Question

Find the value of \[x\] for which the determinant\[\left| {\begin{array}{*{20}{c}}
{2x - 3}&{x - 2}&{x - 1} \\
{x - 2}&{2x - 2}&x \\
{x - 1}&x&{2x - 1}
\end{array}} \right|\] vanishes if

Answer 1

Hint:The value of a determinant\[x\]is given by the formula\[
\det \left( x \right) = \left| {\begin{array}{*{20}{c}}
a&b&c \\
d&e&f \\
g&h&i
\end{array}} \right| \\
= a\left[ {\left( e \right)\left( i \right) - \left( h \right)\left( f \right)} \right] - \left( b \right)\left[ {\left( d
\right)\left( i \right) - \left( g \right)\left( f \right)} \right] + c\left[ {\left( d \right)\left( h \right) - \left( g
\right)\left( e \right)} \right] \\
\]

In this question we will find the value of \[x\]for which the value of\[\det \]becomes zero, simply equating the value of determinant to zero and then by solving.

Complete step by step solution:
Given the determinant\[\left| {\begin{array}{*{20}{c}}
{2x - 3}&{x - 2}&{x - 1} \\
{x - 2}&{2x - 2}&x \\
{x - 1}&x&{2x - 1}
\end{array}} \right|\]
Let us assume\[A = \left| {\begin{array}{*{20}{c}}
{2x - 3}&{x - 2}&{x - 1} \\
{x - 2}&{2x - 2}&x \\

{x - 1}&x&{2x - 1}
\end{array}} \right|\]

We can see the give determinant \[A\]is a \[3X3\]since the determinant has 3 rows and 3 column to it, now to find the value of \[x\] for which the determinant becomes zero we will find the \[\det \]of the
determinant and since the solution will be length so we will try to reduce the element of the
determinant by\[{R_1} \to {R_1} - \left( {{R_2} + {R_3}} \right)\], we get
\[
A = \left| {\begin{array}{*{20}{c}}
{\left( {2x - 3} \right) - \left( {x - 2 - x - 1} \right)}&{\left( {x - 2} \right) - \left( {2x - 2 + 2} \right)}&{\left(
{x - 1} \right) - \left( {x + 2x - 1} \right)} \\
{x - 2}&{2x - 2}&x \\
{x - 1}&x&{2x - 1}
\end{array}} \right| \\
= \left| {\begin{array}{*{20}{c}}
0&{ - 2x}&{ - 2x} \\
{x - 2}&{2x - 2}&x \\
{x - 1}&x&{x - 1}
\end{array}} \right| \\
\]

Now find the \[\det \left( A \right)\]of the \[3 \times 3\]since the determinant to find the value of\[x\],
we can write
\[
\det \left( A \right) = \left| {\begin{array}{*{20}{c}}
0&{ - 2x}&{ - 2x} \\
{x - 2}&{2x - 2}&x \\
{x - 1}&x&{x - 1}
\end{array}} \right| \\

= 0\left[ {\left( {2x - 2} \right)\left( {x - 1} \right) - {x^2}} \right] - \left( { - 2x} \right)\left[ {\left( {x - 2}
\right)\left( {x - 1} \right) - \left( x \right)\left( {x - 1} \right)} \right] - 2x\left[ {\left( {x - 2} \right)\left( x
\right) - \left( {x - 1} \right)\left( {2x - 2} \right)} \right] \\
= 0 + \left( {2x} \right)\left[ {{x^2} - 2x - x + 2 - {x^2} + x} \right] - \left( {2x} \right)\left[ {{x^2} - 2x -
2{x^2} + 2x + 2x - 2} \right] \\
= \left( {2x} \right)\left[ { - 2x + 2} \right] - \left( {2x} \right)\left[ { - {x^2} + 2x - 2} \right] \\
= - 4{x^2} + 4x + 2{x^3} - 4{x^2} + 4x \\
= 2{x^3} - 8{x^2} + 8x \\
\]

