Let A=[\beginarraylll 1 & a & a 0 & 1 & b 0 & 0 & 1 \endarray], a, b \in \mathbbR . If for some n \in \mathbbN, A^n=[\beginarrayccc 1 & 48 & 2160 0 & 1 & 96 0 & 0 & 1 \endarray] then n+a+b is equal to \underline\hspace2cm .

Matrices and Determinants PYQ — JEE Main 2022 · 25 Jul · Shift 2 · Q39 with Solution

View written solutionFree

Correct answer: 24

Write $A$ as $I+N$

We have

A=\begin{bmatrix}1&a&a\\0&1&b\\0&0&1\end{bmatrix} = I+N,

where

N=\begin{bmatrix}0&a&a\\0&0&b\\0&0&0\end{bmatrix}.

Since $N$ is strictly upper triangular of order $3$ , we know $N^3=0.$

So,

A^n=(I+N)^n=I+nN+\binom{n}{2}N^2.

Compute $N^2$

Multiply $N$ by itself:

N^2=\begin{bmatrix}0&a&a\\0&0&b\\0&0&0\end{bmatrix} \begin{bmatrix}0&a&a\\0&0&b\\0&0&0\end{bmatrix} =\begin{bmatrix}0&0&ab\\0&0&0\\0&0&0\end{bmatrix}.

Therefore,

A^n=I+n\begin{bmatrix}0&a&a\\0&0&b\\0&0&0\end{bmatrix} +\binom{n}{2}\begin{bmatrix}0&0&ab\\0&0&0\\0&0&0\end{bmatrix}.

So,

A^n= \begin{bmatrix} 1 & na & na+\binom{n}{2}ab \\ 0 & 1 & nb \\ 0 & 0 & 1 \end{bmatrix}.

Compare with the given matrix

Given

A^n=\begin{bmatrix}1&48&2160\\0&1&96\\0&0&1\end{bmatrix}.

Thus,

na=48 \quad ...(1)

nb=96 \quad ...(2)

na+\binom{n}{2}ab=2160 \quad ...(3)

From (1) and (2),

a=\frac{48}{n}, \qquad b=\frac{96}{n}.

Substitute into (3):

48+\binom{n}{2}\left(\frac{48}{n}\cdot \frac{96}{n}\right)=2160.

Now,

\binom{n}{2}=\frac{n(n-1)}{2},

48+\frac{n(n-1)}{2}\cdot \frac{4608}{n^2}=2160.

Simplify:

48+2304\cdot \frac{n-1}{n}=2160.

Thus,

2304\cdot \frac{n-1}{n}=2112.

Divide by $192$ :

12\cdot \frac{n-1}{n}=11.

So,

12n-12=11n \Rightarrow n=12.

Find $a$ and $b$

From (1):

a=\frac{48}{12}=4.

From (2):

b=\frac{96}{12}=8.

Hence,

n+a+b=12+4+8=24.

Comparison with stored answer

Derived answer is $24$ , which matches the stored correct answer.

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View written solutionFree

Correct answer: 24

Write $A$ as $I+N$

We have

A=\begin{bmatrix}1&a&a\\0&1&b\\0&0&1\end{bmatrix} = I+N,

where

N=\begin{bmatrix}0&a&a\\0&0&b\\0&0&0\end{bmatrix}.

Since $N$ is strictly upper triangular of order $3$ , we know $N^3=0.$

So,

A^n=(I+N)^n=I+nN+\binom{n}{2}N^2.

Compute $N^2$

Multiply $N$ by itself:

N^2=\begin{bmatrix}0&a&a\\0&0&b\\0&0&0\end{bmatrix} \begin{bmatrix}0&a&a\\0&0&b\\0&0&0\end{bmatrix} =\begin{bmatrix}0&0&ab\\0&0&0\\0&0&0\end{bmatrix}.

Therefore,

A^n=I+n\begin{bmatrix}0&a&a\\0&0&b\\0&0&0\end{bmatrix} +\binom{n}{2}\begin{bmatrix}0&0&ab\\0&0&0\\0&0&0\end{bmatrix}.

So,

A^n= \begin{bmatrix} 1 & na & na+\binom{n}{2}ab \\ 0 & 1 & nb \\ 0 & 0 & 1 \end{bmatrix}.

Compare with the given matrix

Given

A^n=\begin{bmatrix}1&48&2160\\0&1&96\\0&0&1\end{bmatrix}.

Thus,

na=48 \quad ...(1)

nb=96 \quad ...(2)

na+\binom{n}{2}ab=2160 \quad ...(3)

From (1) and (2),

a=\frac{48}{n}, \qquad b=\frac{96}{n}.

Substitute into (3):

48+\binom{n}{2}\left(\frac{48}{n}\cdot \frac{96}{n}\right)=2160.

Now,

\binom{n}{2}=\frac{n(n-1)}{2},

48+\frac{n(n-1)}{2}\cdot \frac{4608}{n^2}=2160.

Simplify:

48+2304\cdot \frac{n-1}{n}=2160.

Thus,

2304\cdot \frac{n-1}{n}=2112.

Divide by $192$ :

12\cdot \frac{n-1}{n}=11.

So,

12n-12=11n \Rightarrow n=12.

Find $a$ and $b$

From (1):

a=\frac{48}{12}=4.

From (2):

b=\frac{96}{12}=8.

Hence,

n+a+b=12+4+8=24.

Comparison with stored answer

Derived answer is $24$ , which matches the stored correct answer.

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Matrices and Determinants question

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