If $A=\left( \begin{matrix}
x & 1 \\
1 & 0 \\
\end{matrix} \right)$ and ${{A}^{2}}$ is the identity matrix, then x is equal to ?
A . 1
B. 2
C. 3
D. 0
Answer
300.3k+ views
Hint: We are given a question which is based on matrices. We are given a matrix and we have to find the value of x. A square matrix is a matrix that has the same number of rows and columns. For example – The given matrix is $2\times 2$ matrix. To Find the value of x, we multiply A with A, and given that ${{A}^{2}}$ is an identity matrix and equating both equations and solving it, we get the value of x.
Complete step by step Solution:
We are given the matrix $A=\left( \begin{matrix}
x & 1 \\
1 & 0 \\
\end{matrix} \right)$
And ${{A}^{2}}$ is the identity matrix.
That means $A\times A=I$
Then $\left( \begin{matrix}
x & 1 \\
1 & 0 \\
\end{matrix} \right)\times $$\left( \begin{matrix}
x & 1 \\
1 & 0 \\
\end{matrix} \right)$ = $\left( \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right)$ ………………………………….. (1)
Now we equate the above equation and simplify it, we get
First, we multiply and open the brackets of L.H.S, we get
$\left( \begin{matrix}
x & 1 \\
1 & 0 \\
\end{matrix} \right)\times $$\left( \begin{matrix}
x & 1 \\
1 & 0 \\
\end{matrix} \right)$ = $\left( \begin{matrix}
x\times x+1\times 1 & x\times 1+1\times 0 \\
1\times x+0\times 1 & 1\times 1+0\times 0 \\
\end{matrix} \right)$
Simplifying it, we get
$\left( \begin{matrix}
x & 1 \\
1 & 0 \\
\end{matrix} \right)\times $$\left( \begin{matrix}
x & 1 \\
1 & 0 \\
\end{matrix} \right)$ = $\left( \begin{matrix}
{{x}^{2}}+1 & x \\
x & 1 \\
\end{matrix} \right)$
From equation (1), we get
$\left( \begin{matrix}
{{x}^{2}}+1 & x \\
x & 1 \\
\end{matrix} \right)$ = $\left( \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right)$
That is ${{x}^{2}}+1=1$
That is ${{x}^{2}}=0$ or $x=0$
Therefore, the correct option is (D).
Note: We know that the given question is in matrix form. A matrix is a set of numbers that are arranged in rows and columns to make a rectangular array. In the matrix, the numbers are called the entries or entities of the matrix.
In Multiplication matrices, the number of column of the first matrix match the number of rows of the second matrix. When we want to multiply the matrices, then the parts of the rows in the first matrix are multiplied by the columns in the second matrix.
Complete step by step Solution:
We are given the matrix $A=\left( \begin{matrix}
x & 1 \\
1 & 0 \\
\end{matrix} \right)$
And ${{A}^{2}}$ is the identity matrix.
That means $A\times A=I$
Then $\left( \begin{matrix}
x & 1 \\
1 & 0 \\
\end{matrix} \right)\times $$\left( \begin{matrix}
x & 1 \\
1 & 0 \\
\end{matrix} \right)$ = $\left( \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right)$ ………………………………….. (1)
Now we equate the above equation and simplify it, we get
First, we multiply and open the brackets of L.H.S, we get
$\left( \begin{matrix}
x & 1 \\
1 & 0 \\
\end{matrix} \right)\times $$\left( \begin{matrix}
x & 1 \\
1 & 0 \\
\end{matrix} \right)$ = $\left( \begin{matrix}
x\times x+1\times 1 & x\times 1+1\times 0 \\
1\times x+0\times 1 & 1\times 1+0\times 0 \\
\end{matrix} \right)$
Simplifying it, we get
$\left( \begin{matrix}
x & 1 \\
1 & 0 \\
\end{matrix} \right)\times $$\left( \begin{matrix}
x & 1 \\
1 & 0 \\
\end{matrix} \right)$ = $\left( \begin{matrix}
{{x}^{2}}+1 & x \\
x & 1 \\
\end{matrix} \right)$
From equation (1), we get
$\left( \begin{matrix}
{{x}^{2}}+1 & x \\
x & 1 \\
\end{matrix} \right)$ = $\left( \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right)$
That is ${{x}^{2}}+1=1$
That is ${{x}^{2}}=0$ or $x=0$
Therefore, the correct option is (D).
Note: We know that the given question is in matrix form. A matrix is a set of numbers that are arranged in rows and columns to make a rectangular array. In the matrix, the numbers are called the entries or entities of the matrix.
In Multiplication matrices, the number of column of the first matrix match the number of rows of the second matrix. When we want to multiply the matrices, then the parts of the rows in the first matrix are multiplied by the columns in the second matrix.
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