Let A = [ \beginmatrix 1 & 0 & 0 1 & 1 & 0 1 & 1 & 1 \endmatrix ] and B = A20. Then the sum of the elements of the first column of B is :

Matrices and Determinants PYQ — JEE Main 2018 · 16 Apr · Shift 1 · Q25 with Solution

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Correct answer: C

We are given

A=\begin{bmatrix} 1&0&0\\ 1&1&0\\ 1&1&1 \end{bmatrix}

and $B=A^{20}$ .

We need the sum of the elements of the first column of $B$ .

Observe that the first column of $A^{20}$ is obtained by multiplying $A^{20}$ with the first standard basis vector

e_1=\begin{bmatrix}1\\0\\0\end{bmatrix}.

So we need

A^{20}e_1.

Write $A$ as

A=I+N,

where

N=\begin{bmatrix} 0&0&0\\ 1&0&0\\ 1&1&0 \end{bmatrix}.

Since $N$ is strictly lower triangular of order $3$ , we have

N^3=0.

Hence by the binomial theorem,

A^{20}=(I+N)^{20}=I+\binom{20}{1}N+\binom{20}{2}N^2.

Compute $Ne_1$ and $N^2e_1$ .

First,

Ne_1= \begin{bmatrix} 0&0&0\\ 1&0&0\\ 1&1&0 \end{bmatrix} \begin{bmatrix}1\\0\\0\end{bmatrix} = \begin{bmatrix}0\\1\\1\end{bmatrix}.

Now compute $N^2e_1=N(Ne_1)$ :

N^2e_1= \begin{bmatrix} 0&0&0\\ 1&0&0\\ 1&1&0 \end{bmatrix} \begin{bmatrix}0\\1\\1\end{bmatrix} = \begin{bmatrix}0\\0\\1\end{bmatrix}.

Also,

Ie_1=e_1= \begin{bmatrix}1\\0\\0\end{bmatrix}.

Therefore,

A^{20}e_1=Ie_1+20\,Ne_1+\binom{20}{2}N^2e_1.

A^{20}e_1= \begin{bmatrix}1\\0\\0\end{bmatrix} +20\begin{bmatrix}0\\1\\1\end{bmatrix} +190\begin{bmatrix}0\\0\\1\end{bmatrix}.

Thus the first column of $B=A^{20}$ is

\begin{bmatrix}1\\20\\210\end{bmatrix}.

Sum of the elements of the first column:

1+20+210=231.

Checking options:

A: $210$
B: $211$
C: $231$ ✓
D: $251$

So the correct option is C.

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View written solutionFree

Correct answer: C

We are given

A=\begin{bmatrix} 1&0&0\\ 1&1&0\\ 1&1&1 \end{bmatrix}

and $B=A^{20}$ .

We need the sum of the elements of the first column of $B$ .

Observe that the first column of $A^{20}$ is obtained by multiplying $A^{20}$ with the first standard basis vector

e_1=\begin{bmatrix}1\\0\\0\end{bmatrix}.

So we need

A^{20}e_1.

Write $A$ as

A=I+N,

where

N=\begin{bmatrix} 0&0&0\\ 1&0&0\\ 1&1&0 \end{bmatrix}.

Since $N$ is strictly lower triangular of order $3$ , we have

N^3=0.

Hence by the binomial theorem,

A^{20}=(I+N)^{20}=I+\binom{20}{1}N+\binom{20}{2}N^2.

Compute $Ne_1$ and $N^2e_1$ .

First,

Ne_1= \begin{bmatrix} 0&0&0\\ 1&0&0\\ 1&1&0 \end{bmatrix} \begin{bmatrix}1\\0\\0\end{bmatrix} = \begin{bmatrix}0\\1\\1\end{bmatrix}.

Now compute $N^2e_1=N(Ne_1)$ :

N^2e_1= \begin{bmatrix} 0&0&0\\ 1&0&0\\ 1&1&0 \end{bmatrix} \begin{bmatrix}0\\1\\1\end{bmatrix} = \begin{bmatrix}0\\0\\1\end{bmatrix}.

Also,

Ie_1=e_1= \begin{bmatrix}1\\0\\0\end{bmatrix}.

Therefore,

A^{20}e_1=Ie_1+20\,Ne_1+\binom{20}{2}N^2e_1.

A^{20}e_1= \begin{bmatrix}1\\0\\0\end{bmatrix} +20\begin{bmatrix}0\\1\\1\end{bmatrix} +190\begin{bmatrix}0\\0\\1\end{bmatrix}.

Thus the first column of $B=A^{20}$ is

\begin{bmatrix}1\\20\\210\end{bmatrix}.

Sum of the elements of the first column:

1+20+210=231.

Checking options:

A: $210$
B: $211$
C: $231$ ✓
D: $251$

So the correct option is C.

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