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A solution is the intersection of all the hyperplanes. #### R 2 \\colorOne{\\reals^2} R2 : Intersection of Lines - No intersection - Unique point of intersection (2 lines cross) - Infinitely many points of intersection - Line → \\rightarrow → 1-parameter family of solutions (same line) Example { − x \+ y \= 2 y \= 1 ⟹ \[ − 1 1 2 0 1 1 \] ⟹ x ⃗ \= \[ − 1 1 \] P O I \\begin{cases} -x&+\&y&=&2\\\\ &\&y&=&1 \\end{cases} \\quad \\implies \\quad \\left\[ \\begin{array}{rr\|r} -1 & 1 & 2\\\\ 0 & 1 & 1\\\\ \\end{array} \\right\] \\quad \\implies \\quad \\overset{\\normalsize POI}{ \\boxed{ \\vec x = \\left\[ \\begin{array}{r} -1\\\\ 1 \\end{array} \\right\] }} {−x​\+​yy​\=\=​21​⟹\[−10​11​21​\]⟹ x \=\[−11​\] ​ POI ​  PAGE BREAK #### R 3 \\colorOne{\\reals^3} R3 : Intersection of Planes - No intersection - Unique point of intersection (requires 3 planes) - Infinitely many points of intersection - Line → \\rightarrow → 1-parameter family of solutions (2 planes intersect) - Plane → \\rightarrow → 2-parameter family of solutions (2 of the same plane) Example { − x \+ y \= 2 y \= 1 ⟹ \[ − 1 1 0 2 0 1 0 1 \] ⟹ x ⃗ \= \[ − 1 1 0 \] \+ t \[ 0 0 1 \] line of intersection \\begin{cases} -x&+\&y&=&2\\\\ &\&y&=&1 \\end{cases} \\quad \\implies \\quad \\left\[ \\begin{array}{rrr\|r} -1 & 1 & 0 & 2\\\\ 0 & 1 & 0 & 1\\\\ \\end{array} \\right\] \\quad \\implies \\quad \\overset{\\text{\\normalsize line of intersection}}{ \\boxed{ \\vec x = \\left\[ \\begin{array}{r} -1\\\\ 1\\\\ 0 \\end{array} \\right\] +t \\left\[ \\begin{array}{r} 0\\\\ 0\\\\ 1 \\end{array} \\right\] }} {−x​\+​yy​\=\=​21​⟹\[−10​11​00​21​\]⟹ x \= ​ −110​ ​ \+t ​ 001​ ​ ​ line of intersection ​   10 10 0:00 / 0:00 1X Show Solutions ## Example: Geometric Interpretation of a Linear System Solve the system of linear equations and interpret the solution geometrically: 2 x \+ 2 y \= 6 − x − y \= − 3 \\begin{array}{rl} 2x+2y&=6\\\\ -x-y&=-3\\\\ \\end{array} 2x\+2y−x−y​\=6\=−3​ Steps 1. Let's first write this in augmented matrix form: 2 x \+ 2 y \= 6 − x − y \= − 3 → \[ 2 2 6 − 1 − 1 − 3 \] \\begin{array}{rcr} 2x+2y&=&6\\\\\[0.2em\] -x-y&=&-3\\\\\[0.2em\] \\end{array} \\quad \\rightarrow \\quad \\left\[\\begin{array}{rr\|r} 2&2&6\\\\ -1&-1&-3\\\\ \\end{array}\\right\] 2x\+2y−x−y​\=\=​6−3​→\[2−1​2−1​6−3​\] 1. Turn the matrix into RREF using EROs: \[ 2 2 6 − 1 − 1 − 3 \] 1 2 R 1 ⟶ \[ 1 1 3 − 1 − 1 − 3 \] R 2 \+ R 1 ⟶ \[ 1 1 3 0 0 0 \] \\begin{aligned} &\\left\[\\begin{array}{rr\|r} 2&2&6\\\\ -1&-1&-3\\\\ \\end{array}\\right\] \\begin{array}{l} \\dfrac{1}{2}R\_1\\\\ \\\\ \\\\ \\end{array}\\\\\[2.5em\] \\longrightarrow &\\left\[\\begin{array}{rr\|r} 1&1&3\\\\ -1&-1&-3\\\\ \\end{array}\\right\] \\begin{array}{l} \\\\ R\_2 + R\_1\\\\ \\end{array}\\\\\[2.5em\] \\longrightarrow &\\left\[\\begin{array}{rr\|r} 1&1&3\\\\ 0&0&0\\\\ \\end{array}\\right\] \\end{aligned} ⟶⟶​\[2−1​2−1​6−3​\]21​R1​​\[1−1​1−1​3−3​\]R2​\+R1​​\[10​10​30​\]​ 1. Find all solutions to the SLE: