
Let $A = \begin{bmatrix}x&1\\1&0\end{bmatrix},x \in \mathbb{R}$ and ${A^4} = \left[ {{a_{ij}}} \right]$. If ${a_{11}} = 109$, then ${a_{22}}$ is equal to:
Answer
217.5k+ views
Hint: In this question, we have to find the value of ${a_{22}}$ by using the given matrix $A = \begin{bmatrix}x&1\\1&0\end{bmatrix},x \in \mathbb{R}$. Firstly, we will find the value of ${A^4}$. We can rewrite the ${A^4}$ as ${A^2}{A^2}$ and find the value of ${A^2}$ then the result obtained will be multiplied twice to arrive the value of ${A^4}$. Then compare the values of ${a_{11}}$ and ${a_{22}}$ with ${A^4}$.
Formula Used:
The basic matrix multiplication formula are given as
1. $\begin{bmatrix}{{a_1}}&{{b_1}}\\{{c_1}}&{{d_1}}\end{bmatrix}\times \begin{bmatrix}{{a_2}}&{{b_2}}\\{{c_2}}&{{d_2}}\end{bmatrix}= \begin{bmatrix}{{a_1}{a_2} + {b_1}{c_2}}&{{a_1}{b_2} + {b_1}{d_2}}\\{{c_1}{a_2} + {d_1}{c_2}}&{{c_1}{b_2} + {d_1}{d_2}}\end{bmatrix}$
Complete step by step solution:
Given that the matrix is$A = \begin{bmatrix}x&1\\1&0\end{bmatrix},x \in \mathbb{R}$
Firstly, we will find the value of ${A^4}$.
We try to rewrite the expression ${A^4}$ in such a way that the simplification becomes easier.
So, ${A^4}$ can be rewritten as ${A^2}{A^2}$.
Now, we will find the value of ${A^2}$ by using the matrix multiplication formula $\begin{bmatrix}{{a_1}}&{{b_1}}\\{{c_1}}&{{d_1}}\end{bmatrix} \times \begin{bmatrix}{{a_2}}&{{b_2}}\\{{c_2}}&{{d_2}}\end{bmatrix} = \begin{bmatrix}{{a_1}{a_2} + {b_1}{c_2}}&{{a_1}{b_2} + {b_1}{d_2}}\\{{c_1}{a_2} + {d_1}{c_2}}&{{c_1}{b_2} + {d_1}{d_2}}\end{bmatrix}$, we get
${A^2} = \begin{bmatrix}x&1\\1&0\end{bmatrix} \times \begin{bmatrix}x&1\\1&0\end{bmatrix} \\ {A^2} = \begin{bmatrix}{x\left( x \right) + 1\left( 1 \right)}&{x\left( 1 \right) + 1\left( 0 \right)}\\{1\left( x \right) + 0\left( 1 \right)}&{1\left( 1 \right) + 0\left( 0 \right)}\end{bmatrix}$
Further, we will simplify the above matrix, we get
${A^2} = \begin{bmatrix}{{x^2} + 1}&{x + 0}\\{x + 0}&{1 + 0}\end{bmatrix} \\ {A^2} = \begin{bmatrix}{{x^2} + 1}&x\\x&1\end{bmatrix}$
Furthermore, we will find the value of ${A^4}$ by using the matrix multiplication formula $\begin{bmatrix}{{a_1}}&{{b_1}}\\{{c_1}}&{{d_1}}\end{bmatrix} \times \begin{bmatrix}{{a_2}}&{{b_2}}\\{{c_2}}&{{d_2}}\end{bmatrix} = \begin{bmatrix}{{a_1}{a_2} + {b_1}{c_2}}&{{a_1}{b_2} + {b_1}{d_2}}\\{{c_1}{a_2} + {d_1}{c_2}}&{{c_1}{b_2} + {d_1}{d_2}}\end{bmatrix}$, we get
${A^4} = {A^2}{A^2}\\{A^4} = \begin{bmatrix}{{x^2} + 1}&x\\x&1\end{bmatrix} \times \begin{bmatrix}{{x^2} + 1}&x\\x&1\end{bmatrix}\\{A^4} = \begin{bmatrix}{\left( {{x^2} + 1} \right)\left( {{x^2} + 1} \right) + x\left( x \right)}&{\left( {{x^2} + 1} \right)x + x\left( 1 \right)}\\{x\left( {{x^2} + 1} \right) + 1\left( x \right)}&{x\left( x \right) + 1\left( 1 \right)}\end{bmatrix}$
Now, we will simplify the above matrix, we get
