Question

Find the value of x in the value , if $\left| \begin{matrix} x+1 & x-1 \\\ x-3 & x+2 \\\ \end{matrix} \right|=\left| \begin{matrix} 4 & -1 \\\ 1 & 3 \\\ \end{matrix} \right|$.

Answer 1

Hint: Reduce the determinant in its simplified form by elementary row\column transformation methods. Find the value of x from the reduced equation.

Complete step-by-step answer:

The given determinant equation of unknown x is $\left| \begin{matrix} x+1 & x-1 \\\ x-3 & x+2 \\\ \end{matrix} \right|=\left| \begin{matrix} 4 & -1 \\\ 1 & 3 \\\ \end{matrix} \right|$.

We know from the properties of elementary row/column transformation that if $\left| \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right|$ is a second order determinant then $\left| \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right|=\left| \begin{matrix} a & b \\\ c-a & d-b \\\ \end{matrix} \right|$.

This is an elementary row transformation.

The second row has been replaced by the difference between first and second row.
So, using the above property from $\left| \begin{matrix} x+1 & x-1 \\\ x-3 & x+2 \\\ \end{matrix} \right|=\left| \begin{matrix} 4 & -1 \\\ 1 & 3 \\\ \end{matrix} \right|$ we get,
$\Rightarrow$ $\left| \begin{matrix} x+1 & x-1 \\\ (x-3)-(x+1) & (x+2)-(x-1) \\\ \end{matrix} \right|=\left| \begin{matrix} 4 & -1 \\\ 1 & 3 \\\ \end{matrix} \right|$ [From the properties of determinants]
$\Rightarrow$ $\left| \begin{matrix} x+1 & x-1 \\\ -4 & 3 \\\ \end{matrix} \right|=\left| \begin{matrix} 4 & -1 \\\ 1 & 3 \\\ \end{matrix} \right|$ ……………….. (1)

We know that that if $\left| \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right|$ is a second order determinant then the expanded form of the determinant is $ad-bc$.

Hence,$\left| \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right|=ad-bc$.

So, from (1) we get, $3(x+1)-(-4)(x-1)=4\cdot 3-1\cdot (-1)$ [Applying the above rule]
$\Rightarrow$ $3x+3+4x-4=12+1$
$\Rightarrow$ $7x-1=13$
$\Rightarrow$ $7x=14$
$\Rightarrow$ $x=2$

Therefore, the only value of x that satisfies the given equation is 2. The value of x=2 can be easily verified by putting the value in place of x and getting equality in both sides.

Hence, the required value of x is 2.

Note: Elementary row transformation has been used in the above solution. We can also use elementary column transformations and both would give the same result. Even if we had not used elementary row/column transformations then also it would give the same result but we have to give some labor to get the result. Elementary transformations have been used here just to simplify the calculation.