For \alpha, \beta \in \mathbbR , suppose the system of linear equations \beginaligned & x-y+z=5 & 2 x+2 y+\alpha z=8 & 3 x-y+4 z=\beta \endaligned has infinitely many solutions. Then \alpha and \beta are the roots of :

Matrices and Determinants PYQ — JEE Main 2023 · 30 Jan · Shift 2 · Q24 with Solution

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Correct answer: D

For a system of $3$ linear equations in $3$ variables to have infinitely many solutions, the equations must be dependent.

So the third equation must be a linear combination of the first two, and the augmented matrix must satisfy $\operatorname{rank}(A)=\operatorname{rank}(A|B)<3.$
Write the equations:
$x-y+z&=5 \[2pt] 2x+2y+\alpha z&=8 \[2pt] 3x-y+4z&=\beta \end{aligned}$$ Coefficient matrix: $$A=\begin{pmatrix} 1 & -1 & 1\\ 2 & 2 & \alpha\\ 3 & -1 & 4 \end{pmatrix}$$$
First, impose the condition that the determinant is zero: $\det(A)=0.$

Compute:
$1\begin{vmatrix}2 & \alpha\\ -1 & 4\end{vmatrix} -(-1)\begin{vmatrix}2 & \alpha\\ 3 & 4\end{vmatrix} +1\begin{vmatrix}2 & 2\\ 3 & -1\end{vmatrix}$$ $$=1(8+\alpha)+1(8-3\alpha)+1(-2-6)$$ $$=8+\alpha+8-3\alpha-8=8-2\alpha$$ Hence, $$8-2\alpha=0 \implies \alpha=4.$$$
Now substitute $\alpha=4$ into the equations:
$x-y+z&=5 \quad ...(1)\\ 2x+2y+4z&=8 \quad ...(2)\\ 3x-y+4z&=\beta \quad ...(3) \end{aligned}$$$
Check whether equation (3) is a linear combination of (1) and (2).

Let $a(x-y+z)+b(2x+2y+4z)=3x-y+4z.$

Comparing coefficients: $a+2b=3,$ $-a+2b=-1,$ $a+4b=4.$

From the first two equations:
$\quad -a+2b=-1.$$ Adding, $$4b=2 \implies b=\frac12.$$ Then $$a+1=3 \implies a=2.$$ Check the $z$-coefficient: $$a+4b=2+2=4,$$ which is correct. So indeed, $$\text{(3)} = 2\times\text{(1)}+\frac12\times\text{(2)}.$$$
Therefore the constant term must also satisfy the same relation: $\beta=2\cdot 5+\frac12\cdot 8=10+4=14.$

Hence, $\alpha=4,\quad \beta=14.$
The required quadratic having roots $4$ and $14$ is $x^2-(4+14)x+(4)(14)=0$ $x^2-18x+56=0.$
Therefore the correct option is: $\boxed{\text{D}}$

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View written solutionFree

Correct answer: D

For a system of $3$ linear equations in $3$ variables to have infinitely many solutions, the equations must be dependent.

So the third equation must be a linear combination of the first two, and the augmented matrix must satisfy $\operatorname{rank}(A)=\operatorname{rank}(A|B)<3.$
Write the equations:
$x-y+z&=5 \[2pt] 2x+2y+\alpha z&=8 \[2pt] 3x-y+4z&=\beta \end{aligned}$$ Coefficient matrix: $$A=\begin{pmatrix} 1 & -1 & 1\\ 2 & 2 & \alpha\\ 3 & -1 & 4 \end{pmatrix}$$$
First, impose the condition that the determinant is zero: $\det(A)=0.$

Compute:
$1\begin{vmatrix}2 & \alpha\\ -1 & 4\end{vmatrix} -(-1)\begin{vmatrix}2 & \alpha\\ 3 & 4\end{vmatrix} +1\begin{vmatrix}2 & 2\\ 3 & -1\end{vmatrix}$$ $$=1(8+\alpha)+1(8-3\alpha)+1(-2-6)$$ $$=8+\alpha+8-3\alpha-8=8-2\alpha$$ Hence, $$8-2\alpha=0 \implies \alpha=4.$$$
Now substitute $\alpha=4$ into the equations:
$x-y+z&=5 \quad ...(1)\\ 2x+2y+4z&=8 \quad ...(2)\\ 3x-y+4z&=\beta \quad ...(3) \end{aligned}$$$
Check whether equation (3) is a linear combination of (1) and (2).

Let $a(x-y+z)+b(2x+2y+4z)=3x-y+4z.$

Comparing coefficients: $a+2b=3,$ $-a+2b=-1,$ $a+4b=4.$

From the first two equations:
$\quad -a+2b=-1.$$ Adding, $$4b=2 \implies b=\frac12.$$ Then $$a+1=3 \implies a=2.$$ Check the $z$-coefficient: $$a+4b=2+2=4,$$ which is correct. So indeed, $$\text{(3)} = 2\times\text{(1)}+\frac12\times\text{(2)}.$$$
Therefore the constant term must also satisfy the same relation: $\beta=2\cdot 5+\frac12\cdot 8=10+4=14.$

Hence, $\alpha=4,\quad \beta=14.$
The required quadratic having roots $4$ and $14$ is $x^2-(4+14)x+(4)(14)=0$ $x^2-18x+56=0.$
Therefore the correct option is: $\boxed{\text{D}}$

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