# Find the value of x if \[\left| {\begin{array}{*{20}{c}}

a&a&x \\

m&m&m \\

b&x&b

\end{array}} \right| = 0\]

A) a

B) b

C) a-b

D) m

This question has multiple correct answers.

Last updated date: 29th Mar 2023

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Answer

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Hint: In these types of questions you first have to see what elements of the matrix are the same. Then see what elements you can take common. Then apply some operations on the determinant like subtracting two rows, adding two columns etc. Simplify the determinant and then expand.

Complete step-by-step answer:

Det. \[\left| {\begin{array}{*{20}{c}}

a&a&x \\

m&m&m \\

b&x&b

\end{array}} \right| = 0\]

Here , we have one complete row of m.

∴ taking m common, we get

\[m\left| {\begin{array}{*{20}{c}}

a&a&x \\

1&1&1 \\

b&x&b

\end{array}} \right| = 0\]

Now\[{c_2} \to {c_2} - {c_1}\] and \[{c_3} \to {c_3} - {c_1}\]

\[m\left| {\begin{array}{*{20}{c}}

a&0&{x - a} \\

1&0&0 \\

b&{x - b}&0

\end{array}} \right| = 0\]

Now expand along row 2, we get

\[m\left[ { - 1\left| {\begin{array}{*{20}{c}}

0&{x - a} \\

{x - b}&0

\end{array}} \right| + 0 + 0} \right] = 0\]

\[m\left[ { - \left( { - \left( {x - a} \right)\left( {x - b} \right)} \right)} \right] = 0\]

\[m\left( {x - a} \right)\left( {x - b} \right) = 0\]

This means m=0, (x-a) = 0 and (x-b) = 0

∴ x can have two values x = a and x = b.

∴ we have two correct options ‘A’ and ‘B’.

Note: For simplifying the determinant, we have to make as many elements of the determinant zero as possible. While applying the operations we can multiply some scalar quantities with the rows and columns to make the elements zero. Then expand through the row or the columns whose maximum no. of elements are zero for easy calculations.

Complete step-by-step answer:

Det. \[\left| {\begin{array}{*{20}{c}}

a&a&x \\

m&m&m \\

b&x&b

\end{array}} \right| = 0\]

Here , we have one complete row of m.

∴ taking m common, we get

\[m\left| {\begin{array}{*{20}{c}}

a&a&x \\

1&1&1 \\

b&x&b

\end{array}} \right| = 0\]

Now\[{c_2} \to {c_2} - {c_1}\] and \[{c_3} \to {c_3} - {c_1}\]

\[m\left| {\begin{array}{*{20}{c}}

a&0&{x - a} \\

1&0&0 \\

b&{x - b}&0

\end{array}} \right| = 0\]

Now expand along row 2, we get

\[m\left[ { - 1\left| {\begin{array}{*{20}{c}}

0&{x - a} \\

{x - b}&0

\end{array}} \right| + 0 + 0} \right] = 0\]

\[m\left[ { - \left( { - \left( {x - a} \right)\left( {x - b} \right)} \right)} \right] = 0\]

\[m\left( {x - a} \right)\left( {x - b} \right) = 0\]

This means m=0, (x-a) = 0 and (x-b) = 0

∴ x can have two values x = a and x = b.

∴ we have two correct options ‘A’ and ‘B’.

Note: For simplifying the determinant, we have to make as many elements of the determinant zero as possible. While applying the operations we can multiply some scalar quantities with the rows and columns to make the elements zero. Then expand through the row or the columns whose maximum no. of elements are zero for easy calculations.

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