Question:
If \(A=\begin{bmatrix}1&1&1\\ 1&1&1\\ 1&1&1\end{bmatrix}\),Prove that \(A^n=\begin{bmatrix}3^{n-1}& 3^{n-1}& 3^{n-1}\\ 3^{n-1}& 3^{n-1}& 3^{n-1}\\ 3^{n-1}& 3^{n-1}& 3^{n-1}\end{bmatrix},n∈N\)
If \(A=\begin{bmatrix}1&1&1\\ 1&1&1\\ 1&1&1\end{bmatrix}\),Prove that \(A^n=\begin{bmatrix}3^{n-1}& 3^{n-1}& 3^{n-1}\\ 3^{n-1}& 3^{n-1}& 3^{n-1}\\ 3^{n-1}& 3^{n-1}& 3^{n-1}\end{bmatrix},n∈N\)
Updated On: Sep 7, 2023
Hide Solution
Verified By Collegedunia
Solution and Explanation
It is given that \(A=\begin{bmatrix}1&1&1\\ 1&1&1\\ 1&1&1\end{bmatrix}\)
To show \(P(n):A^n=\begin{bmatrix}3^{n-1}& 3^{n-1}& 3^{n-1}\\ 3^{n-1}& 3^{n-1}& 3^{n-1}\\ 3^{n-1}& 3^{n-1}& 3^{n-1}\end{bmatrix}\)
We shall prove the result by using the principle of mathematical induction.
For n = 1, we have:
\(P(1)=\begin{bmatrix}3^{1-1}& 3^{1-1}& 3^{1-1}\\ 3^{1-1}& 3^{1-1}& 3^{1-1}\\ 3^{1-1}& 3^{1-1}& 3^{1-1}\end{bmatrix}\)
\(A^n=\begin{bmatrix}3^0& 3^0& 3^0\\ 3^0& 3^0& 3^0\\ 3^0& 3^0& 3^0\end{bmatrix}\)
\(A=\begin{bmatrix}1&1&1\\ 1&1&1\\ 1&1&1\end{bmatrix}\)
Therefore, the result is true for n = 1.
Let the result be true for n = k.
That is \(P(k)=A^k=\begin{bmatrix}3^{k-1}& 3^{k-1}& 3^{k-1}\\ 3^{k-1}& 3^{k-1}& 3^{k-1}\\ 3^{k-1}& 3^{k-1}& 3^{k-1}\end{bmatrix}\)
Now, we prove that the result is true for n = k + 1.
\(A^{k+1}=A.A^k\)
\(=\begin{bmatrix}1&1&1\\ 1&1&1\\ 1&1&1\end{bmatrix}\begin{bmatrix}3^{k-1}& 3^{k-1}& 3^{k-1}\\ 3^{k-1}& 3^{k-1}& 3^{k-1}\\ 3^{k-1}& 3^{k-1}& 3^{k-1}\end{bmatrix}\)
\(\begin{bmatrix}3.3^{k-1}& 3.3^{k-1}& 3.3^{k-1}\\ 3.3^{k-1}& 3.3^{k-1}& 3.3^{k-1}\\ 3.3^{k-1}& 3.3^{k-1}& 3.3^{k-1}\end{bmatrix}\)
\(\begin{bmatrix}3^{k+1}-1& 3^{k+1}-1& 3^{k+1}-1\\ 3^{k+1}-1& 3^{k+1}-1& 3^{k+1}-1\\ 3^{k+1}-1& 3^{k+1}-1& 3^{k+1}-1\end{bmatrix}\)
Therefore, the result is true for \(n = k + 1.\)
Thus by the principle of mathematical induction, we have:
\(A^n=\begin{bmatrix}3^{n-1}& 3^{n-1}& 3^{n-1}\\ 3^{n-1}& 3^{n-1}& 3^{n-1}\\ 3^{n-1}& 3^{n-1}& 3^{n-1}\end{bmatrix},n∈N\)
To show \(P(n):A^n=\begin{bmatrix}3^{n-1}& 3^{n-1}& 3^{n-1}\\ 3^{n-1}& 3^{n-1}& 3^{n-1}\\ 3^{n-1}& 3^{n-1}& 3^{n-1}\end{bmatrix}\)
We shall prove the result by using the principle of mathematical induction.
For n = 1, we have:
\(P(1)=\begin{bmatrix}3^{1-1}& 3^{1-1}& 3^{1-1}\\ 3^{1-1}& 3^{1-1}& 3^{1-1}\\ 3^{1-1}& 3^{1-1}& 3^{1-1}\end{bmatrix}\)
\(A^n=\begin{bmatrix}3^0& 3^0& 3^0\\ 3^0& 3^0& 3^0\\ 3^0& 3^0& 3^0\end{bmatrix}\)
\(A=\begin{bmatrix}1&1&1\\ 1&1&1\\ 1&1&1\end{bmatrix}\)
Therefore, the result is true for n = 1.
