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Linear System of Equations
Linear System of Equations
System of equations of form
a1x + b1y + c1z = d1
a2x + b2y + c2z = d2
a3x + b3y + c3z = d3
is called a linear system of equation.
In the matrix form above system of equation is written as:
Linear System of Equations
Vector X is such that on linear transformation with matrix M, it lands on vector D.
Here X is unknown while D and M are known. So if we perform inverse transformation of M on D, we can find X.
Learn Basics:
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Consistency of Linear System of Equations
Case 1:
For M ≠ 0, there always exists a unique vector X which after linear transformation lands on vector D. In this case, there exists a single unique solution of system of equations.
Case 2:
For M = 0, unit vector i and j become collinear after transformation and 2D plane is reduced to a line.

If vector D lies on the same line as iT and jT, then system of equation has infinitely many solution.
Consistency of system

If vector D does not lie on the same line as iT and jT, then system of equation is inconsistent amd no solution exists.
For a zero determinant matrix, whether solution exists or not is determined by the concept of rank.
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Rank of a Matrix
For a linear transformation matrix, rank is defined as number of dimensions in the output of transformation.
Consider a 2x2 square matrix. It may transform a 2D plane such that:

iT and jT (unit vector i and j after transformation) remain in one plane but are not collinear. Number of dimension in the output of transformation is 2, hence rank is 2.
Rank

iT and jT become collinear after transformation. No of dimension in the output is reduced to 1, hence rank is 1.