The coefficient matrix has just one leading 1, but 2 columns/variables: r a n k ( A ) \= 1 \< n \= 2 {\\rm rank}(A)=1 \\ \\ \\bm\< \\ \\ \\ n=2 rank(A)\=1 \< n\=2 Thus, there is 2 − 1 \= 1 2-1 = 1 2−1\=1 free variable ( y y y ). Let y \= t y=t y\=t . Rewrite the augmented matrix back into a system of linear equations: { x \+ t \= 3 y \= t ⟹ { x \= 3 − t y \= t \\begin{cases} x+t=3\\\\ y=t \\end{cases} \\quad\\implies\\quad \\begin{cases} x=3-t\\\\ y=t \\end{cases} {x\+t\=3y\=t​⟹{x\=3−ty\=t​ Therefore, the solution is x ⃗ \= \[ 3 0 \] \+ t \[ − 1 1 \] \\vec x = \\begin{bmatrix} 3\\\\ 0\\\\ \\end{bmatrix} +t \\begin{bmatrix} -1\\\\ 1\\\\ \\end{bmatrix} x \= \[30​\]\+t\[−11​\] . 1. Interpret the solution geometrically: We were given a system of linear equations in 2 variables ⟹ R 2 \\implies \\reals^2 ⟹R2 . Each equation is a hyperplane: in R 2 \\reals^2 R2 , hyperplanes are simply lines (in general, a hyperplane in R n \\reals^n Rn is n − 1 n-1 n−1 dimensional). So we have 2 lines, but both lines are identical (hence reducing to a row of 0s). This means the lines intersect everywhere on the line defined by x ⃗ \= \[ 3 0 \] \+ t \[ − 1 1 \] \\vec x = \\begin{bmatrix} 3\\\\ 0\\\\ \\end{bmatrix} +t \\begin{bmatrix} -1\\\\ 1\\\\ \\end{bmatrix} x \= \[30​\]\+t\[−11​\] . Consider the following augmented matrix in RREF: \[ 1 − 1 0 0 − 5 4 0 0 1 0 − 2 2 0 0 0 1 4 − 3 \] \\left\[\\begin{array}{rrrrr\|r} 1&-1&0&0&-5&4\\\\\[0.5em\] 0&0&1&0&-2&2\\\\\[0.5em\] 0&0&0&1&4&-3 \\end{array}\\right\] ​ 100​−100​010​001​−5−24​42−3​ ​ If the unknowns in this system are x 1 , x 2 , x 3 , x 4 , x 5 x\_1,\\ x\_2,\\ x\_3,\\ x\_4,\\ x\_5 x1​, x2​, x3​, x4​, x5​ , find the solution(s) to the linear system. Then, determine which of the following is the correct geometrical interpretation of the solution. A) The five hyperplanes intersect at a single point in R 5 \\reals^5 R5 B) The three hyperplanes are non-intersecting C) The three hyperplanes intersect in a line in R 5 \\reals^5 R5 D) The three hyperplanes intersect in a plane in R 5 \\reals^5 R5 E) None of the above I don't know Check Submission ##### Extra Practice [What is the reduced row echelon form of an augmented matrix with three equations and three unknowns, a) If all solutions lie on the line \[ x y z \] = ( 3 , 1 , 0 ) + s ( 1 , 2 , 3 ) \[x\\ y\\ z\] =(3,1,0)+s(1,2,3) \[x y z\]=(3,1,0)+s(1,2,3) b) If all solutions lie on the plane x + y + 2 z = 1 x+y+2z=1 x+y+2z=1](https://www.wizeprep.com/practice-questions/125521) [For what value of p does the following system of equations have no solution? What is the geometric interpretation of these equations? x + 3 y = 1 x+3y=1 x+3y=1 p x − 6 y = − 4 px-6y=-4 px−6y=−4](https://www.wizeprep.com/practice-questions/125513) [Practice Question 6: Solving Systems of Linear EquationsPractice Question: Solving Systems of Linear Equations The following equations represent three planes in ℝ3. Describe the regions of intersection. a) x 1 − x 2 + 2 x 3 = 0 x\_1-x\_2+2x\_3=0 