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Matrices and Determinants question

2021 · Shift 1 · Q26
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  5. /2021 · Shift 1 · Q26

Matrices and Determinants question

2021 · Shift 1 · Q26

JEE AdvancedMathematicsMatrices and DeterminantsNumerical+2 / −1
Let α\alphaα, β\betaβ and γ\gammaγ be real numbers such that the system of linear equations x + 2y + 3z =α\alphaα 4x + 5y + 6z =β\betaβ 7x + 8y + 9z =γ−\gamma -γ− 1 is consistent. Let | M | represent the determinant of the matrix M=[α2γβ10−101]M = \left[ {\begin{matrix} \alpha & 2 & \gamma \\ \beta & 1 & 0 \\ { - 1} & 0 & 1 \\ \end{matrix} } \right]M=​αβ−1​210​γ01​​ Let P be the plane containing all those (α\alphaα, β\betaβ, γ\gammaγ) for which the above system of linear equations is consistent, and D be the square of the distance of the point (0, 1, 0) from the plane P. The value of | M | is ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 1

  1. Write the given system in matrix form

The system is

{x+2y+3z=α4x+5y+6z=β7x+8y+9z=γ−1\begin{cases} x+2y+3z=\alpha \\ 4x+5y+6z=\beta \\ 7x+8y+9z=\gamma-1 \end{cases}⎩⎨⎧​x+2y+3z=α4x+5y+6z=β7x+8y+9z=γ−1​

Its coefficient matrix is

A=(123456789)A=\begin{pmatrix} 1&2&3\\ 4&5&6\\ 7&8&9 \end{pmatrix}A=​147​258​369​​

and the right-hand side vector is

(αβγ−1).\begin{pmatrix}\alpha\\ \beta\\ \gamma-1\end{pmatrix}.​αβγ−1​​.

For consistency, the RHS must satisfy the same linear relation as the rows of AAA.

  1. Find the dependence relation among the rows

Since

R1−2R2+R3=0,R_1-2R_2+R_3=0,R1​−2R2​+R3​=0,

because

(1,2,3)−2(4,5,6)+(7,8,9)=(0,0,0),(1,2,3)-2(4,5,6)+(7,8,9)=(0,0,0),(1,2,3)−2(4,5,6)+(7,8,9)=(0,0,0),

the system is consistent iff the constants also satisfy

α−2β+(γ−1)=0.\alpha-2\beta+(\gamma-1)=0.α−2β+(γ−1)=0.

So,

α−2β+γ=1.\alpha-2\beta+\gamma=1.α−2β+γ=1.

Hence the plane PPP containing all such (α,β,γ)(\alpha,\beta,\gamma)(α,β,γ) is

P: α−2β+γ=1.P:\ \alpha-2\beta+\gamma=1.P: α−2β+γ=1.
  1. Find the square of the distance from (0,1,0)(0,1,0)(0,1,0) to the plane

The plane is

α−2β+γ−1=0.\alpha-2\beta+\gamma-1=0.α−2β+γ−1=0.

Treating (α,β,γ)(\alpha,\beta,\gamma)(α,β,γ) as coordinates (x,y,z)(x,y,z)(x,y,z), distance from point (0,1,0)(0,1,0)(0,1,0) is

d=∣1⋅0+(−2)⋅1+1⋅0−1∣12+(−2)2+12=∣−3∣6=36.d=\frac{|1\cdot 0+(-2)\cdot 1+1\cdot 0-1|}{\sqrt{1^2+(-2)^2+1^2}} =\frac{|-3|}{\sqrt{6}}=\frac{3}{\sqrt6}.d=12+(−2)2+12​∣1⋅0+(−2)⋅1+1⋅0−1∣​=6​∣−3∣​=6​3​.

Thus

D=d2=96=32.D=d^2=\frac{9}{6}=\frac32.D=d2=69​=23​.
  1. Compute ∣M∣|M|∣M∣

Given

M=(α2γβ10−101).M=\begin{pmatrix} \alpha&2&\gamma\\ \beta&1&0\\ -1&0&1 \end{pmatrix}.M=​αβ−1​210​γ01​​.

Its determinant is

∣M∣=α∣1001∣−2∣β0−11∣+γ∣β1−10∣.|M|=\alpha\begin{vmatrix}1&0\\0&1\end{vmatrix} -2\begin{vmatrix}\beta&0\\-1&1\end{vmatrix} +\gamma\begin{vmatrix}\beta&1\\-1&0\end{vmatrix}.∣M∣=α​10​01​​−2​β−1​01​​+γ​β−1​10​​.

So,

∣M∣=α(1)−2(β)+γ(1)=α−2β+γ.|M|=\alpha(1)-2(\beta)+\gamma(1)=\alpha-2\beta+\gamma.∣M∣=α(1)−2(β)+γ(1)=α−2β+γ.

But from consistency,

α−2β+γ=1.\alpha-2\beta+\gamma=1.α−2β+γ=1.

Therefore,

∣M∣=1.|M|=1.∣M∣=1.
  1. Final answer
1\boxed{1}1​

The computed value matches the stored correct answer.

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