iT and jT are reduced to a point after linear transformation. Since origin remain fixed in linear transformation, hence iT and jT both shift to origin. For this case rank of the matrix is zero.
Similarly for a 3x3 matrix rank may be 3 (output dimension after transformation is 3), 2 (3D space is reduced to a 2D plane), 1 (3D space is reduced to a line after transformation) or 0 (space is reduced to a point after transformation).
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To find rank, matrix is converted into echelon form. No of nonzero rows (rows having minimum one element as nonzero) in echelon form is called rank of the matrix.
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* For a nonsingular square matrix (M ≠ 0), rank is always equal to number of columns of the matrix.
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Consistency of Linear System of Equation
Consider a linear system of equation MX = D. For this system augmented matrix MA is defined as
A) System of equation is consistent if r(MA) = r(M)
i) If r(MA) = r(M) = number of unknows, system of equation has unique solution.
ii) If r(MA) = r(M) < number of unknows, system of equation has infinite number of solution.
B) System of equation is inconsistent if r(MA) ≠ r(M): No solution exists.
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Example 1:
GATE 2018: The rank of matrix M (as given below) is
A) 1
B) 2
C) 3
D) 4
Previous years questions
Solution:
M = 4(13) 1(1+7) 1(3+7) = 0
Since determinant of matrix is zero, hence this is not a full rank matrix. Linear transformation will reduce the space either to a plane or to a line.
r(M)<3
Unit vector i, j and k lands to iT, jT and kT after linear transformation:
iT = column vector C1 of matrix = 4i j +7k
jT = column vector C2 of matrix = i  j 3k
kT = column vector C3 of matrix = i j +k
From the above, it is clear that all vectors lie in a plane y = 1. Hence after transformation ouput dimensions are 2 only.
So, rank of matrix r(M) = 2
Option B is the correct answer.
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Example 2:
GATE 2005: A is a 3x4 real matrix and Ax = b is an inconsistent system of equation. The highest possible rank of A is:
A) 1
B) 2
C) 3
D) 4
Solution: Consider a 3x4 real matrix A as:
For system of equation Ax = b, augmented matrix of A:b will be a 3 x 5 matrix. In echelon form maximum number of nonzero rows for augmented matrix may be 3. Hence, maximum rank of augmented matrix is 3.
Now since the system of equation is inconsistent:
r(A) ≠ r(A:b)
So maximum rank of A may be 2.
Option B is the correct answer.
Example 3:
GATE 2001: The rank of a 3x3 matrix C=(AB), formed by multiplying a nonzero column matrix A of size 3x1 and a nonzero row matrix B of size 1x3 is
A) 0
B) 1
C) 2
D) 3
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Solution: Consider matrix A and B as
For matrix C, transformed i, j and k vectors are given as:
iT = d*(ai + bj + ck)
jT = e*(ai + bj + ck)
kT = f*(ai + bj + ck)
It is observed from the above, that all three unit vectors become collinear after transformation. Hence, output dimension after transformation is 1.
So r(C) = 1
Option B is the correct answer.
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Example 4:
GATE 1999: Rank of the matrix M (given below) is:
A) 1
B) 2
C) 3
D) √2
Solution: For matrix M column vector is given as:
iT = 3i  6j + 12k = 3(i  2j + 4k)
jT = 2i  4j + 8k = 2(i  2j + 4k)
kT = 9i + 18j 36k = 9(i  2j + 4k)
It is observed from the above that all three unit vectors become collinear after transformation. Hence, output dimension after transformation is 1.
So r(M) = 1
Option A is the correct answer.
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Example 5:
GATE 2019: The set of equations
x + y + z = 1
ax  ay + 3z = 5
5x  3y +az = 6
has infinite solution if a =
A) 3
B) 3
C) 4
D) 4
Solution: System of equation may be written in matrix form as AX = D
where;
Since system of equation has infinite solution, determinant of A must be zero.
A = 1(a*a + 9) 1(a*a  15) + 1(3a + 5a) = 0
 a^2 + 9  a^2 + 15 + 2a = 0
2a^2  2a  24 = 0
a^2  a  12 = 0
(a4)(a+3) = 0
a = 4, 3
For a = 4, augmented matrix A:D is
Convert to echelon form
r(A) = r(A:D) = 2
Hence system of equation has infinitely many solution for a = 4.
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Similarly for a = 3, augmented matrix A:D in echelon form is
r(A) = 2 but r(A:D) = 3
Hence system of equation is inconsistent for a = 3.
Option C is the correct answer.
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Example 6:
GATE 2016: The solution to the system of equations as given below is
A) 6,2
B) 6,2
C) 6,2
D) 6,2
Solution: For system of equation Ax = B, x is found out by
Matrix of cofactors of A
Adjoint of A
Inverse of A
Hence correct answer is option D.
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Example 7:
GATE 2012: The system of algebraic equations given below has
x + 2y + z = 4
2x + y + 2z = 5
x  y + z = 1
A) a unique solution of x = 1, y = 1, z = 1
B) only two solutions of (x = 1, y = 1 & z = 1) and (x = 2, y = 1, z = 0)
C) infinite number of solution
D) no feasible solution
Solution: System of equation may be written in matrix form as Ax = B
where;
A = 1(1+2) 2(22) +1(21) = 0
Since determinant of A is zero, hence system of equation will have either infinite solution or no solution.
Augmented matrix A:B is
Echelon form of A:B is
r(A) = r(A:B)
Hence system of equation has infinite solution.
Option C is the correct answer.
Example 8:
GATE 2011: Consider the following system of equations
2x1 + x2 + x3 = 0
x2  x3 = 0
x1 + x2 = 0
This system has
A) A unique solution
B) No solution
C) Infinite number of solutions
D) Five solutions
Solution: System of equation in matrix form is written as
A = 2(0+1)  1(0+1) + 1(01) = 0
Since this is a homogeneous system of equation and A = 0, hence it has infinite no of solutions.
Option C is the correct answer.
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*Note: Homogeneous system of equations either have unique solution (A ≠ 0) are infinite number of solution (A = 0).
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Example 9:
GATE 2008: For what value of a, if any, will the following system of equations in x, y and z have a solution
2x + 3y = 4
x + y + z = 4
x + 2y − z = a
(A) Any real number
(B) 0
(C) 1
(D) There is no such value
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Solution: System of equation in matrix form Ax = B is written as:
A = 0
Augmented matrix A:B is written as
Augmented matrix in echelon form is
For solution to exist rank of A and A:B should be equal.
Since r(A) = 2, hence r(A:B) = 2 for which a = 0
Option B is the correct answer.
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