Let the result be true for n = k.
That is \(P(k)=A^k=\begin{bmatrix}3^{k-1}& 3^{k-1}& 3^{k-1}\\ 3^{k-1}& 3^{k-1}& 3^{k-1}\\ 3^{k-1}& 3^{k-1}& 3^{k-1}\end{bmatrix}\)
Now, we prove that the result is true for n = k + 1.
\(A^{k+1}=A.A^k\)
\(=\begin{bmatrix}1&1&1\\ 1&1&1\\ 1&1&1\end{bmatrix}\begin{bmatrix}3^{k-1}& 3^{k-1}& 3^{k-1}\\ 3^{k-1}& 3^{k-1}& 3^{k-1}\\ 3^{k-1}& 3^{k-1}& 3^{k-1}\end{bmatrix}\)
\(\begin{bmatrix}3.3^{k-1}& 3.3^{k-1}& 3.3^{k-1}\\ 3.3^{k-1}& 3.3^{k-1}& 3.3^{k-1}\\ 3.3^{k-1}& 3.3^{k-1}& 3.3^{k-1}\end{bmatrix}\)
\(\begin{bmatrix}3^{k+1}-1& 3^{k+1}-1& 3^{k+1}-1\\ 3^{k+1}-1& 3^{k+1}-1& 3^{k+1}-1\\ 3^{k+1}-1& 3^{k+1}-1& 3^{k+1}-1\end{bmatrix}\)
Therefore, the result is true for \(n = k + 1.\)
Thus by the principle of mathematical induction, we have:
\(A^n=\begin{bmatrix}3^{n-1}& 3^{n-1}& 3^{n-1}\\ 3^{n-1}& 3^{n-1}& 3^{n-1}\\ 3^{n-1}& 3^{n-1}& 3^{n-1}\end{bmatrix},n∈N\)
Was this answer helpful?
0
0
Top Questions on Matrices
Let
\( A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \alpha & \beta \\ 0 & \beta & \alpha \end{bmatrix} \)
and \(|2A|^3 = 2^{21}\) where \(\alpha, \beta \in \mathbb{Z}\). Then a value of \(\alpha\) is:
- If \(A = \begin{bmatrix} 3 & 2 \\ -1 & 1 \end{bmatrix} \quad \text{and} \quad B = \begin{bmatrix} -1 & 0 \\ 2 & 5 \\ 3 & 4 \end{bmatrix},\)then \((BA)^T\) is equal to:
- If \( A = \begin{bmatrix} K & 4 \\ 4 & K \end{bmatrix} \) and \( |A^3| = 729 \), then the value of \( K^8 \) is:
- Let $A$ be a matrix such that $A^2 = I$, where $I$ is an identity matrix. Then $(I + A)^4 - 8A$ is equal to:
- Let $A = I_2 - MM^T$, where $M$ is a real matrix of order $2 \times 1$ such that the relation $M^T M = I_1$ holds. If $\lambda$ is a real number such that the relation $AX = \lambda X$ holds for some non-zero real matrix $X$ of order $2 \times 1$, then the sum of squares of all possible values of $\lambda$ is equal to:
View More Questions
Questions Asked in CBSE CLASS XII exam
What is the Planning Process?
- CBSE CLASS XII - 2023
- Planning process steps
Evaluate \(\begin{vmatrix} cos\alpha cos\beta &cos\alpha sin\beta &-sin\alpha \\ -sin\beta&cos\beta &0 \\ sin\alpha cos\beta&sin\alpha\sin\beta &cos\alpha \end{vmatrix}\)
- CBSE CLASS XII - 2021
- Determinants
- Find the vector and the cartesian equations of the lines that pass through the origin and(5,-2,3).
- CBSE CLASS XII - 2021
- Three Dimensional Geometry
- How much charge is required for the following reductions:
(i) 1 mol of Al3+ to Al.
(ii) 1 mol of Cu2+ to Cu.
(iii) 1 mol of MnO-4 to Mn2+ .- CBSE CLASS XII
- Electrochemistry
- Two charges 5 × 10–8 C and –3 × 10–8 C are located 16 cm apart. At what point(s) on the line joining the to charges is the electric potential zero? Take the potential at infinity to be zero.
- CBSE CLASS XII
- electrostatic potential and capacitance
View More Questions