${A^4} = \begin{bmatrix}{{x^4} + {x^2} + {x^2} + 1 + {x^2}}&{{x^3} + x + x}\\{{x^3} + x + x}&{{x^2} + 1}\end{bmatrix}\\{A^4} = \begin{bmatrix}{{x^4}3 + {x^2} + 1}&{{x^3} + 2x}\\{{x^3} + 2x}&{{x^2} + 1}\end{bmatrix}$
As we know that the general terms of the matrix can be written as $A = \begin{bmatrix}{{a_{11}}}&{{a_{12}}}\\{{a_{21}}}&{{a_{22}}}\end{bmatrix}$ and we are given that ${a_{11}} = 109$.
So, we will compare the value of ${a_{11}}$ with matrix ${A^4}$, we get
${a_{11}} = {x^4} + 3{x^2} + 1\\109 = {x^4} + 3{x^2} + 1$
Now, we will simplify the above equation using factorization method, we get
${x^4} + 3{x^2} - 108 = 0\\{x^4} + 12{x^2} - 9{x^2} - 108 = 0\\{x^2}\left( {{x^2} + 12} \right) - 9\left( {{x^2} + 12} \right) = 0$
Further, we will factor out the common factor, we get
$\left( {{x^2} + 12} \right)\left( {{x^2} - 9} \right) = 0\\{x^2} = - 12,9$
As it is given that $x \in R$ so ${x^2} = - 12$ cannot be possible because the value under root cannot be negative.
Thus, $x = \pm 3$.
Now, we will find the value of ${a_{22}}$ by comparing it with matrix ${A^4}$, we get
${a_{22}} = {x^2} + 1$
Further, substitute ${x^2} = 9$, we get
${a_{22}} = 9 + 1\\{a_{22}} = 10$
Hence, the value of ${a_{22}}$ is $10$.
Note: Try to reduce the given terms so that we can solve them easily. Otherwise find the terms like ${A^2},{A^3}$ and ${A^4}$ to reduce the chances of mistakes.
Formula Used:
The basic matrix multiplication formula are given as
1. $\begin{bmatrix}{{a_1}}&{{b_1}}\\{{c_1}}&{{d_1}}\end{bmatrix}\times \begin{bmatrix}{{a_2}}&{{b_2}}\\{{c_2}}&{{d_2}}\end{bmatrix}= \begin{bmatrix}{{a_1}{a_2} + {b_1}{c_2}}&{{a_1}{b_2} + {b_1}{d_2}}\\{{c_1}{a_2} + {d_1}{c_2}}&{{c_1}{b_2} + {d_1}{d_2}}\end{bmatrix}$
Complete step by step solution:
Given that the matrix is$A = \begin{bmatrix}x&1\\1&0\end{bmatrix},x \in \mathbb{R}$
Firstly, we will find the value of ${A^4}$.
We try to rewrite the expression ${A^4}$ in such a way that the simplification becomes easier.
So, ${A^4}$ can be rewritten as ${A^2}{A^2}$.
Now, we will find the value of ${A^2}$ by using the matrix multiplication formula $\begin{bmatrix}{{a_1}}&{{b_1}}\\{{c_1}}&{{d_1}}\end{bmatrix} \times \begin{bmatrix}{{a_2}}&{{b_2}}\\{{c_2}}&{{d_2}}\end{bmatrix} = \begin{bmatrix}{{a_1}{a_2} + {b_1}{c_2}}&{{a_1}{b_2} + {b_1}{d_2}}\\{{c_1}{a_2} + {d_1}{c_2}}&{{c_1}{b_2} + {d_1}{d_2}}\end{bmatrix}$, we get
${A^2} = \begin{bmatrix}x&1\\1&0\end{bmatrix} \times \begin{bmatrix}x&1\\1&0\end{bmatrix} \\ {A^2} = \begin{bmatrix}{x\left( x \right) + 1\left( 1 \right)}&{x\left( 1 \right) + 1\left( 0 \right)}\\{1\left( x \right) + 0\left( 1 \right)}&{1\left( 1 \right) + 0\left( 0 \right)}\end{bmatrix}$
Further, we will simplify the above matrix, we get
${A^2} = \begin{bmatrix}{{x^2} + 1}&{x + 0}\\{x + 0}&{1 + 0}\end{bmatrix} \\ {A^2} = \begin{bmatrix}{{x^2} + 1}&x\\x&1\end{bmatrix}$
