Matrices & Determinants
System of linear equations
Grade Class 12
Question:
If system of equation <i>a</i><sub>1</sub><i>x</i> + <i>b</i><sub>1</sub><i>y</i> = <i>c</i><sub>1</sub> & <i>a</i><sub>2</sub><i>x</i> + <i>b</i><sub>2</sub><i>y</i> = <i>c</i><sub>2</sub> (where <i>a</i><sub>1</sub>, <i>b</i><sub>1</sub>, <i>c</i><sub>1</sub>, <i>a</i><sub>2</sub>, <i>b</i><sub>2</sub>, <i>c</i><sub>2</sub> ≠ 0) has infinite solutions, then-<br>(A) <sup><i>a</i><sub>1</sub></sup>/<sub><i>a</i><sub>2</sub></sub> = <sup><i>b</i><sub>1</sub></sup>/<sub><i>b</i><sub>2</sub></sub> = <sup><i>c</i><sub>1</sub></sup>/<sub><i>c</i><sub>2</sub></sub><br>(B) <sup><i>a</i><sub>1</sub> + <i>a</i><sub>2</sub></sup>/<sub><i>a</i><sub>1</sub> - <i>a</i><sub>2</sub></sub> = <sup><i>b</i><sub>1</sub> + <i>b</i><sub>2</sub></sup>/<sub><i>b</i><sub>1</sub> - <i>b</i><sub>2</sub></sub> = <sup><i>c</i><sub>1</sub> + <i>c</i><sub>2</sub></sup>/<sub><i>c</i><sub>1</sub> - <i>c</i><sub>2</sub></sub><br>(C) the quadratic equations <i>a</i><sub>1</sub><i>x</i><sup>2</sup> + <i>b</i><sub>1</sub><i>x</i> + <i>c</i><sub>1</sub> = 0 & <i>a</i><sub>2</sub><i>x</i><sup>2</sup> + <i>b</i><sub>2</sub><i>x</i> + <i>c</i><sub>2</sub> = 0 have no common root<br>(D) system of equation <i>a</i><sub>1</sub><sup>2</sup> <i>a</i><sub>2</sub><i>x</i> + <i>b</i><sub>1</sub><sup>2</sup> <i>b</i><sub>2</sub><i>y</i> = <i>c</i><sub>1</sub><sup>2</sup> <i>c</i><sub>2</sub> & <i>a</i><sub>1</sub> <i>a</i><sub>2</sub><sup>2</sup> <i>x</i> + <i>b</i><sub>1</sub> <i>b</i><sub>2</sub><sup>2</sup> <i>y</i> = <i>c</i><sub>1</sub> <i>c</i><sub>2</sub><sup>2</sup> will also have infinite number of solutions
(A) <sup><i>a</i><sub>1</sub></sup>/<sub><i>a</i><sub>2</sub></sub> = <sup><i>b</i><sub>1</sub></sup>/<sub><i>b</i><sub>2</sub></sub> = <sup><i>c</i><sub>1</sub></sup>/<sub><i>c</i><sub>2</sub></sub>
(B) <sup><i>a</i><sub>1</sub> + <i>a</i><sub>2</sub></sup>/<sub><i>a</i><sub>1</sub> - <i>a</i><sub>2</sub></sub> = <sup><i>b</i><sub>1</sub> + <i>b</i><sub>2</sub></sup>/<sub><i>b</i><sub>1</sub> - <i>b</i><sub>2</sub></sub> = <sup><i>c</i><sub>1</sub> + <i>c</i><sub>2</sub></sup>/<sub><i>c</i><sub>1</sub> - <i>c</i><sub>2</sub></sub>
(C) the quadratic equations <i>a</i><sub>1</sub><i>x</i><sup>2</sup> + <i>b</i><sub>1</sub><i>x</i> + <i>c</i><sub>1</sub> = 0 & <i>a</i><sub>2</sub><i>x</i><sup>2</sup> + <i>b</i><sub>2</sub><i>x</i> + <i>c</i><sub>2</sub> = 0 have no common root
(D) system of equation <i>a</i><sub>1</sub><sup>2</sup> <i>a</i><sub>2</sub><i>x</i> + <i>b</i><sub>1</sub><sup>2</sup> <i>b</i><sub>2</sub><i>y</i> = <i>c</i><sub>1</sub><sup>2</sup> <i>c</i><sub>2</sub> & <i>a</i><sub>1</sub> <i>a</i><sub>2</sub><sup>2</sup> <i>x</i> + <i>b</i><sub>1</sub> <i>b</i><sub>2</sub><sup>2</sup> <i>y</i> = <i>c</i><sub>1</sub> <i>c</i><sub>2</sub><sup>2</sup> will also have infinite number of solutions
Step-by-Step Solution
Key Concept: For a system of linear equations to have infinite solutions, the ratios of coefficients must be equal. Applying componendo and dividendo to the ratio a1/a2 = b1/b2 = c1/c2 = k yields (a1+a2)/(a1-a2) = (b1+b2)/(b1-b2) = (c1+c2)/(c1-c2).
For infinite solutions, a1/a2 = b1/b2 = c1/c2 = k. By componendo and dividendo, (a1+a2)/(a1-a2) = (b1+b2)/(b1-b2) = (c1+c2)/(c1-c2). Thus, option (B) is correct.
Correct Answer: 3