Let A = [[1, 2], [0, 1]] and B = I + adj(A) + (adj A)² + ... + (adj A)¹⁰. Then, the sum of all the e

Q: Wait, why do we sum from $k=0$ to $10$? That is $11$ terms, not $10$. Is that a mistake?

That is an excellent catch! It is a classic trap in JEE problems. When we sum from $k=0$ to $n$, we are including the $k=0$ term, which is the identity matrix $I$. So, yes, there are $n+1$ terms. In this case, $10+1 = 11$ terms. Always double-check your summation limits!

Q: Can I use the formula for a geometric series of matrices, like $S = (I - M)^{-1}$?

While that formula is mathematically elegant, it is dangerous here. That formula is valid only if the series converges and $(I - M)$ is invertible. In competitive exams, it is much safer and more reliable to identify the pattern of powers and sum the elements individually, as we did here.

Q: Why does the bottom-left element of $M^k$ always stay zero?

Great observation! This is a property of upper triangular matrices. When you multiply two upper triangular matrices, the resulting matrix is also upper triangular. Since our initial matrix $M$ is upper triangular with zeros in the bottom-left, all its powers will maintain that zero in the bottom-left position.

Question

Let $A = \begin{bmatrix} 1 & 2 \ 0 & 1 \end{bmatrix}$ and $B = I + 	ext{adj}(A) + (	ext{adj } A)^2 + \dots + (	ext{adj } A)^{10}$. Then, the sum of all the elements of the matrix $B$ is:

Options:

(A) -124

(B) 22

(C) -88

(D) -110

Sigma Percentile · Accepted Answer

### The Elegance of Matrix Patterns

Welcome, future engineers! Today, we are going to peel back the layers of a seemingly intimidating matrix problem.

When you first look at a matrix $A = \begin{bmatrix} 1 & 2 \ 0 & 1 \end{bmatrix}$ and a summation of its adjoint powers $B = I + 	ext{adj}(A) + (	ext{adj } A)^2 + \dots + (	ext{adj } A)^{10}$, it is natural to feel a bit overwhelmed.

But remember, in the world of JEE Advanced, complexity is often just a mask for a beautiful, underlying pattern. Let us embark on this journey to uncover it.

### Phase 1: The Adjoint Transformation

Our first step is to tame the adjoint. For any $2 	imes 2$ matrix $P = \begin{bmatrix} a & b \ c & d \end{bmatrix}$, the adjoint is simply $	ext{adj}(P) = \begin{bmatrix} d & -b \ -c & a \end{bmatrix}$.

Applying this to our matrix $A$, we swap the diagonal elements and negate the off-diagonal ones.

Thus, we find that $M = 	ext{adj}(A) = \begin{bmatrix} 1 & -2 \ 0 & 1 \end{bmatrix}$. This matrix $M$ is our building block.

### Phase 2: The Power of Patterns

Now, we need to find the powers of $M$. Let us calculate $M^2$ first:

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

If we continue to $M^3$, we get:

$$M^3 = M^2 \cdot M = \begin{bmatrix} 1 & -4 \ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & -2 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -6 \ 0 & 1 \end{bmatrix}$$

Do you see the rhythm? The diagonal elements remain $1$, the bottom-left remains $0$, and the top-right element is simply $-2$ times the power $k$.

By the power of mathematical induction, we can confidently state that for any power $k$:

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

This is the key that unlocks the entire problem.

### Phase 3: The Final Summation

With our general form $M^k$ in hand, we can express $B$ as a summation of matrices:

$$B = \sum_{k=0}^{10} \begin{bmatrix} 1 & -2k \ 0 & 1 \end{bmatrix} = \begin{bmatrix} \sum_{k=0}^{10} 1 & \sum_{k=0}^{10} -2k \ \sum_{k=0}^{10} 0 & \sum_{k=0}^{10} 1 \end{bmatrix}$$

Now, we evaluate each component. The summation $\sum_{k=0}^{10} 1$ represents adding $1$ eleven times (remember, $k=0$ to $10$ is $11$ terms!), which gives us $11$.

The top-right element is $-2 \sum_{k=0}^{10} k$. Since the $k=0$ term is $0$, this is:

$$-2 	imes \frac{10(11)}{2} = -110$$

Putting it all together, we get:

$$B = \begin{bmatrix} 11 & -110 \ 0 & 11 \end{bmatrix}$$

The final step is to sum all the elements of $B$: $11 + (-110) + 0 + 11 = -88$.

And there you have it! The final answer is **$-88$**. We have navigated through the matrix powers and arrived at the solution with precision and clarity.

Animated Solution for Mathematics - Matrices and Determinants: Let $A = [1021]$ and $B = I + adj (A) + (adj A)^{2} + \dots + (adj A)^{10}$ . Then, the sum of all the elements of the matrix $B$ is:

Select Answer:

Visualized Solution

Introduction to Matrix $A$

Finding the Adjoint of $A$

Calculating $M^{2}$

Generalizing to $M^{k}$

Setting up the Sum for $B$

Calculating Diagonal Elements

Summing the Top-Right Element

Constructing Matrix $B$

Final Sum of Elements

The Sigma Insight: Adjoint and Inverse of a Matrix

The Elegance of Matrix Patterns

Phase 1

Phase 2

Phase 3

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Animated Solution for Mathematics - Matrices and Determinants: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A=[10​21​] and B=I+adj(A)+(adj A)2+⋯+(adj A)10. Then, the sum of all the elements of the matrix B is:

Select Answer:

Visualized Solution

Introduction to Matrix .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Finding the Adjoint of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M2

Generalizing to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Mk

Setting up the Sum for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B