If [1 1; 0 1] [1 2; 0 1] [1 3; 0 1] ... [1 n-1; 0 1] = [1 78; 0 1], then the inverse of [1 n; 0 1] i

Q: Wait, why can't we just multiply them one by one?

In JEE Advanced, time is your most precious currency. If you try to multiply these matrices sequentially, you will be trapped in an algebraic nightmare. Always look for a pattern or a general term first. The structure of these matrices is a 'shearing' transformation, and their composition is additive.

Q: How did you find the inverse so quickly without using the determinant formula?

You absolutely could use the standard formula $A^{-1} = \frac{1}{\det(A)} \text{adj}(A)$, but for this specific class of matrices, there is a geometric intuition. Since the matrix adds $k$ to the top-right, the inverse must subtract $k$. It is an 'undo' operation. Recognizing this saves you precious seconds.

Q: Why do we reject the negative root for $n$?

In the context of this problem, $n$ represents the index of the last matrix in the product sequence. A sequence index cannot be negative. Always check the physical or logical constraints of your variables after solving a quadratic equation.

Question

If $\begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 \ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 3 \ 0 & 1 \end{bmatrix} \dots \begin{bmatrix} 1 & n-1 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 78 \ 0 & 1 \end{bmatrix}$, then the inverse of $\begin{bmatrix} 1 & n \ 0 & 1 \end{bmatrix}$ is

Options:

(A) $\begin{bmatrix} 1 & -13 \ 0 & 1 \end{bmatrix}$

(B) $\begin{bmatrix} 1 & 0 \ 12 & 1 \end{bmatrix}$

(C) $\begin{bmatrix} 1 & -12 \ 0 & 1 \end{bmatrix}$

(D) $\begin{bmatrix} 1 & 0 \ 13 & 1 \end{bmatrix}$

Sigma Percentile · Accepted Answer

### The Hidden Elegance of Matrix Chains

Welcome, warriors of JEE Advanced. Today, we are going to dismantle a problem that, at first glance, looks like a tedious exercise in matrix multiplication.

You see a long chain of matrices, and your instinct might be to start multiplying the first two, then the third, and so on. But stop! If you do that, you are falling into the trap.

In the arena of competitive exams, we do not brute-force; we observe, we generalize, and we conquer.

### The Pattern Recognition

Let us define our general matrix as $M_k = \begin{bmatrix} 1 & k \ 0 & 1 \end{bmatrix}$.

When you multiply two such matrices, $M_a$ and $M_b$, something magical happens. The product is:

$$M_a \cdot M_b = \begin{bmatrix} 1 & a \ 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} 1 & b \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & a+b \ 0 & 1 \end{bmatrix}$$

Do you see the elegance? The matrix multiplication is simply adding the top-right elements. This is not just algebra; it is a transformation. This property is the key to unlocking the entire problem.

### The Summation Bridge

Now, apply this to our chain. We have the product of matrices from $k=1$ to $n-1$.

Because of our discovery, the final matrix is simply:

$$\begin{bmatrix} 1 & \sum_{k=1}^{n-1} k \ 0 & 1 \end{bmatrix}$$

We know this sum is 78. The sum of the first $N$ natural numbers is given by the formula $\frac{N(N+1)}{2}$.

Here, our number of terms is $N = n-1$. So, we set the sum equal to 78:

$$\frac{(n-1)((n-1)+1)}{2} = 78$$

This simplifies beautifully to:

$$\frac{n(n-1)}{2} = 78$$

### The Quadratic Moment of Truth

Now, we are left with a simple quadratic equation. Multiplying by 2 gives $n(n-1) = 156$, which expands to $n^2 - n - 156 = 0$.

Solving this quadratic, we look for factors of $-156$ that add up to $-1$. Those are $-13$ and $+12$.

Thus, $(n-13)(n+12) = 0$. We get two roots: $n=13$ and $n=-12$.

Since $n$ represents the count of terms in our sequence, it must be positive. Therefore, **$n=13$**.

### The Final Twist

Finally, we need the inverse of $M_{13} = \begin{bmatrix} 1 & 13 \ 0 & 1 \end{bmatrix}$.

Since $M_k$ adds $k$, its inverse must subtract $k$. It is the inverse operation.

Thus, the inverse is:

$$\begin{bmatrix} 1 & -13 \ 0 & 1 \end{bmatrix}$$

You have mastered the pattern, solved the quadratic, and found the inverse. This is the power of mathematical insight. Keep practicing, keep observing, and never let the complexity of a problem intimidate you. You are capable of seeing the beauty beneath the surface.

Animated Solution for Mathematics - Matrices and Determinants: If $[1011] [1021] [1031] \dots [10 n - 1 1] = [10781]$ , then the inverse of $[10 n 1]$ is

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Visualized Solution

Identify the Pattern of $M_{k}$

Property of $M_{a} \cdot M_{b}$

Computing the Product $M_{a} \cdot M_{b}$

Generalizing to $\prod M_{k}$

Equating to the Given Matrix

Formula for $\sum N$

Substituting $N = n - 1$

Forming the Quadratic Equation

Solving for $n$

Target Matrix Inverse

Shortcut for $M_{k}^{- 1}$

Final Answer

The Sigma Insight: Algebraic Operations on Matrices

The Hidden Elegance of Matrix Chains

The Pattern Recognition

The Summation Bridge

The Quadratic Moment of Truth

The Final Twist

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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; } If [10​11​][10​21​][10​31​]…[10​n−11​]=[10​781​], then the inverse of [10​n1​] is

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Visualized Solution

Identify the Pattern of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Mk​

Property of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ma​⋅Mb​

Computing the Product .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ma​⋅Mb​

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

Equating to the Given Matrix

Formula for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∑N

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } N=n−1