x1​−x2​+2x3​=0](https://www.wizeprep.com/practice-questions/68880) [Solving Linear SystemsPractice: Geometric Interpretation of a Linear System Consider the following augmented matrix in RREF: \[ 1 − 1 0 0 − 5 4 0 0 1 0 − 2 2 0 0 0 1 4 − 3 \] \\left\[\\begin{array}{rrrrr\|r} 1&-1&0&0&-5&4\\\\\[0.5em\] 0&0&1&0&-2&2\\\\\[0.5em\] 0&0&0&1&4&-3 \\end{array}\\right\] ​ 100​−100​010​001​−5−24​42−3​ ​](https://www.wizeprep.com/practice-questions/115516) [Geometric Interpretation of Linear SystemsExample: Geometric Interpretation of a Linear System Solve the system of linear equations and interpret the solution geometrically: 2 x + 2 y = 6 − x − y = − 3 \\begin{array}{rl} 2x+2y&=6\\\\ -x-y&=-3\\\\ \\end{array} 2x+2y−x−y​=6=−3​](https://www.wizeprep.com/practice-questions/122521) [Homogeneous equationsFind the solution(s) of Ax=0, if the solution of Ax=b is given as follows: x p = \[ 1 0 3 0 \] + s \[ − 1 1 0 0 \] + t \[ 0 0 − 2 3 \] x\_p=\\begin{bmatrix} 1\\\\ 0\\\\ 3\\\\ 0 \\end{bmatrix}+s\\begin{bmatrix} -1\\\\ 1\\\\ 0\\\\ 0 \\end{bmatrix}+t\\begin{bmatrix} 0\\\\ 0\\\\ -2\\\\ 3 \\end{bmatrix} xp​= ​ 1030​ ​ + s ​ −1100​ ​ + t ​ 00−23​ ​](https://www.wizeprep.com/practice-questions/125514) [Interpretation of Linear SystemsWhich of the following describes the intersection between the following 3 hyperplanes in R 4 \\reals^4 R4 ? 2 x 1 + 3 x 3 + x 4 = 1 x 2 − x 4 = 0 x 2 = − 3 \\begin{array}{rcl} 2x\_1+3x\_3+x\_4&=&1\\\\ x\_2-x\_4&=&0\\\\ x\_2&=&-3 \\end{array} 2x1​+3x3​+x4​x2​−x4​x2​​===​10−3​](https://www.wizeprep.com/practice-questions/125504) [The following linear equations represent planes in R 3 R^3 R3 . Describe the intersection of these planes (if any). 2 x + y − z = 3 x − y − z = 3 3 x + z = 6 \\begin{array}{c} 2x+y-z=3\\\\ x-y-z=3\\\\ 3x+z=6 \\end{array} 2x+y−z=3x−y−z=33x+z=6​](https://www.wizeprep.com/practice-questions/19846) [Homogeneous equationsFind the solution(s) of Ax=0, if the solution of Ax=b is given as follows: x p = \[ 1 0 3 0 \] + s \[ − 1 1 0 0 \] + t \[ 0 0 − 2 3 \] x\_p=\\begin{bmatrix} 1\\\\ 0\\\\ 3\\\\ 0 \\end{bmatrix}+s\\begin{bmatrix} -1\\\\ 1\\\\ 0\\\\ 0 \\end{bmatrix}+t\\begin{bmatrix} 0\\\\ 0\\\\ -2\\\\ 3 \\end{bmatrix} xp​= ​ 1030​ ​ + s ​ −1100​ ​ + t ​ 00−23​ ​](https://www.wizeprep.com/practice-questions/42057) [Practice: Solutions to a System of Equations (Step 3)Practice: Solutions to a System of Equations (Step 3) Find the solution(s) to the systems of linear equations whose augmented matrices are given: a.) \[ 1 − 1 0 3 0 0 1 2 \] \\left\[ \\begin{array}{ccc\|c} 1&-1&0&3\\\\ 0&0&1&2\\\\ \\end{array}\\right\] \[10​−10​01​32​\]](https://www.wizeprep.com/practice-questions/102016) [System of Equationse.g. True or False: if a linear system has more unknowns than equations, then it must have infinitely many solutions. Justify your answer with a proof or counter-example.](https://www.wizeprep.com/practice-questions/52126) [System of EquationsTrue or False: if a linear system has more unknowns than equations, then it must