The values of \lambda and \mu for which the system of linear equations x + y + z = 2 x + 2y + 3z = 5 x + 3y + \lambda z = \mu has infinitely many solutions are, respectively:

Matrices and Determinants PYQ — JEE Main 2020 · 6 Sep · Shift 1 · Q36 with Solution

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Correct answer: B

Write the system in augmented matrix form

The system is:

\begin{cases} x+y+z=2 \\ x+2y+3z=5 \\ x+3y+\lambda z=\mu \end{cases}

Its augmented matrix is

\left[ \begin{array}{ccc|c} 1&1&1&2\\ 1&2&3&5\\ 1&3&\lambda&\mu \end{array} \right].

For infinitely many solutions, the system must be consistent and the rank of coefficient matrix must be less than 3:

\operatorname{rank}(A)=\operatorname{rank}([A|B])<3.

Perform row operations on the coefficient matrix

Start with

A=\begin{pmatrix} 1&1&1\\ 1&2&3\\ 1&3&\lambda \end{pmatrix}.

Apply:

R_2 \to R_2-R_1, \qquad R_3 \to R_3-R_1

Then

\begin{pmatrix} 1&1&1\\ 0&1&2\\ 0&2&\lambda-1 \end{pmatrix}.

Now apply:

R_3 \to R_3-2R_2

So we get

\begin{pmatrix} 1&1&1\\ 0&1&2\\ 0&0&\lambda-5 \end{pmatrix}.

For rank to be less than 3, we need

\lambda-5=0 \implies \lambda=5.

Now check consistency using the augmented matrix

Take the augmented matrix and perform the same operations:

\left[ \begin{array}{ccc|c} 1&1&1&2\\ 1&2&3&5\\ 1&3&\lambda&\mu \end{array} \right].

After

R_2 \to R_2-R_1, \qquad R_3 \to R_3-R_1,

we get

\left[ \begin{array}{ccc|c} 1&1&1&2\\ 0&1&2&3\\ 0&2&\lambda-1&\mu-2 \end{array} \right].

Now apply

R_3 \to R_3-2R_2:

\left[ \begin{array}{ccc|c} 1&1&1&2\\ 0&1&2&3\\ 0&0&\lambda-5&\mu-8 \end{array} \right].

For infinitely many solutions, the last row must become all zeros:

\lambda-5=0 \quad \text{and} \quad \mu-8=0.

Hence,

\lambda=5, \qquad \mu=8.

Match with the options

This corresponds to:

Option B: $5$ and $8$

Comparison with stored correct answer

Stored correct answer is B, which matches our result.

More from Matrices and Determinants

View written solutionFree

Correct answer: B

Write the system in augmented matrix form

The system is:

\begin{cases} x+y+z=2 \\ x+2y+3z=5 \\ x+3y+\lambda z=\mu \end{cases}

Its augmented matrix is

\left[ \begin{array}{ccc|c} 1&1&1&2\\ 1&2&3&5\\ 1&3&\lambda&\mu \end{array} \right].

For infinitely many solutions, the system must be consistent and the rank of coefficient matrix must be less than 3:

\operatorname{rank}(A)=\operatorname{rank}([A|B])<3.

Perform row operations on the coefficient matrix

Start with

A=\begin{pmatrix} 1&1&1\\ 1&2&3\\ 1&3&\lambda \end{pmatrix}.

Apply:

R_2 \to R_2-R_1, \qquad R_3 \to R_3-R_1

Then

\begin{pmatrix} 1&1&1\\ 0&1&2\\ 0&2&\lambda-1 \end{pmatrix}.

Now apply:

R_3 \to R_3-2R_2

So we get

\begin{pmatrix} 1&1&1\\ 0&1&2\\ 0&0&\lambda-5 \end{pmatrix}.

For rank to be less than 3, we need

\lambda-5=0 \implies \lambda=5.

Now check consistency using the augmented matrix

Take the augmented matrix and perform the same operations:

\left[ \begin{array}{ccc|c} 1&1&1&2\\ 1&2&3&5\\ 1&3&\lambda&\mu \end{array} \right].

After

R_2 \to R_2-R_1, \qquad R_3 \to R_3-R_1,

we get

\left[ \begin{array}{ccc|c} 1&1&1&2\\ 0&1&2&3\\ 0&2&\lambda-1&\mu-2 \end{array} \right].

Now apply

R_3 \to R_3-2R_2:

\left[ \begin{array}{ccc|c} 1&1&1&2\\ 0&1&2&3\\ 0&0&\lambda-5&\mu-8 \end{array} \right].

For infinitely many solutions, the last row must become all zeros:

\lambda-5=0 \quad \text{and} \quad \mu-8=0.

Hence,

\lambda=5, \qquad \mu=8.

Match with the options

This corresponds to:

Option B: $5$ and $8$

Comparison with stored correct answer

Stored correct answer is B, which matches our result.

Matrices and Determinants question

Matrices and Determinants question

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Matrices and Determinants question

Matrices and Determinants question

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