If A = [ \beginmatrix 1 & 1 & 1 0 & 1 & 1 0 & 0 & 1 \endmatrix ] and M = A + A2 + A3 + ....... + A20, then the sum of all the elements of the matrix M is equal to \underline\hspace2cm .

Matrices and Determinants PYQ — JEE Main 2021 · 27 Jul · Shift 2 · Q44 with Solution

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Correct answer: 2020

Write the matrix in a useful form

Given

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

Notice that

A=I+N,

where

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

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

N^3=0.

Compute $N^2$

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

Find a formula for $A^k$

Using

A^k=(I+N)^k=I+kN+\binom{k}{2}N^2

because $N^3=0$ .

Thus,

A^k= \begin{bmatrix} 1 & k & k+\binom{k}{2}\\ 0 & 1 & k\\ 0 & 0 & 1 \end{bmatrix}.

Now simplify the $(1,3)$ entry:

k+\binom{k}{2}=k+\frac{k(k-1)}{2}=\frac{k(k+1)}{2}.

So,

A^k= \begin{bmatrix} 1 & k & \dfrac{k(k+1)}{2}\\ 0 & 1 & k\\ 0 & 0 & 1 \end{bmatrix}.

Find the sum of all entries of $A^k$

The nonzero entries are:

1,\; k,\; \frac{k(k+1)}{2},\; 1,\; k,\; 1.

Hence sum of all elements of $A^k$ is

S_k=3+2k+\frac{k(k+1)}{2}.

Simplify:

S_k=3+2k+\frac{k^2+k}{2}= rac{k^2+5k+6}{2}= rac{(k+2)(k+3)}{2}.

Now sum from $k=1$ to $20$

We need

M=A+A^2+A^3+\cdots+A^{20}.

So the sum of all elements of $M$ is

\sum_{k=1}^{20} S_k =\sum_{k=1}^{20} \frac{k^2+5k+6}{2}.

Thus,

\sum_{k=1}^{20} S_k =\frac{1}{2}\left(\sum_{k=1}^{20}k^2+5\sum_{k=1}^{20}k+6\cdot 20\right).

Use standard formulas:

\sum_{k=1}^{20}k=\frac{20\cdot 21}{2}=210,

\sum_{k=1}^{20}k^2=\frac{20\cdot 21\cdot 41}{6}=2870.

Therefore,

\sum_{k=1}^{20} S_k =\frac{1}{2}\left(2870+5(210)+120\right)

=\frac{1}{2}(2870+1050+120) =\frac{4040}{2}=2020.

Final answer

The sum of all elements of the matrix $M$ is

\boxed{2020}.

Comparison with stored answer

Stored correct answer = $2020$ .

Our derived answer matches the stored answer.

More from Matrices and Determinants

View written solutionFree

Correct answer: 2020

Write the matrix in a useful form

Given

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

Notice that

A=I+N,

where

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

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

N^3=0.

Compute $N^2$

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

Find a formula for $A^k$

Using

A^k=(I+N)^k=I+kN+\binom{k}{2}N^2

because $N^3=0$ .

Thus,

A^k= \begin{bmatrix} 1 & k & k+\binom{k}{2}\\ 0 & 1 & k\\ 0 & 0 & 1 \end{bmatrix}.

Now simplify the $(1,3)$ entry:

k+\binom{k}{2}=k+\frac{k(k-1)}{2}=\frac{k(k+1)}{2}.

So,

A^k= \begin{bmatrix} 1 & k & \dfrac{k(k+1)}{2}\\ 0 & 1 & k\\ 0 & 0 & 1 \end{bmatrix}.

Find the sum of all entries of $A^k$

The nonzero entries are:

1,\; k,\; \frac{k(k+1)}{2},\; 1,\; k,\; 1.

Hence sum of all elements of $A^k$ is

S_k=3+2k+\frac{k(k+1)}{2}.

Simplify:

S_k=3+2k+\frac{k^2+k}{2}= rac{k^2+5k+6}{2}= rac{(k+2)(k+3)}{2}.

Now sum from $k=1$ to $20$

We need

M=A+A^2+A^3+\cdots+A^{20}.

So the sum of all elements of $M$ is

\sum_{k=1}^{20} S_k =\sum_{k=1}^{20} \frac{k^2+5k+6}{2}.

Thus,

\sum_{k=1}^{20} S_k =\frac{1}{2}\left(\sum_{k=1}^{20}k^2+5\sum_{k=1}^{20}k+6\cdot 20\right).

Use standard formulas:

\sum_{k=1}^{20}k=\frac{20\cdot 21}{2}=210,

\sum_{k=1}^{20}k^2=\frac{20\cdot 21\cdot 41}{6}=2870.

Therefore,

\sum_{k=1}^{20} S_k =\frac{1}{2}\left(2870+5(210)+120\right)

=\frac{1}{2}(2870+1050+120) =\frac{4040}{2}=2020.

Final answer

The sum of all elements of the matrix $M$ is

\boxed{2020}.

Comparison with stored answer

Stored correct answer = $2020$ .

Our derived answer matches the stored answer.

Matrices and Determinants question

Matrices and Determinants question

More from Matrices and Determinants

Matrices and Determinants question

Matrices and Determinants question

More from Matrices and Determinants