have infinitely many solutions. Justify your answer with a proof or counter-example.](https://www.wizeprep.com/practice-questions/55869) [Linear combinationsIf u ⃗ = \[ 5 − 2 3 \] \\vec{u}=\\begin{bmatrix}5\\\\-2\\\\3\\end{bmatrix} u = ​ 5−23​ ​ is a linear combination of v ⃗ = \[ − 2 k 3 \] \\vec{v}=\\begin{bmatrix}-2\\\\k\\\\3\\end{bmatrix} v = ​ −2k3​ ​ and w ⃗ = \[ 3 2 − 1 \] \\vec{w}=\\begin{bmatrix}3\\\\2\\\\-1\\end{bmatrix} w = ​ 32−1​ ​ , what is the value of k k k ?](https://www.wizeprep.com/practice-questions/167) [Solving Linear SystemsConsider the following augmented matrix in RREF: \[ 1 − 1 0 0 − 5 4 0 0 1 0 − 2 2 0 0 0 1 4 − 3 \] \\left\[\\begin{array}{rrrrr\|r} 1&-1&0&0&-5&4\\\\\[0.5em\] 0&0&1&0&-2&2\\\\\[0.5em\] 0&0&0&1&4&-3 \\end{array}\\right\] ​ 100​−100​010​001​−5−24​42−3​ ​ If the unknowns in this system are x 1 , x 2 , x 3 , x 4 , x 5 x\_1,\\ x\_2,\\ x\_3,\\ x\_4,\\ x\_5 x1​, x2​, x3​, x4​, x5​ , find the solution(s) to the linear system.](https://www.wizeprep.com/practice-questions/125415) [Solving Linear SystemsConsider the following augmented matrix in RREF: \[ 1 − 1 0 0 − 5 4 0 0 1 0 − 2 2 0 0 0 1 4 − 3 \] \\left\[\\begin{array}{rrrrr\|r} 1&-1&0&0&-5&4\\\\\[0.5em\] 0&0&1&0&-2&2\\\\\[0.5em\] 0&0&0&1&4&-3 \\end{array}\\right\] ​ 100​−100​010​001​−5−24​42−3​ ​ If the unknowns in this system are x 1 , x 2 , x 3 , x 4 , x 5 x\_1,\\ x\_2,\\ x\_3,\\ x\_4,\\ x\_5 x1​, x2​, x3​, x4​, x5​ , find the solution(s) to the linear system.](https://www.wizeprep.com/practice-questions/106324) [Basics of Systems of Linear EquationsDetermine which vectors are solutions to the following SLE \[select all that apply\]: x 1 − 5 x 2 + x 3 = 2 − x 1 + 5 x 2 − 3 x 3 = − 6 \\begin{alignedat}{8} x\_1\\ &-&\\ 5x\_2\\ &+& x\_3 &=& 2\\\\ -x\_1\\ &+& 5x\_2\\ &-& 3x\_3 &=&\\ -6 \\end{alignedat} x1​ −x1​ ​−+​ 5x2​ 5x2​ ​+−​x3​3x3​​==​2 −6​](https://www.wizeprep.com/practice-questions/106132) [Interpretation of Linear SystemsWhich of the following describes the intersection between the following 3 hyperplanes in R 4 \\reals^4 R4 ? 2 x 1 + 3 x 3 + x 4 = 1 x 2 − x 4 = 0 x 2 = − 3 \\begin{array}{rcl} 2x\_1+3x\_3+x\_4&=&1\\\\ x\_2-x\_4&=&0\\\\ x\_2&=&-3 \\end{array} 2x1​+3x3​+x4​x2​−x4​x2​​===​10−3​](https://www.wizeprep.com/practice-questions/114614) ###### Company [About](https://www.wizeprep.com/about)[Careers](https://www.wizeprep.com/careers)[Blog](https://www.wizeprep.com/blog)[Free Resources](https://www.wizeprep.com/blog?category=free_resources)[Pricing](https://www.wizeprep.com/pricing)[Help Center](https://help.wizeprep.com/)[Scholarships](https://www.wizeprep.com/scholarships) ###### University [Calculus](https://www.wizeprep.com/courses?school_type=undergrad&subject=mathematics)[Chemistry](https://www.wizeprep.com/courses?school_type=undergrad&subject=chemistry)[Biology](https://www.wizeprep.com/courses?school_type=undergrad&subject=biology)[Statistics](https://www.wizeprep.com/courses?school_type=undergrad&subject=statistics)[Textbooks](https://www.wizeprep.com/textbooks/undergrad)[See