Now since we need to find the value of \[x\] for which the determinant A vanishes, hence we can write\[\det \left( A \right) = 0\], so we get the determinant as
\[
\det \left( A \right) = 0 \\
2{x^3} - 8{x^2} + 8x = 0 \\
2x\left( {{x^2} - 4x + 4} \right) = 0 \\
2x{\left( {x - 2} \right)^2} = 0 \\
\]
Hence we get the value of\[x = 0,2,2\]

If we substitute this value of \[x\]in the determinant A then the determinant becomes 0.
To check: When\[x = 0\], we get
\[A = \left| {\begin{array}{*{20}{c}}
{2x - 3}&{x - 2}&{x - 1} \\
{x - 2}&{2x - 2}&x \\
{x - 1}&x&{2x - 1}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
{\left( {2 \times 0} \right) - 3}&{0 - 2}&{0 - 1} \\

{0 - 2}&{\left( {2 \times 0} \right) - 2}&0 \\
{0 - 1}&0&{\left( {2 \times 0} \right) - 1}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
{ - 3}&{ - 2}&{ - 1} \\
{ - 2}&{ - 2}&0 \\
{ - 1}&0&{ - 1}
\end{array}} \right|\]
Hence we get
\[
\det \left( A \right) = \left| {\begin{array}{*{20}{c}}
{ - 3}&{ - 2}&{ - 1} \\
{ - 2}&{ - 2}&0 \\
{ - 1}&0&{ - 1}
\end{array}} \right| \\
= \left( { - 3} \right)\left\{ {\left( { - 2} \right)\left( { - 1} \right) - 0} \right\} - \left( { - 2} \right)\left\{
{\left( { - 2} \right)\left( { - 1} \right) - 0} \right\} + \left( { - 1} \right)\left\{ { - \left( { - 2} \right)\left( { - 1}
\right)} \right\} \\
= \left( { - 3} \right)\left\{ {2 - 0} \right\} - \left( { - 2} \right)\left\{ {2 - 0} \right\} + \left( { - 1}
\right)\left\{ { - 2} \right\} \\
= - 6 + 2\left( 2 \right) + 2 \\
= - 6 + 4 + 2 \\
= 0 \\
\]

Similarly when\[x = 2\], we get
\[A = \left| {\begin{array}{*{20}{c}}
{2x - 3}&{x - 2}&{x - 1} \\
{x - 2}&{2x - 2}&x \\

{x - 1}&x&{2x - 1}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
{\left( {2 \times 2} \right) - 3}&{2 - 2}&{2 - 1} \\
{2 - 2}&{\left( {2 \times 2} \right) - 2}&2 \\
{2 - 1}&2&{\left( {2 \times 2} \right) - 1}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
1&0&1 \\
0&2&2 \\
1&2&3
\end{array}} \right|\]
Hence we get
\[
\det \left( A \right) = \left| {\begin{array}{*{20}{c}}
1&0&1 \\
0&2&2 \\
1&2&3
\end{array}} \right| \\
= \left( 1 \right)\left\{ {\left( 2 \right)\left( 3 \right) - \left( 2 \right)\left( 2 \right)} \right\} - 0\left\{ {0 -
\left( 1 \right)\left( 2 \right)} \right\} + \left( 1 \right)\left\{ {0 - \left( 2 \right)\left( 1 \right)} \right\} \\
= \left( 1 \right)\left\{ {6 - 4} \right\} - 0 + \left( 1 \right)\left\{ { - 2} \right\} \\
= 2 - 2 \\
= 0 \\
\]

Hence we can say if we substitute this value of \[x = 0,2,2\]in the determinant A then the value of the determinant becomes 0.

Note:In determinant if one of the rows is entirely zero then the determinant will also be zero.Also if one of the columns is entirely zero then the determinant will also be zero.

So to equate a determinant to zero always try to simplify any of the row or the column of the
determinant to make it zero and then determine the value of determinant.