Let A = ( \beginmatrix 1 & - 1 & 0 0 & 1 & - 1 0 & 0 & 1 \endmatrix ) and B = 7A20 - 20A7 + 2I, where I is an identity matrix of order 3 \times 3. If B = [bij], then b13is equal to \underline\hspace2cm .

Matrices and Determinants PYQ — JEE Main 2021 · 20 Jul · Shift 1 · Q38 with Solution

View written solutionFree

Correct answer: 910

We are given

A=\begin{pmatrix} 1&-1&0\\ 0&1&-1\\ 0&0&1 \end{pmatrix}

and

B=7A^{20}-20A^7+2I.

We need to find $b_{13}$ , the $(1,3)$ -entry of $B$ .

1. Write $A$ in a convenient form

Observe that

A=I+N,

where

N=\begin{pmatrix} 0&-1&0\\ 0&0&-1\\ 0&0&0 \end{pmatrix}.

Now compute powers of $N$ :

N^2=\begin{pmatrix} 0&0&1\\ 0&0&0\\ 0&0&0 \end{pmatrix}, y N^3=0.

So $N$ is nilpotent of index $3$ .

2. Find a general formula for $A^n$

Since $A=I+N$ and $N^3=0$ , by binomial expansion:

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

Thus,

A^n= \begin{pmatrix} 1&-n&\binom{n}{2}\\ 0&1&-n\\ 0&0&1 \end{pmatrix}.

Therefore, the $(1,3)$ entry of $A^n$ is

(A^n)_{13}=\binom{n}{2}.

3. Compute the required entries

For $n=20$ :

(A^{20})_{13}=\binom{20}{2}=\frac{20\cdot 19}{2}=190.

For $n=7$ :

(A^7)_{13}=\binom{7}{2}=\frac{7\cdot 6}{2}=21.

Also, the $(1,3)$ entry of $I$ is $0$ .

4. Compute $b_{13}$

From

B=7A^{20}-20A^7+2I,

we get

b_{13}=7(A^{20})_{13}-20(A^7)_{13}+2(I)_{13}.

So,

b_{13}=7(190)-20(21)+2(0).

Now,

7\cdot 190=1330, \qquad 20\cdot 21=420.

Hence,

b_{13}=1330-420=910.

5. Final answer

\boxed{910}

The derived answer matches the stored correct answer.

More from Matrices and Determinants

View written solutionFree

Correct answer: 910

We are given

A=\begin{pmatrix} 1&-1&0\\ 0&1&-1\\ 0&0&1 \end{pmatrix}

and

B=7A^{20}-20A^7+2I.

We need to find $b_{13}$ , the $(1,3)$ -entry of $B$ .

1. Write $A$ in a convenient form

Observe that

A=I+N,

where

N=\begin{pmatrix} 0&-1&0\\ 0&0&-1\\ 0&0&0 \end{pmatrix}.

Now compute powers of $N$ :

N^2=\begin{pmatrix} 0&0&1\\ 0&0&0\\ 0&0&0 \end{pmatrix}, y N^3=0.

So $N$ is nilpotent of index $3$ .

2. Find a general formula for $A^n$

Since $A=I+N$ and $N^3=0$ , by binomial expansion:

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

Thus,

A^n= \begin{pmatrix} 1&-n&\binom{n}{2}\\ 0&1&-n\\ 0&0&1 \end{pmatrix}.

Therefore, the $(1,3)$ entry of $A^n$ is

(A^n)_{13}=\binom{n}{2}.

3. Compute the required entries

For $n=20$ :

(A^{20})_{13}=\binom{20}{2}=\frac{20\cdot 19}{2}=190.

For $n=7$ :

(A^7)_{13}=\binom{7}{2}=\frac{7\cdot 6}{2}=21.

Also, the $(1,3)$ entry of $I$ is $0$ .

4. Compute $b_{13}$

From

B=7A^{20}-20A^7+2I,

we get

b_{13}=7(A^{20})_{13}-20(A^7)_{13}+2(I)_{13}.

So,

b_{13}=7(190)-20(21)+2(0).

Now,

7\cdot 190=1330, \qquad 20\cdot 21=420.

Hence,

b_{13}=1330-420=910.

5. Final answer

\boxed{910}

The derived answer matches the stored correct answer.

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Matrices and Determinants question

1. Write $A$ in a convenient form

2. Find a general formula for $A^n$

3. Compute the required entries

4. Compute $b_{13}$

5. Final answer

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2. Find a general formula for $A^n$

3. Compute the required entries

4. Compute $b_{13}$

5. Final answer

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2. Find a general formula for $A^n$

4. Compute $b_{13}$