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Matrices and Determinants question

2020 · 4 Sep · Shift 1 · Q25
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  5. /2020 · 4 Sep · Shift 1 · Q25

Matrices and Determinants question

2020 · 4 Sep · Shift 1 · Q25

JEE MainMathematicsMatrices and DeterminantsNumerical+4 / −1
If the system of equations x - 2y + 3z = 9 2x + y + z = b x - 7y + az = 24, has infinitely many solutions, then a - b is equal to.........
Numerical answer
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Correct answer: 5

  1. For the system
{x−2y+3z=92x+y+z=bx−7y+az=24\begin{cases} x-2y+3z=9\\ 2x+y+z=b\\ x-7y+az=24 \end{cases}⎩⎨⎧​x−2y+3z=92x+y+z=bx−7y+az=24​

to have infinitely many solutions, the three equations must be dependent and consistent.

That means the third equation must be a linear combination of the first two.

  1. Let
λ(x−2y+3z=9)+μ(2x+y+z=b)\lambda(x-2y+3z=9)+\mu(2x+y+z=b)λ(x−2y+3z=9)+μ(2x+y+z=b)

produce the third equation:

x−7y+az=24.x-7y+az=24.x−7y+az=24.

So we compare coefficients.

From coefficients of x,y,zx,y,zx,y,z:

λ+2μ=1...(1)\lambda+2\mu=1 \quad ...(1)λ+2μ=1...(1) −2λ+μ=−7...(2)-2\lambda+\mu=-7 \quad ...(2)−2λ+μ=−7...(2) 3λ+μ=a...(3)3\lambda+\mu=a \quad ...(3)3λ+μ=a...(3)

And from constants:

9λ+bμ=24....(4)9\lambda+b\mu=24. \quad ...(4)9λ+bμ=24....(4)
  1. Solve (1)(1)(1) and (2)(2)(2):

From (1)(1)(1),

λ=1−2μ.\lambda=1-2\mu.λ=1−2μ.

Substitute into (2)(2)(2):

−2(1−2μ)+μ=−7-2(1-2\mu)+\mu=-7−2(1−2μ)+μ=−7 −2+4μ+μ=−7-2+4\mu+\mu=-7−2+4μ+μ=−7 5μ=−55\mu=-55μ=−5 μ=−1.\mu=-1.μ=−1.

Then

λ=1−2(−1)=3.\lambda=1-2(-1)=3.λ=1−2(−1)=3.
  1. Find aaa using (3)(3)(3):
a=3λ+μ=3(3)+(−1)=9−1=8.a=3\lambda+\mu=3(3)+(-1)=9-1=8.a=3λ+μ=3(3)+(−1)=9−1=8.
  1. Find bbb using (4)(4)(4):
9λ+bμ=249\lambda+b\mu=249λ+bμ=24 9(3)+b(−1)=249(3)+b(-1)=249(3)+b(−1)=24 27−b=2427-b=2427−b=24 b=3.b=3.b=3.
  1. Therefore,
a−b=8−3=5.a-b=8-3=5.a−b=8−3=5.

Hence the required integer is

5.\boxed{5}.5​.

Comparison with stored answer: stored correct answer is 555, which matches our result.

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