Furthermore, we will find the value of ${A^4}$ by using the matrix multiplication formula $\begin{bmatrix}{{a_1}}&{{b_1}}\\{{c_1}}&{{d_1}}\end{bmatrix} \times \begin{bmatrix}{{a_2}}&{{b_2}}\\{{c_2}}&{{d_2}}\end{bmatrix} = \begin{bmatrix}{{a_1}{a_2} + {b_1}{c_2}}&{{a_1}{b_2} + {b_1}{d_2}}\\{{c_1}{a_2} + {d_1}{c_2}}&{{c_1}{b_2} + {d_1}{d_2}}\end{bmatrix}$, we get
${A^4} = {A^2}{A^2}\\{A^4} = \begin{bmatrix}{{x^2} + 1}&x\\x&1\end{bmatrix} \times \begin{bmatrix}{{x^2} + 1}&x\\x&1\end{bmatrix}\\{A^4} = \begin{bmatrix}{\left( {{x^2} + 1} \right)\left( {{x^2} + 1} \right) + x\left( x \right)}&{\left( {{x^2} + 1} \right)x + x\left( 1 \right)}\\{x\left( {{x^2} + 1} \right) + 1\left( x \right)}&{x\left( x \right) + 1\left( 1 \right)}\end{bmatrix}$
Now, we will simplify the above matrix, we get
${A^4} = \begin{bmatrix}{{x^4} + {x^2} + {x^2} + 1 + {x^2}}&{{x^3} + x + x}\\{{x^3} + x + x}&{{x^2} + 1}\end{bmatrix}\\{A^4} = \begin{bmatrix}{{x^4}3 + {x^2} + 1}&{{x^3} + 2x}\\{{x^3} + 2x}&{{x^2} + 1}\end{bmatrix}$
As we know that the general terms of the matrix can be written as $A = \begin{bmatrix}{{a_{11}}}&{{a_{12}}}\\{{a_{21}}}&{{a_{22}}}\end{bmatrix}$ and we are given that ${a_{11}} = 109$.
So, we will compare the value of ${a_{11}}$ with matrix ${A^4}$, we get
${a_{11}} = {x^4} + 3{x^2} + 1\\109 = {x^4} + 3{x^2} + 1$
Now, we will simplify the above equation using factorization method, we get
${x^4} + 3{x^2} - 108 = 0\\{x^4} + 12{x^2} - 9{x^2} - 108 = 0\\{x^2}\left( {{x^2} + 12} \right) - 9\left( {{x^2} + 12} \right) = 0$
Further, we will factor out the common factor, we get
$\left( {{x^2} + 12} \right)\left( {{x^2} - 9} \right) = 0\\{x^2} = - 12,9$
As it is given that $x \in R$ so ${x^2} = - 12$ cannot be possible because the value under root cannot be negative.
Thus, $x = \pm 3$.
Now, we will find the value of ${a_{22}}$ by comparing it with matrix ${A^4}$, we get
${a_{22}} = {x^2} + 1$
Further, substitute ${x^2} = 9$, we get
${a_{22}} = 9 + 1\\{a_{22}} = 10$
Hence, the value of ${a_{22}}$ is $10$.
Note: Try to reduce the given terms so that we can solve them easily. Otherwise find the terms like ${A^2},{A^3}$ and ${A^4}$ to reduce the chances of mistakes.
Recently Updated Pages
Elastic Collision in Two Dimensions Explained Simply

Elastic Collisions in One Dimension Explained

Electric Field of Infinite Line Charge and Cylinders Explained

Electric Flux and Area Vector Explained Simply

Electric Field of a Charged Spherical Shell Explained

Electricity and Magnetism Explained: Key Concepts & Applications

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Collisions: Types and Examples for Students

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Understanding Atomic Structure for Beginners

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Degree of Dissociation: Meaning, Formula, Calculation & Uses

Understanding Electromagnetic Waves and Their Importance

Understanding the Electric Field of a Uniformly Charged Ring

