Let A = [[2, 0, 1], [1, 1, 0], [1, 0, 1]], B = [B1, B2, B3], where B1, B2, B3 are column matrices, a

Q: Wait, why can't I just solve for $B_1, B_2, B_3$ using Gaussian elimination? It seems more straightforward.

You absolutely can! Gaussian elimination is a robust tool. However, in JEE Advanced, time is your most precious currency. Recognizing the matrix structure $AB=C$ allows you to use properties of determinants and inverses, which are often faster and less prone to arithmetic errors than solving three separate systems of linear equations.

Q: Do I really need to find the entire matrix $B$ to get the trace? That seems like a lot of work.

That is a brilliant observation! You absolutely do not. The trace is the sum of the diagonal elements. If you know $B = A^{-1}C$, then the diagonal element $b_{ii}$ is just the dot product of the $i$-th row of $A^{-1}$ and the $i$-th column of $C$. By focusing only on these specific dot products, you bypass the need to calculate the off-diagonal elements entirely.

Q: Why is the determinant of $C$ just the product of its diagonal elements?

Great question! Look at the matrix $C = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 3 & 2 \\ 0 & 0 & 1 \end{bmatrix}$. Because all elements below the main diagonal are zero, it is an upper triangular matrix. For any triangular matrix, the determinant is always the product of the diagonal entries. It is a beautiful property that saves you from the tedious process of cofactor expansion.

Question

Let $A = \begin{bmatrix} 2 & 0 & 1 \ 1 & 1 & 0 \ 1 & 0 & 1 \end{bmatrix}$, $B = [B_1, B_2, B_3]$, where $B_1, B_2, B_3$ are column matrices, and $AB_1 = \begin{bmatrix} 1 \ 0 \ 0 \end{bmatrix}$, $AB_2 = \begin{bmatrix} 2 \ 3 \ 0 \end{bmatrix}$, $AB_3 = \begin{bmatrix} 3 \ 2 \ 1 \end{bmatrix}$. If $\alpha = |B|$ and $\beta$ is the sum of all the diagonal elements of $B$, then $\alpha^3 + \beta^3$ is equal to

Sigma Percentile · Accepted Answer

### The Matrix Dance: A Journey into Elegance

Welcome, future engineers! Today, we are not just solving a matrix problem; we are uncovering the hidden architecture of linear systems.

When you look at a problem like this, it is easy to feel overwhelmed by the sheer number of variables. But I want you to take a deep breath.

In JEE Advanced, the most complex-looking problems often have the most elegant, symmetric solutions. Let us break this down together.

### Phase 1: The Power of Synthesis

We are given a matrix $A$ and three column vectors $B_1, B_2, B_3$ that satisfy the following relations:

$$AB_1 = \begin{bmatrix} 1 \ 0 \ 0 \end{bmatrix}, \quad AB_2 = \begin{bmatrix} 2 \ 3 \ 0 \end{bmatrix}, \quad AB_3 = \begin{bmatrix} 3 \ 2 \ 1 \end{bmatrix}$$

A novice might jump straight into solving three separate systems of equations. But you are not a novice; you are a strategist.

Notice that if we define $B = [B_1, B_2, B_3]$, then the definition of matrix multiplication tells us that $AB = [AB_1, AB_2, AB_3]$.

This transforms our scattered data into a single, beautiful equation: $AB = C$, where:

$$C = \begin{bmatrix} 1 & 2 & 3 \ 0 & 3 & 2 \ 0 & 0 & 1 \end{bmatrix}$$

We have just simplified the entire problem into one equation. This is the power of matrix notation.

### Phase 2: The Determinant Strategy

Now, we need to find $\alpha = |B|$. We know that $|AB| = |A| \cdot |B|$.

Since $AB = C$, we have $|A| \cdot |B| = |C|$, which implies:

$$\alpha = |B| = \frac{|C|}{|A|}$$

Let us calculate $|A|$ first. Expanding along the first row of $A = \begin{bmatrix} 2 & 0 & 1 \ 1 & 1 & 0 \ 1 & 0 & 1 \end{bmatrix}$, we get:

$$|A| = 2(1 - 0) - 0 + 1(0 - 1) = 2 - 1 = 1$$

Now, look at $C$. It is an upper triangular matrix! The determinant of an upper triangular matrix is simply the product of its diagonal elements:

$$|C| = 1 \cdot 3 \cdot 1 = 3$$

Thus, $\alpha = \frac{3}{1} = 3$. We have conquered the first half of the mountain.

### Phase 3: The Trace Insight

Next, we need $\beta$, the sum of the diagonal elements of $B$, also known as the trace of $B$, denoted as $	ext{tr}(B)$. We know $B = A^{-1}C$.

To find $A^{-1}$, we use the adjoint method: $A^{-1} = \frac{1}{|A|} 	ext{adj}(A)$. Since $|A| = 1$, $A^{-1}$ is just the adjoint of $A$.

Calculating the cofactors, we find:

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

Here is the secret: do not multiply the entire matrix $A^{-1}$ by $C$. We only need the diagonal elements $b_{11}, b_{22}, b_{33}$.

The element $b_{ii}$ is the dot product of the $i$-th row of $A^{-1}$ and the $i$-th column of $C$:

*   $b_{11} = (1)(1) + (0)(0) + (-1)(0) = 1$
*   $b_{22} = (-1)(2) + (1)(3) + (1)(0) = 1$
*   $b_{33} = (-1)(3) + (0)(2) + (2)(1) = -1$

Summing these up, $\beta = 1 + 1 - 1 = 1$.

### The Final Victory

We have $\alpha = 3$ and $\beta = 1$. The problem asks for $\alpha^3 + \beta^3$.

Plugging in our values, we get:

$$3^3 + 1^3 = 27 + 1 = 28$$

Look at that! Through logical deduction and the smart application of matrix properties, we arrived at the answer without getting lost in a sea of unnecessary calculations.

**The final result is 28.** Keep this mindset—always look for the structure before you start calculating. You are ready for the next challenge!

Enter Numerical Value:

Visualized Solution

Forming the Matrix Equation $A B = C$

The Determinant Strategy for $α$

Calculating the Determinant of Matrix $A$

Calculating the Determinant of Matrix $C$ and $α$

Strategy to find $β$ (Trace of $B$ )

Calculating Cofactors of $A$ (Rows 1 & 2)

Calculating Cofactors of $A$ (Row 3) & Constructing $A^{- 1}$

Calculating Diagonal Elements of $B$

Finding $β$ and Final Calculation

The Sigma Insight: Solution of System of Linear Equations (Matrix Method and Cramer's Rule)

The Matrix Dance

Phase 1

Phase 2

Phase 3

The Final Victory

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If $A = a 11 1 b b 0 d c, B = a 0 f 1 d g 1 c h, U = f g h, V = a^{2} 00, X = x y z$ and $A X = U$ has infinitely many solutions, prove that $B X = V$ has no unique solution. Also show that if $a f d \neq = 0$ , then $B X = V$ has no solution.

Enter Numerical Value:

Visualized Solution

Forming the Matrix Equation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB=C

The Determinant Strategy for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

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

Calculating the Determinant of Matrix .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Strategy to find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β (Trace of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B)

Calculating Cofactors of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A (Rows 1 & 2)

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

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β and Final Calculation