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Matrices and Determinants question

2024 · 5 Apr · Shift 2 · Q39
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  5. /2024 · 5 Apr · Shift 2 · Q39

Matrices and Determinants question

2024 · 5 Apr · Shift 2 · Q39

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
The values of m,nm, nm,n, for which the system of equations x+y+z=4,2x+5y+5z=17,x+2y+mz=n\begin{aligned} & x+y+z=4, \\ & 2 x+5 y+5 z=17, \\ & x+2 y+\mathrm{m} z=\mathrm{n} \end{aligned}​x+y+z=4,2x+5y+5z=17,x+2y+mz=n​ has infinitely many solutions, satisfy the equation :
  1. A
    m2+n2−m−n=46\mathrm{m}^2+\mathrm{n}^2-\mathrm{m}-\mathrm{n}=46m2+n2−m−n=46
  2. B
    m2+n2+mn=68\mathrm{m}^2+\mathrm{n}^2+\mathrm{mn}=68m2+n2+mn=68
  3. C
    m2+n2−mn=39\mathrm{m}^2+\mathrm{n}^2-\mathrm{mn}=39m2+n2−mn=39
  4. D
    m2+n2+m+n=64\mathrm{m}^2+\mathrm{n}^2+\mathrm{m}+\mathrm{n}=64m2+n2+m+n=64
View written solutionFree

Correct answer: C

  1. For the system to have infinitely many solutions, the third equation must be a linear combination of the first two, and the coefficient matrix must be singular in a consistent way.

The equations are:

x+y+z=4...(1)2x+5y+5z=17...(2)x+2y+mz=n...(3)\begin{aligned} &x+y+z=4 \quad ...(1)\\ &2x+5y+5z=17 \quad ...(2)\\ &x+2y+mz=n \quad ...(3) \end{aligned}​x+y+z=4...(1)2x+5y+5z=17...(2)x+2y+mz=n...(3)​
  1. Let equation (3) be a linear combination of (1) and (2):
(3)=α(1)+β(2)(3)=\alpha(1)+\beta(2)(3)=α(1)+β(2)

Then comparing coefficients of x,y,zx,y,zx,y,z and constants:

α+2β=1\alpha+2\beta=1α+2β=1 α+5β=2\alpha+5\beta=2α+5β=2 α+5β=m\alpha+5\beta=mα+5β=m 4α+17β=n4\alpha+17\beta=n4α+17β=n
  1. Solve for α,β\alpha,\betaα,β using the first two equations: Subtract:
(α+5β)−(α+2β)=2−1(\alpha+5\beta)-(\alpha+2\beta)=2-1(α+5β)−(α+2β)=2−1 3β=1⇒β=133\beta=1 \Rightarrow \beta=\frac133β=1⇒β=31​

Then

α+2(13)=1\alpha+2\left(\frac13\right)=1α+2(31​)=1 α=13\alpha=\frac13α=31​
  1. Now find mmm and nnn: From
α+5β=m\alpha+5\beta=mα+5β=m

we get

m=13+5⋅13=63=2m=\frac13+5\cdot\frac13=\frac63=2m=31​+5⋅31​=36​=2

From

4α+17β=n4\alpha+17\beta=n4α+17β=n

we get

n=4⋅13+17⋅13=213=7n=4\cdot\frac13+17\cdot\frac13=\frac{21}{3}=7n=4⋅31​+17⋅31​=321​=7

Thus,

m=2,n=7m=2,\quad n=7m=2,n=7
  1. Check the options.

Option A:

m2+n2−m−n=22+72−2−7=4+49−9=44m^2+n^2-m-n=2^2+7^2-2-7=4+49-9=44m2+n2−m−n=22+72−2−7=4+49−9=44

Not correct.

Option B:

m2+n2+mn=4+49+14=67m^2+n^2+mn=4+49+14=67m2+n2+mn=4+49+14=67

Not correct.

Option C:

m2+n2−mn=4+49−14=39m^2+n^2-mn=4+49-14=39m2+n2−mn=4+49−14=39

Correct.

Option D:

m2+n2+m+n=4+49+2+7=62m^2+n^2+m+n=4+49+2+7=62m2+n2+m+n=4+49+2+7=62

Not correct.

Therefore, the correct option is:

C\boxed{\text{C}}C​
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