \\\\ \n1 & 2 & -1 & 0 \\\\ \n \\end{array}\n\\right]"


"\\mathbf{R_3\\to R_3-2R_1\\;gives-}"


"\\left[\n\\begin{array}{ccc|c}\n0 & 1 & 2 & 0 \\\\ \n0 & -7 & 5 & 0 \\\\ \n1 & 0 & -5 & 0 \\\\ \n \\end{array}\n\\right]"


"\\mathbf{R_2\\to R_2+7R_1\\;gives-}"


"\\left[\n\\begin{array}{ccc|c}\n0 & 1 & 2 & 0 \\\\ \n0 & 0 & 19 & 0 \\\\ \n1 & 0 & -5 & 0 \\\\ \n \\end{array}\n\\right]"


"\\mathbf{R_2\\to \\dfrac{1}{19} R_2\\;gives-}"


"\\left[\n\\begin{array}{ccc|c}\n0 & 1 & 2 & 0 \\\\ \n0 & 0 & 1 & 0 \\\\ \n1 & 0 & -5 & 0 \\\\ \n \\end{array}\n\\right]"


"\\mathbf{So,\\;the\\;system\\;of\\;equations\\;gives-}"


"\\mathbf{y+2z=0 \\implies y=-2z}"

"\\mathbf{z=0}"

"\\mathbf{x-5z=0 \\implies x=5z \\implies x=0}"

"\\mathbf{Similarly, y=-2z => y=0}"


"\\mathbf{Hence,\\;(0,0,0)\\;is\\;the\\;only\\;solution\\;to\\;this\\;system.}"


"\\mathbf{The\\;coefficient\\;matrix\\;is\\;A-}"


"\\left[\n\\begin{array}{ccc}\n4 & 2 & 3\\\\ \n3 & -1 & 2\\\\ \n1 & 2 & -1\\\\ \n \\end{array}\n\\right]"


"\\mathbf{and\\;the\\;determinant\\;is\\;given\\;by-}"


"\\mathbf{det(A)=4}\\begin{vmatrix}\n -1 & 2 \\\\\n 2 & -1\n\\end{vmatrix}-2\\begin{vmatrix}\n 3 & 2 \\\\\n 1 & -1\n\\end{vmatrix}+3\\begin{vmatrix}\n 3 & -1 \\\\\n 1 & 2\n\\end{vmatrix}"


"\\mathbf{=4(1-4)-2(-3-2)+3(6+1)=4(-4)-2(-5)+3(7)}"


"\\mathbf{=-16+10+21=15}"


"\\mathbf{Observe\\;that\\;det(A) \\neq 0 \\;which \\;implies\\;that\\;the\\;solution }"

"\\mathbf{ \\;of\\;this\\;system\\;is\\;unique.}"


Result :- A n x n non-homogeneous system of linear equations has a unique non-trivial solution if and only if its determinant is non-zero.











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