Test: Solving Simultaneous Equations - Commerce MCQ

 2x + 3y = 5 
Multiply equation with ‘5’, we get
10x + 15y = 25……………………(1)
5x – 2y = 3
Multiply equation with ‘2’, we get
10x - 4y = 6………………………(2)
Subtracting (1) from (2), we get
19y = 19
y = 1
Put the value of y in eq(1)
10x + 15(1) = 25
10x = 10
 x = 1

Solution:-
Let the two angles be 'x' and 'y'.
So, according to the question,
One third of the sum of two angles :- 1/3(x+y) = 60  
x/3 + y/3 = 60    ......(1)
Quarter of their difference :- 1/4(x-y) = 28  
x/4 - y/4 = 28...........(2)
Multiplying the equation (1) by 3 and equation (2) by 4, we get
x + y =180  .......(3)   x - y = 112  ......(4)
Subtracting (4) from (3),
 x + y = 180
 x - y = 112
-   +     -
_________
    2y = 68
_________
2y = 68
y = 34
Putting the value of y = 34 in the equation (3)
x + y = 180
x + 34 = 180
x = 180 - 34
x = 146
The two angles are 146° and 34°.

Given, AX = B, where A is a square matrix.

If A is invertible (i.e., A 0), then there exists a unique solution for X.

Explanation: When A is invertible, it means that there exists a unique matrix A-1 such that A-1A = I, where I is the identity matrix.

Now, if we multiply both sides of the given equation by A-1, we get: A-1AX = A-1B ⇒ IX = A-1B (using A-1A = I) ⇒ X = A-1B

Hence, we get a unique solution for X, which is X = A-1B. This is because the inverse of a matrix is unique, and so there can be only one solution for X.

Therefore, the correct option is (A) - There exists a unique solution.
 

Let A = {(1,3,3) (1,4,3) (1,3,4)}
|A| = 1(16-9) -3(4-3) +3(3-4)
|A| = 1(7) -3(1) +3(-1)
= 7 - 3 - 3
= 1
Therefore, A is not equal to zero, it has unique solution.

Inverse of matrix (A^-1) = (adj A)/|A| 

The inverse of matrix \( A \), which is the identity matrix \( I \), is itself. So, the answer is ( A^{-1} = A ), which corresponds to option C.

This system of equations has a unique solution, if

If (adj A) B ≠ 0 (zero matrix), then the solution does not exist. The system of equations is inconsistent. Else, if (adj A) B = 0 then the system will either have infinitely many solutions (consistent system) or no solution (inconsistent system).

A = {(6,7) (8,9)}
|A| = (6 * 9) - (8 * 7)
= 54 - 56 
|A| = -2
A^-1 = -½{(9,-7) (-8,6)}
A^-1 = {(-9/2, 7/2) (4,-3)}

A linear system is said to be consistent if it has at least one solution; and is said to be inconsistent if it has no solution. have no solution, a unique solution, and infinitely many solutions, respectively.