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NIMCET 2016 Mathematics PYQ — The system of the equations x + y + 2z = a x + z = b 2x + y + 3z … | Mathem Solvex

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NIMCET 2016 Mathematics PYQ — The system of the equations x + y + 2z = a x + z = b 2x + y + 3z … | Mathem Solvex | Mathem Solvex

Explanation

Concept:
Consider the system of m linear equations
$a_{11} x_{1} + a_{12} x_{2} + \dots + a_{1 n} x_{n} = b_{1}$
$a_{21} x_{1} + a_{22} x_{2} + \dots + a_{2 n} x_{n} = b_{2}$
$\dots$
$a_{m 1} x_{1} + a_{m 2} x_{2} + \dots + a_{mn} x_{n} = b_{m}$

The above equations containing the n unknowns $x_{1}$ , $x_{2}$ , ..., $x_{n}$ . To determine whether the above system of equations is consistent or not, we need to find the rank of the following matrices.

$A = a_{11} a_{21} \dots a_{m 1} am p; a_{12} am p; a_{22} am p; \dots am p; a_{n 2} am p; \dots am p; \dots am p; \dots am p; \dots am p; a_{1 n} am p; a_{2 n} am p; \dots am p; a_{mn} and [A ∣ B] = a_{11} a_{21} \dots a_{m 1} am p; a_{12} am p; a_{22} am p; \dots am p; a_{n 2} am p; \dots am p; \dots am p; \dots am p; \dots am p; a_{1 n} am p; a_{2 n} am p; \dots am p; a_{mn} am p; b_{1} am p; b_{2} am p; \dots am p; b_{m}$

A is the coefficient matrix and [A|B] is called an augmented matrix of the given system of equations.

We can find the consistency of the given system of equations as follows:

(i) If the rank of matrix A is equal to the rank of an augmented matrix and it is equal to the number of unknowns, then the system is consistent and there is a unique solution.

Rank of A = Rank of augmented matrix = n

(ii) If the rank of matrix A is equal to the rank of an augmented matrix and it is less than the number of unknowns, then the system is consistent and there are an infinite number of solutions.

Rank of A = Rank of augmented matrix < n

(iii) If the rank of matrix A is not equal to the rank of the augmented matrix, then the system is inconsistent, and it has no solution.

Rank of A ≠ Rank of the augmented matrix

Calculation:

From the given equations:
$x + y + 2 z = a$
$x + z = b$
$2 x + y + 3 z = c$

The coefficient matrix is:
A = $112 am p; 1 am p; 0 am p; 1 am p; 2 am p; 1 am p; 3$ and the augmented matrix, [A][B] = $112 am p; 1 am p; 0 am p; 1 am p; 2 am p; 1 am p; 3 am p; a am p; b am p; c$

The determinant of the matrix [A] = $112 am p; 1 am p; 0 am p; 1 am p; 2 am p; 1 am p; 3$

R3 = R3 - R1 - R2
$\Rightarrow ∣ A ∣ = 110 am p; 1 am p; 0 am p; 0 am p; 2 am p; 1 am p; 0 = 0$

The rank of matrix is 2
$So, the solution if the rank of matrix A is equal to rank of augmented matrix A|B = 2$
$\Rightarrow [A ∣ B] = 112 am p; 1 am p; 0 am p; 1 am p; 2 am p; 1 am p; 3 am p; a am p; b am p; c$
$R_{3} = R_{3} - R_{1} - R_{2}$
$[A ∣ B] = 110 am p; 1 am p; 0 am p; 0 am p; 2 am p; 1 am p; 0 am p; a am p; b am p; c - a - b$
$For rank of [A|B] to be 2$
$c - a - b = 0$
$c = a + b$

Explanation

A is the coefficient matrix and [A|B] is called an augmented matrix of the given system of equations.

We can find the consistency of the given system of equations as follows:

(i) If the rank of matrix A is equal to the rank of an augmented matrix and it is equal to the number of unknowns, then the system is consistent and there is a unique solution.

Rank of A = Rank of augmented matrix = n

(ii) If the rank of matrix A is equal to the rank of an augmented matrix and it is less than the number of unknowns, then the system is consistent and there are an infinite number of solutions.

Rank of A = Rank of augmented matrix < n

(iii) If the rank of matrix A is not equal to the rank of the augmented matrix, then the system is inconsistent, and it has no solution.

Rank of A ≠ Rank of the augmented matrix

Calculation:

From the given equations:
$x + y + 2 z = a$
$x + z = b$
$2 x + y + 3 z = c$

The coefficient matrix is:
A = $112 am p; 1 am p; 0 am p; 1 am p; 2 am p; 1 am p; 3$ and the augmented matrix, [A][B] = $112 am p; 1 am p; 0 am p; 1 am p; 2 am p; 1 am p; 3 am p; a am p; b am p; c$

The determinant of the matrix [A] = $112 am p; 1 am p; 0 am p; 1 am p; 2 am p; 1 am p; 3$

R3 = R3 - R1 - R2
$\Rightarrow ∣ A ∣ = 110 am p; 1 am p; 0 am p; 0 am p; 2 am p; 1 am p; 0 = 0$

NIMCET 2016 — Mathematics PYQ

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