
If $A=\left[ \begin{matrix}
{}^{1}/{}_{3} & 2 \\
0 & 2x-3\text{ }\!\!~\!\!\text{ } \\
\end{matrix} \right],B=\left[ \begin{matrix}
3 & 6 \\
0 & -1 \\
\end{matrix} \right]$ and $AB = I$ , then what is the value of$x$ ?
A. $- 1$
B. $1$
C. $0$
D. $2$
Answer
164.1k+ views
Hint: In the above question, we are provided two matrices $A$ and $B$ such that $AB = I$ . Perform Matrix Multiplication and calculate the product $AB$ . Then, equate each of its elements to the corresponding elements of the identity matrix and calculate the value of $x$ .
Complete step by step Solution:
We are given two matrices $A$ and $B$ such that $A=\left[ \begin{matrix}
{}^{1}/{}_{3} & 2 \\
0 & 2x-3\text{ }\!\!~\!\!\text{ } \\
\end{matrix} \right],B=\left[ \begin{matrix}
3 & 6 \\
0 & -1 \\
\end{matrix} \right]$ and $AB = I$.
Substituting the values of $A$ and $B$ in $AB = I$,
$\left[ \begin{matrix}
\text{ }~\text{ }{}^{1}/{}_{3} & 2 \\
0 & 2x-3 \\
\end{matrix} \right]\times \left[ \begin{matrix}
\text{ }~\text{ }3 & 6 \\
0 & -1 \\
\end{matrix} \right]=I$
Performing Matrix Multiplication,
$\left[ \begin{matrix}
\frac{1}{3}\times 3+2\times 0 & \frac{1}{3}\times 6+2\times \left( -1 \right) \\
\text{ }\!\!~\!\!\text{ }0\times 3+\left( 2x-3 \right)\times 0 & 0\times 6+\left( 2x-3 \right)\times \left( -1 \right) \\
\end{matrix} \right]=I$Simplifying further,
$\left[ \begin{matrix}
1 & 0 \\
\text{ }\!\!~\!\!\text{ }0 & -2x+3 \\
\end{matrix} \right]=I$We know that the Identity Matrix of order $2$ is $I=\left[ \begin{matrix}
\text{ }\!\!~\!\!\text{ }1 & 0 \\
0 & 1 \\
\end{matrix} \right]$ .
Therefore, substituting this,
$\left[ \begin{matrix}
1 & 0 \\
\text{ }\!\!~\!\!\text{ }0 & -2x+3 \\
\end{matrix} \right]=\left[ \begin{matrix}
\text{ }\!\!~\!\!\text{ }1 & 0 \\
0 & 1 \\
\end{matrix} \right]$Comparing the elements of the matrix on the Left-Hand side with the one on the Right-Hand side,
$- 2x + 3 = 1$
This gives:
$x = 1$
Hence, the value of $x$ is $1$.
Therefore, the correct option is (B).
Note: Two matrices $A$ and $B$ are said to be equal to each other when they both are of the same order and when every element of matrix $A$ is equal to the corresponding elements of matrix $B$ .
Complete step by step Solution:
We are given two matrices $A$ and $B$ such that $A=\left[ \begin{matrix}
{}^{1}/{}_{3} & 2 \\
0 & 2x-3\text{ }\!\!~\!\!\text{ } \\
\end{matrix} \right],B=\left[ \begin{matrix}
3 & 6 \\
0 & -1 \\
\end{matrix} \right]$ and $AB = I$.
Substituting the values of $A$ and $B$ in $AB = I$,
$\left[ \begin{matrix}
\text{ }~\text{ }{}^{1}/{}_{3} & 2 \\
0 & 2x-3 \\
\end{matrix} \right]\times \left[ \begin{matrix}
\text{ }~\text{ }3 & 6 \\
0 & -1 \\
\end{matrix} \right]=I$
Performing Matrix Multiplication,
$\left[ \begin{matrix}
\frac{1}{3}\times 3+2\times 0 & \frac{1}{3}\times 6+2\times \left( -1 \right) \\
\text{ }\!\!~\!\!\text{ }0\times 3+\left( 2x-3 \right)\times 0 & 0\times 6+\left( 2x-3 \right)\times \left( -1 \right) \\
\end{matrix} \right]=I$Simplifying further,
$\left[ \begin{matrix}
1 & 0 \\
\text{ }\!\!~\!\!\text{ }0 & -2x+3 \\
\end{matrix} \right]=I$We know that the Identity Matrix of order $2$ is $I=\left[ \begin{matrix}
\text{ }\!\!~\!\!\text{ }1 & 0 \\
0 & 1 \\
\end{matrix} \right]$ .
Therefore, substituting this,
$\left[ \begin{matrix}
1 & 0 \\
\text{ }\!\!~\!\!\text{ }0 & -2x+3 \\
\end{matrix} \right]=\left[ \begin{matrix}
\text{ }\!\!~\!\!\text{ }1 & 0 \\
0 & 1 \\
\end{matrix} \right]$Comparing the elements of the matrix on the Left-Hand side with the one on the Right-Hand side,
$- 2x + 3 = 1$
This gives:
$x = 1$
Hence, the value of $x$ is $1$.
Therefore, the correct option is (B).
Note: Two matrices $A$ and $B$ are said to be equal to each other when they both are of the same order and when every element of matrix $A$ is equal to the corresponding elements of matrix $B$ .
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