Let a, b, c, d in arithmetic progression with common difference \lambda . If | \beginmatrix x + a - c & x + b & x + a x - 1 & x + c & x + b x - b + d & x + d & x + c \endmatrix | = 2 , then value of \lambda 2 is equal to \underline\hspace2cm .

Matrices and Determinants PYQ — JEE Main 2021 · 20 Jul · Shift 1 · Q42 with Solution

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Correct answer: 1

Since $a,b,c,d$ are in arithmetic progression with common difference $\lambda$ , we can write $b=a+\lambda,\quad c=a+2\lambda,\quad d=a+3\lambda.$
Substitute these into the determinant entries:

So the determinant becomes

\Delta= \begin{vmatrix} x-2\lambda & x+a+\lambda & x+a \\ x-1 & x+a+2\lambda & x+a+\lambda \\ x+2\lambda & x+a+3\lambda & x+a+2\lambda \end{vmatrix}.

Given $\Delta=2$ .

Now perform column operations to simplify: Let $C_2 \to C_2-C_3, \qquad C_3 \to C_3.$ Then

\begin{pmatrix} \lambda\\ \lambda\\ \lambda \end{pmatrix} =\lambda \begin{pmatrix} 1\\1\\1 \end{pmatrix}.

Thus

\Delta= \begin{vmatrix} x-2\lambda & \lambda & x+a \\ x-1 & \lambda & x+a+\lambda \\ x+2\lambda & \lambda & x+a+2\lambda \end{vmatrix}.

Factor $\lambda$ out of the second column:

\Delta=\lambda \begin{vmatrix} x-2\lambda & 1 & x+a \\ x-1 & 1 & x+a+\lambda \\ x+2\lambda & 1 & x+a+2\lambda \end{vmatrix}.

Now use row operations: $R_2\to R_2-R_1, \qquad R_3\to R_3-R_2\text{(original)}.$ Using the current matrix inside,

\Delta=\lambda \begin{vmatrix} x-2\lambda & 1 & x+a \\ 2\lambda-1 & 0 & \lambda \\ 2\lambda+1 & 0 & \lambda \end{vmatrix}.

\Delta=\lambda\cdot (-1)^{1+2} \begin{vmatrix} 2\lambda-1 & \lambda\\ 2\lambda+1 & \lambda \end{vmatrix}.

Thus

\Delta=-\lambda\left[(2\lambda-1)\lambda-(2\lambda+1)\lambda\right].

Simplify the bracket:

(2\lambda-1)\lambda-(2\lambda+1)\lambda =\lambda[(2\lambda-1)-(2\lambda+1)] =\lambda(-2)=-2\lambda.

Hence

\Delta=-\lambda(-2\lambda)=2\lambda^2.

Given $\Delta=2$ , we get $2\lambda^2=2 \implies \lambda^2=1.$ Therefore, $\boxed{\lambda^2=1}.$

So the required integer is $1$ .

Comparison with stored answer: Stored correct answer = $1$ . Our derived answer = $1$ . Hence they agree.

View written solutionFree

Correct answer: 1

Since $a,b,c,d$ are in arithmetic progression with common difference $\lambda$ , we can write $b=a+\lambda,\quad c=a+2\lambda,\quad d=a+3\lambda.$
Substitute these into the determinant entries:

So the determinant becomes

\Delta= \begin{vmatrix} x-2\lambda & x+a+\lambda & x+a \\ x-1 & x+a+2\lambda & x+a+\lambda \\ x+2\lambda & x+a+3\lambda & x+a+2\lambda \end{vmatrix}.

Given $\Delta=2$ .

Now perform column operations to simplify: Let $C_2 \to C_2-C_3, \qquad C_3 \to C_3.$ Then

\begin{pmatrix} \lambda\\ \lambda\\ \lambda \end{pmatrix} =\lambda \begin{pmatrix} 1\\1\\1 \end{pmatrix}.

Thus

\Delta= \begin{vmatrix} x-2\lambda & \lambda & x+a \\ x-1 & \lambda & x+a+\lambda \\ x+2\lambda & \lambda & x+a+2\lambda \end{vmatrix}.

Factor $\lambda$ out of the second column:

\Delta=\lambda \begin{vmatrix} x-2\lambda & 1 & x+a \\ x-1 & 1 & x+a+\lambda \\ x+2\lambda & 1 & x+a+2\lambda \end{vmatrix}.

Now use row operations: $R_2\to R_2-R_1, \qquad R_3\to R_3-R_2\text{(original)}.$ Using the current matrix inside,

\Delta=\lambda \begin{vmatrix} x-2\lambda & 1 & x+a \\ 2\lambda-1 & 0 & \lambda \\ 2\lambda+1 & 0 & \lambda \end{vmatrix}.

\Delta=\lambda\cdot (-1)^{1+2} \begin{vmatrix} 2\lambda-1 & \lambda\\ 2\lambda+1 & \lambda \end{vmatrix}.

Thus

\Delta=-\lambda\left[(2\lambda-1)\lambda-(2\lambda+1)\lambda\right].

Simplify the bracket:

(2\lambda-1)\lambda-(2\lambda+1)\lambda =\lambda[(2\lambda-1)-(2\lambda+1)] =\lambda(-2)=-2\lambda.

Hence

\Delta=-\lambda(-2\lambda)=2\lambda^2.

Given $\Delta=2$ , we get $2\lambda^2=2 \implies \lambda^2=1.$ Therefore, $\boxed{\lambda^2=1}.$

So the required integer is $1$ .

Comparison with stored answer: Stored correct answer = $1$ . Our derived answer = $1$ . Hence they agree.