Let A=[\beginarrayll1 & 2 0 & 1\endarray] and B=I+\operatornameadj(A)+(\operatornameadj A)^2+\ldots+(\operatornameadj A)^10 . Then, the sum of all the elements of the matrix B is:

Matrices and Determinants PYQ — JEE Main 2024 · 4 Apr · Shift 2 · Q49 with Solution

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Correct answer: D

Given matrix

$A=\begin{bmatrix}1&2\\0&1\end{bmatrix}$

We need to find

$B=I+\operatorname{adj}(A)+(\operatorname{adj}A)^2+\cdots +(\operatorname{adj}A)^{10}.$

Then compute the sum of all elements of $B$ .

Find $\operatorname{adj}(A)$

For a $2\times 2$ matrix

$\begin{bmatrix}a&b\\c&d\end{bmatrix},$

the adjugate is

$\operatorname{adj}(A)=\begin{bmatrix}d&-b\\-c&a\end{bmatrix}.$

So for

$A=\begin{bmatrix}1&2\\0&1\end{bmatrix},$

we get

$\operatorname{adj}(A)=\begin{bmatrix}1&-2\\0&1\end{bmatrix}.$

Let

$M=\operatorname{adj}(A)=\begin{bmatrix}1&-2\\0&1\end{bmatrix}.$

Find the general power $M^n$

Write

$M=I+N,$

where

$N=\begin{bmatrix}0&-2\\0&0\end{bmatrix}.$

Now,

$N^2=0$

because multiplying $N$ by itself gives the zero matrix.

Hence,

$(I+N)^n = I+nN$

for any positive integer $n$ .

Therefore,

$M^n=\begin{bmatrix}1&-2n\\0&1\end{bmatrix}.$

Check quickly:

$M^1=\begin{bmatrix}1&-2\\0&1\end{bmatrix}$
$M^2=\begin{bmatrix}1&-4\\0&1\end{bmatrix}$

correct.

Sum the series

We need

$B=\sum_{k=0}^{10} M^k.$

Using

$M^k=\begin{bmatrix}1&-2k\\0&1\end{bmatrix},$

we get

=\begin{bmatrix} \sum_{k=0}^{10}1 & \sum_{k=0}^{10}(-2k)\\[4pt] \sum_{k=0}^{10}0 & \sum_{k=0}^{10}1 \end{bmatrix}.$$ Now, $$\sum_{k=0}^{10}1=11,$$ and $$\sum_{k=0}^{10}k=\frac{10\cdot 11}{2}=55.$$ So, $$\sum_{k=0}^{10}(-2k)=-2\cdot 55=-110.$$ Thus, $$B=\begin{bmatrix}11&-110\\0&11\end{bmatrix}.$$ --- 5. **Sum of all elements of $B$** $$11+(-110)+0+11=-88.$$ --- 6. **Compare with options** The sum of all elements is $$\boxed{-88}$$ So the correct option is **D**. --- 7. **Comparison with stored answer** Stored correct answer: **D** Our derived answer: **D** They agree.

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View written solutionFree

Correct answer: D

Given matrix

$A=\begin{bmatrix}1&2\\0&1\end{bmatrix}$

We need to find

$B=I+\operatorname{adj}(A)+(\operatorname{adj}A)^2+\cdots +(\operatorname{adj}A)^{10}.$

Then compute the sum of all elements of $B$ .

Find $\operatorname{adj}(A)$

For a $2\times 2$ matrix

$\begin{bmatrix}a&b\\c&d\end{bmatrix},$

the adjugate is

$\operatorname{adj}(A)=\begin{bmatrix}d&-b\\-c&a\end{bmatrix}.$

So for

$A=\begin{bmatrix}1&2\\0&1\end{bmatrix},$

we get

$\operatorname{adj}(A)=\begin{bmatrix}1&-2\\0&1\end{bmatrix}.$

Let

$M=\operatorname{adj}(A)=\begin{bmatrix}1&-2\\0&1\end{bmatrix}.$

Find the general power $M^n$

Write

$M=I+N,$

where

$N=\begin{bmatrix}0&-2\\0&0\end{bmatrix}.$

Now,

$N^2=0$

because multiplying $N$ by itself gives the zero matrix.

Hence,

$(I+N)^n = I+nN$

for any positive integer $n$ .

Therefore,

$M^n=\begin{bmatrix}1&-2n\\0&1\end{bmatrix}.$

Check quickly:

$M^1=\begin{bmatrix}1&-2\\0&1\end{bmatrix}$
$M^2=\begin{bmatrix}1&-4\\0&1\end{bmatrix}$

correct.

Sum the series

We need

$B=\sum_{k=0}^{10} M^k.$

Using

$M^k=\begin{bmatrix}1&-2k\\0&1\end{bmatrix},$

we get

=\begin{bmatrix} \sum_{k=0}^{10}1 & \sum_{k=0}^{10}(-2k)\\[4pt] \sum_{k=0}^{10}0 & \sum_{k=0}^{10}1 \end{bmatrix}.$$ Now, $$\sum_{k=0}^{10}1=11,$$ and $$\sum_{k=0}^{10}k=\frac{10\cdot 11}{2}=55.$$ So, $$\sum_{k=0}^{10}(-2k)=-2\cdot 55=-110.$$ Thus, $$B=\begin{bmatrix}11&-110\\0&11\end{bmatrix}.$$ --- 5. **Sum of all elements of $B$** $$11+(-110)+0+11=-88.$$ --- 6. **Compare with options** The sum of all elements is $$\boxed{-88}$$ So the correct option is **D**. --- 7. **Comparison with stored answer** Stored correct answer: **D** Our derived answer: **D** They agree.

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