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System of the linear equation: ISI MMA 2018 Question 11

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Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2" _i="1" _address="0.0.0.1"]The value of  \(\lambda\)  for which the system of linear equations 2x - y - z = 12 , x -2y +z = -4 , x +y + \(\lambda\)z = 4 has no solution is [/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.27" _i="1" _address="0.1"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.1.0"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.27" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" _i="0" _address="0.1.0.0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.27" _i="0" _address="0.1.0.0.0"]Sample Questions (MMA) :2019[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.27" _i="1" _address="0.1.0.0.1" open="off"]Linear Algebra[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.27" _i="2" _address="0.1.0.0.2" open="off"]Medium[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.27" _i="3" _address="0.1.0.0.3" open="off"]Schaums Outline of Linear Algebra [/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" _i="1" _address="0.1.0.1"]

Start with hints

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.27" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff" _i="2" _address="0.1.0.2"][et_pb_tab title="Hint 0" _builder_version="3.22.4" _i="0" _address="0.1.0.2.0"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.27" _i="1" _address="0.1.0.2.1"]Write down the case in the matrix form & try to use the knowledge of system of linear cases 2x  -y -z =12, x -2y+ z =-4,x + y + \(\lambda\) z  =4[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.27" _i="2" _address="0.1.0.2.2"]The corresponding matrix format that you will get is \begin{pmatrix}  2&-1&-1\\  1&-2&1\\  1&1&\lambda\\  \end{pmatrix} \begin{pmatrix} x \\ y\\ z \end{pmatrix}=\begin{pmatrix} 12\\-4\\4\\ \end{pmatrix}   The  system has no solution if  \begin{pmatrix} 12\\-4\\4\\ \end{pmatrix}  does not belong to the Sp { \begin{pmatrix} 2 \\ 1\\ 1 \end{pmatrix},\begin{pmatrix} -1 \\ -2\\1 \end{pmatrix},\begin{pmatrix} -1 \\ 1\\\lambda \end{pmatrix}} -----------(1) So, what is the possible value of \(\lambda\)? Can you think the condition (1) in some other ways say in terms of determinants?

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.27" _i="3" _address="0.1.0.2.3"]

If the determinant of matrix  A=\begin{pmatrix}  2&-1&-1\\  1&-2&1\\  1&1&\lambda\\  \end{pmatrix}

is zero then it has no solution

[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.27" _i="4" _address="0.1.0.2.4"]det A = 0 => \(\lambda\) = -2[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" _i="3" _address="0.1.0.3"]

Watch the video

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Similar Problems

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