Question

Write the value of x 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: It is a question of determinants. Here, we have to compute the determinants which are given in the question. The determinant of a matrix \[A=\left[ \begin{matrix}
   a & b \\
   c & d \\
\end{matrix} \right]\] is given by $\left| A \right|=\left| \begin{matrix}
   a & b \\
   c & d \\
\end{matrix} \right|=ad-bc$ . Then we will equate both the determinants and will find the value of x from it.

Complete step-by-step solution:
We have to solve for x where we have been given $\left| \begin{matrix}
   x+1 & x-1 \\
   x-3 & x+2 \\
\end{matrix} \right|=\left| \begin{matrix}
   4 & -1 \\
   1 & 3 \\
\end{matrix} \right|$ .
So, we will find both the determinants one by one.
So, first we will compute the first determinant $\left| \begin{matrix}
   x+1 & x-1 \\
   x-3 & x+2 \\
\end{matrix} \right|$ .
We know determinant is the scalar value that can be computed from the elements of the square matrix. The determinant of non-square matrix does not exist, only determinants of square matrices are defined mathematically. It is denoted by det (A) or $\left| A \right|$ .
The formula for the determinant of a matrix is stated as below:
If we have given matrix $A=\left[ \begin{matrix}
   a & b \\
   c & d \\
\end{matrix} \right]$ , then its determinant is given by $\left| A \right|=\left| \begin{matrix}
   a & b \\
   c & d \\
\end{matrix} \right|=ad-bc$.
We will apply this formula to both the determinants, we get
$\begin{align}
  & \left| \begin{matrix}
   x+1 & x-1 \\
   x-3 & x+2 \\
\end{matrix} \right|=(x+1).(x+2)-(x-1)(x-3) \\
 & \Rightarrow ({{x}^{2}}+x+2x+2)-({{x}^{2}}-3x-x+3) \\
 & \Rightarrow {{x}^{2}}+3x+2-{{x}^{2}}+4x-3 \\
 & \Rightarrow 7x-1......(1) \\
\end{align}$
Now we will find determinant of $\left| \begin{matrix}
   4 & -1 \\
   1 & 3 \\
\end{matrix} \right|$ . By applying above determinant formula, we get
$\begin{align}
  & \left| \begin{matrix}
   4 & -1 \\
   1 & 3 \\
\end{matrix} \right|=4(3)-1(-1) \\
 & \Rightarrow 12+1 \\
 & \Rightarrow 13...(2) \\
\end{align}$
We get the value of both the determinants here.
Now, we have been given that $\left| \begin{matrix}
   x+1 & x-1 \\
   x-3 & x+2 \\
\end{matrix} \right|=\left| \begin{matrix}
   4 & -1 \\
   1 & 3 \\
\end{matrix} \right|$.
We can equate the values we found in (1) and (2). Hence, from (1) and (2), we get
$\begin{align}
  & 7x-1=13 \\
 & \therefore 7x=14 \\
 & \therefore x=2 \\
\end{align}$
Hence we get the required solution here,
The answer of the given question is $x=2$.

Note: There is no alternate way to solve or simplify the calculations. So, students must continue with this approach and perform the calculations step by step and carefully. There is a chance that students might directly equate the elements and form equations like x+1=4, then they will get x=3. But, they must realize that this is not valid, since we have been given determinants and not matrices and we cannot equate terms.