Let A = [[2, -1], [0, 2]]. If B = I - 5C1(adj A) + 5C2(adj A)^2 - ... - 5C5(adj A)^5, then the sum o

Q: Wait, why can we use the binomial theorem for matrices? I thought matrix multiplication wasn't commutative?

That is a brilliant observation! You are absolutely right that matrix multiplication is generally not commutative. However, the binomial theorem works here because the identity matrix $I$ commutes with every square matrix $X$ (i.e., $IX = XI = X$). Since $I$ and $\text{adj} A$ commute, the expansion holds perfectly.

Q: Is there a faster way to find the adjoint of a $2 \times 2$ matrix?

Yes! For any $2 \times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the adjoint is simply $\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$. You just swap the diagonal elements and negate the off-diagonal ones. It is a lifesaver in competitive exams!

Q: How do I know the pattern for $K^n$ is $\begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix}$?

You can verify this using mathematical induction. If $K = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$, then $K^2 = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$. By continuing this multiplication, you will see the top-right element increments by 1 each time. It is a standard result for nilpotent-like matrices!

Question

Let $A = \begin{bmatrix} 2 & -1 \ 0 & 2 \end{bmatrix}$. If $B = I - {}^5C_1 (	ext{adj} A) + {}^5C_2 (	ext{adj} A)^2 - \dots - {}^5C_5 (	ext{adj} A)^5$, then the sum of all elements of the matrix $B$ is

Options:

(A) -5

(B) -6

(C) -7

(D) -8

Sigma Percentile · Accepted Answer

### The Beauty of Binomial Symmetry

Welcome, future engineers! Today, we are going to dismantle a problem that looks like a monster but is actually a masterpiece of algebraic elegance.

We are given a matrix $A = \begin{bmatrix} 2 & -1 \ 0 & 2 \end{bmatrix}$ and asked to find the sum of elements of a matrix $B$ defined by the series:
$$B = I - {}^{5}C_{1} (	ext{adj} A) + {}^{5}C_{2} (	ext{adj} A)^{2} - \dots - {}^{5}C_{5} (	ext{adj} A)^{5}$$

If you try to calculate $(	ext{adj} A)^{2}, (	ext{adj} A)^{3}$, and so on, you will be lost in a sea of arithmetic errors. Instead, let us look at the structure.

This series is the exact expansion of $(I - 	ext{adj} A)^{5}$. Because the identity matrix $I$ commutes with any matrix, we can treat this exactly like the scalar binomial expansion $(x - y)^{n}$. This realization is our first victory.

### The Adjoint Shortcut

Now, we need to find $	ext{adj} A$. For a $2 	imes 2$ matrix, we do not need the long cofactor method. We simply swap the diagonal elements and negate the off-diagonal ones.

Given $A = \begin{bmatrix} 2 & -1 \ 0 & 2 \end{bmatrix}$, swapping the $2$s keeps the diagonal as $2, 2$. Negating the $-1$ gives us $1$, and the $0$ stays $0$.

Thus, we find:
$$	ext{adj} A = \begin{bmatrix} 2 & 1 \ 0 & 2 \end{bmatrix}$$

Now, let us define $M = I - 	ext{adj} A$. Substituting our values, we get:
$$M = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} - \begin{bmatrix} 2 & 1 \ 0 & 2 \end{bmatrix} = \begin{bmatrix} -1 & -1 \ 0 & -1 \end{bmatrix}$$

### The Power of Patterns

We need to compute $M^{5}$. Notice that we can factor out a $-1$ from matrix $M$:
$$M = -1 \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix}$$

Let $K = \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix}$. Then $M^{5} = (-1)^{5} K^{5} = -K^{5}$.

Now, let us look at the powers of $K$:
$$K^{2} = \begin{bmatrix} 1 & 2 \ 0 & 1 \end{bmatrix}, \quad K^{3} = \begin{bmatrix} 1 & 3 \ 0 & 1 \end{bmatrix}$$

The pattern is undeniable:
$$K^{n} = \begin{bmatrix} 1 & n \ 0 & 1 \end{bmatrix}$$

Therefore, we conclude:
$$K^{5} = \begin{bmatrix} 1 & 5 \ 0 & 1 \end{bmatrix}$$

### The Final Summation

We are almost there! Substituting back into our expression for $B$:
$$B = -K^{5} = -\begin{bmatrix} 1 & 5 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} -1 & -5 \ 0 & -1 \end{bmatrix}$$

The question asks for the sum of all elements in $B$. That is:
$$(-1) + (-5) + 0 + (-1) = -7$$

We have navigated the complexity and arrived at the truth. Remember, in JEE Advanced, the most complex-looking expressions often hide the most beautiful, simple patterns.

The final answer is **-7**. Keep your eyes open, trust your fundamentals, and keep pushing forward!

Animated Solution for Mathematics - Matrices and Determinants: Let $A = [20 - 1 2]$ . If $B = I -^{5} C_{1} (adj A) +^{5} C_{2} (adj A)^{2} - \dots -^{5} C_{5} (adj A)^{5}$ , then the sum of all elements of the matrix $B$ is

Select Answer:

Visualized Solution

Observe Matrix $A$

Analyze the Expression for $B$

Apply Binomial Theorem

Calculate Adjoint of $A$

Calculate Matrix $M = I - adj A$

Find the Pattern for $M^{n}$

Calculate $M^{5}$ (Matrix $B$ )

Sum of All Elements

The Sigma Insight: Adjoint and Inverse of a Matrix

The Beauty of Binomial Symmetry

The Adjoint Shortcut

The Power of Patterns

The Final Summation

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Select Answer:

Visualized Solution

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

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

Apply Binomial Theorem

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

Calculate Matrix .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M=I−adjA

Find the Pattern for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Mn

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M5 (Matrix .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B)

Sum of All Elements

The Sigma Insight: Adjoint and Inverse of a Matrix

The Beauty of Binomial Symmetry

The Adjoint Shortcut

The Power of Patterns

The Final Summation

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