All](https://www.wizeprep.com/courses?school_type=undergrad) ###### High School [HS Math](https://www.wizeprep.com/courses?school_type=high_school&subject=mathematics) [HS Biology](https://www.wizeprep.com/courses?school_type=high_school&subject=biology) [HS Chemistry](https://www.wizeprep.com/courses?school_type=high_school&subject=chemistry) [HS English](https://www.wizeprep.com/courses?school_type=high_school&subject=english) [HS Textbooks](https://www.wizeprep.com/textbooks/high-school) [AP Textbooks](https://www.wizeprep.com/textbooks/ap) [See All](https://www.wizeprep.com/courses?school_type=high_school) ###### MCAT [MCAT Programs](https://www.wizeprep.com/mcat)[Free MCAT Resources](https://www.wizeprep.com/mcat/free-resources)[MCAT Events](https://www.wizeprep.com/mcat/events)[Med School Calculator](https://www.wizeprep.com/med-school-calculator)[MCAT Blog](https://www.wizeprep.com/blog?category=mcat) Terms of Use \| Privacy Policy ###### Wizedemy Inc. ©2026

[Wize University Linear Algebra Textbook \> Systems of Linear Equations (SLEs) (Linear Systems)](https://www.wizeprep.com/textbooks/undergrad/mathematics) ###### Popular Courses [Find My Course](https://www.wizeprep.com/courses?school_type=undergrad&subject=mathematics)  0:00 / 0:00 ## Geometric Interpretation of Solutions to SLEs Linear systems are made up of equations of hyperplanes (e.g. lines in R 2 \\reals^2; planes in R 3 \\reals^3). A solution is the intersection of all the hyperplanes. #### R 2 \\colorOne{\\reals^2}: Intersection of Lines - No intersection - Unique point of intersection (2 lines cross) - Infinitely many points of intersection - Line → \\rightarrow 1-parameter family of solutions (same line) Example { − x \+ y \= 2 y \= 1 ⟹ \[ − 1 1 2 0 1 1 \] ⟹ x ⃗ \= \[ − 1 1 \] P O I \\begin{cases} -x&+\&y&=&2\\\\ &\&y&=&1 \\end{cases} \\quad \\implies \\quad \\left\[ \\begin{array}{rr\|r} -1 & 1 & 2\\\\ 0 & 1 & 1\\\\ \\end{array} \\right\] \\quad \\implies \\quad \\overset{\\normalsize POI}{ \\boxed{ \\vec x = \\left\[ \\begin{array}{r} -1\\\\ 1 \\end{array} \\right\] }}  PAGE BREAK #### R 3 \\colorOne{\\reals^3}: Intersection of Planes - No intersection - Unique point of intersection (requires 3 planes) - Infinitely many points of intersection - Line → \\rightarrow 1-parameter family of solutions (2 planes intersect) - Plane → \\rightarrow 2-parameter family of solutions (2 of the same plane) Example { − x \+ y \= 2 y \= 1 ⟹ \[ − 1 1 0 2 0 1 0 1 \] ⟹ x ⃗ \= \[ − 1 1 0 \] \+ t \[ 0 0 1 \] line of intersection \\begin{cases} -x&+\&y&=&2\\\\ &\&y&=&1 \\end{cases} \\quad \\implies \\quad \\left\[ \\begin{array}{rrr\|r} -1 & 1 & 0 & 2\\\\ 0 & 1 & 0 & 1\\\\ \\end{array} \\right\] \\quad \\implies \\quad \\overset{\\text{\\normalsize line of intersection}}{ \\boxed{ \\vec x = \\left\[ \\begin{array}{r} -1\\\\ 1\\\\ 0 \\end{array} \\right\] +t \\left\[ \\begin{array}{r} 0\\\\ 0\\\\ 1 \\end{array} \\right\] }}   0:00 / 0:00 ## Example: Geometric Interpretation of a Linear System Solve the system of linear equations and interpret the solution geometrically: 2 x \+ 2 y \= 6 − x − y \= − 3 \\begin{array}{rl} 2x+2y&=6\\\\ -x-y&=-3\\\\ \\end{array} Steps 1. Let's first write this in augmented matrix form: 2 x \+ 2 y \= 6 − x − y \= − 3 → \[ 2 2 6 − 1 − 1 − 3 \] \\begin{array}{rcr} 2x+2y&=&6\\\\\[0.2em\] -x-y&=&-3\\\\\[0.2em\] \\end{array} \\quad \\rightarrow \\quad \\left\[\\begin{array}{rr\|r} 2&2&6\\\\ -1&-1&-3\\\\ \\end{array}\\right\] 1. Turn the matrix into RREF using EROs: \[ 2 2 6 − 1 − 1 − 3 \] 1 2 R 1 ⟶ \[ 1 1 3 − 1 − 1 − 3 \] R 2 \+ R 1 ⟶ \[ 1 1 3 0 0 0 \] \\begin{aligned} &\\left\[\\begin{array}{rr\|r} 2&2&6\\\\ -1&-1&-3\\\\ \\end{array}\\right\] \\begin{array}{l} \\dfrac{1}{2}R\_1\\\\ \\\\ \\\\ \\end{array}\\\\\[2.5em\] \\longrightarrow &\\left\[\\begin{array}{rr\|r} 1&1&3\\\\ -1&-1&-3\\\\ \\end{array}\\right\] \\begin{array}{l} \\\\ R\_2 + R\_1\\\\ \\end{array}\\\\\[2.5em\] \\longrightarrow &\\left\[\\begin{array}{rr\|r} 1&1&3\\\\ 0&0&0\\\\ \\end{array}\\right\] \\end{aligned} 1. Find all solutions to the SLE: The coefficient matrix has just one leading 1, but 2 columns/variables: r a n k ( A ) \= 1 \< n \= 2 {\\rm rank}(A)=1 \\ \\ \\bm\< \\ \\ \\ n=2 Thus, there is 2 − 1 \= 1 2-1 = 1 free variable (y y). Let y \= t y=t. Rewrite the augmented matrix back into a system of linear equations: { x \+ t \= 3 y \= t ⟹ { x \= 3 − t y \= t \\begin{cases} x+t=3\\\\ y=t \\end{cases} \\quad\\implies\\quad \\begin{cases} x=3-t\\\\ y=t \\end{cases} Therefore, the solution is x ⃗ \= \[ 3 0 \] \+ t \[ − 1 1 \] \\vec x = \\begin{bmatrix} 3\\\\ 0\\\\ \\end{bmatrix} +t \\begin{bmatrix} -1\\\\ 1\\\\ \\end{bmatrix}. 1. Interpret the solution geometrically: We were given a system of linear equations in 2 variables ⟹ R 2 \\implies \\reals^2. Each equation is a hyperplane: in R 2 \\reals^2, hyperplanes are simply lines (in general, a hyperplane in R n \\reals^n is n − 1 n-1 dimensional). So we have 2 lines, but both lines are identical (hence reducing to a row of 0s). This means the lines intersect everywhere on the line defined by x ⃗ \= \[ 3 0 \] \+ t \[ − 1 1 \] \\vec x = \\begin{bmatrix} 3\\\\ 0\\\\ \\end{bmatrix} +t \\begin{bmatrix} -1\\\\ 1\\\\ \\end{bmatrix}. Consider the following augmented matrix in RREF: \[ 1 − 1 0 0 − 5 4 0 0 1 0 − 2 2 0 0 0 1 4 − 3 \] \\left\[\\begin{array}{rrrrr\|r} 1&-1&0&0&-5&4\\\\\[0.5em\] 0&0&1&0&-2&2\\\\\[0.5em\] 0&0&0&1&4&-3 \\end{array}\\right\] If the unknowns in this system are x 1 , x 2 , x 3 , x 4 , x 5 x\_1,\\ x\_2,\\ x\_3,\\ x\_4,\\ x\_5, find the solution(s) to the linear system. Then, determine which of the following is the correct geometrical interpretation of the solution. A) The five hyperplanes intersect at a single point in R 5 \\reals^5 B) The three hyperplanes are non-intersecting C) The three hyperplanes intersect in a line in R 5 \\reals^5 D) The three hyperplanes intersect in a plane in R 5 \\reals^5 E) None of the above I don't know