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Matrices and Determinants question

2019 · 9 Jan · Shift 2 · Q39
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Matrices and Determinants question

2019 · 9 Jan · Shift 2 · Q39

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
If the system of linear equations x −-− 4y + 7z = g 3y −-− 5z = h −-− 2x + 5y −-− 9z = k is consistent, then :
  1. A
    g + 2h + k = 0
  2. B
    g + h + 2k = 0
  3. C
    2g + h + k = 0
  4. D
    g + h + k = 0
View written solutionFree

Correct answer: C

  1. Write the system in standard form

Given: x−4y+7z=gx-4y+7z=gx−4y+7z=g 3y−5z=h3y-5z=h3y−5z=h −2x+5y−9z=k-2x+5y-9z=k−2x+5y−9z=k

Rewrite the second equation including the xxx-coefficient: 0x+3y−5z=h0x+3y-5z=h0x+3y−5z=h

So the coefficient matrix is

1 & -4 & 7\\ 0 & 3 & -5\\ -2 & 5 & -9 \end{pmatrix}$$ and the system is $$A\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}g\\h\\k\end{pmatrix}.$$ 2. **Check when the system is consistent** A linear system is always consistent if $\det(A)\neq 0$. If $\det(A)=0$, then consistency requires a relation among the constants $g,h,k$. Let us compute $\det(A)$: $$\det(A)=\begin{vmatrix} 1 & -4 & 7\\ 0 & 3 & -5\\ -2 & 5 & -9 \end{vmatrix}$$ Expanding along the first row,

\det(A)=1\begin{vmatrix}3 & -5\5 & -9\end{vmatrix}-(-4)\begin{vmatrix}0 & -5\-2 & -9\end{vmatrix}+7\begin{vmatrix}0 & 3\-2 & 5\end{vmatrix}

Now, $$\begin{vmatrix}3 & -5\\5 & -9\end{vmatrix}=3(-9)-(-5)(5)=-27+25=-2$$ $$\begin{vmatrix}0 & -5\\-2 & -9\end{vmatrix}=0(-9)-(-5)(-2)= -10$$ $$\begin{vmatrix}0 & 3\\-2 & 5\end{vmatrix}=0(5)-3(-2)=6$$ Therefore,

\det(A)=1(-2)-(-4)(-10)+7(6)=-2-40+42=0

So the matrix is singular, and consistency requires a condition on $g,h,k$. 3. **Find the linear dependence among the equations** We look for constants $\lambda_1,\lambda_2,\lambda_3$, not all zero, such that $$\lambda_1 R_1+\lambda_2 R_2+\lambda_3 R_3=0$$ for the left-hand sides. Let us test whether the third row is a combination of the first two: $$a(1,-4,7)+b(0,3,-5)=(-2,5,-9).$$ Comparing coefficients: - From $x$: $$a=-2$$ - From $y$: $$-4a+3b=5$$ Substituting $a=-2$: $$8+3b=5 \Rightarrow 3b=-3 \Rightarrow b=-1$$ - Check $z$: $$7a-5b=7(-2)-5(-1)=-14+5=-9$$ which is correct. Hence, $$R_3=-2R_1-R_2$$ or equivalently, $$2R_1+R_2+R_3=0.$$ 4. **Apply the same relation to the constants for consistency** For the system to be consistent, the right-hand sides must satisfy the same relation: $$2g+h+k=0.$$ 5. **Check options** - A: $g+2h+k=0$ — incorrect - B: $g+h+2k=0$ — incorrect - C: $2g+h+k=0$ — correct - D: $g+h+k=0$ — incorrect Therefore, the correct option is **C**.
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