Unlocking Determinants: A Step-by-Step Guide

by Andrew McMorgan 45 views

Hey Plastik Magazine readers! Ever stumbled upon a determinant problem that looked like a tangled mess of variables? Don't sweat it! Today, we're diving deep into a specific determinant, breaking it down step-by-step to make it crystal clear. We're going to prove a fascinating mathematical identity, and trust me, by the end of this, you'll be able to tackle similar challenges with confidence. Let's get started, shall we?

The Problem: Setting the Stage

Alright, guys, let's look at the determinant we're aiming to solve. We're trying to prove that:

∣a+bb+cc+ac+aa+bb+cb+cc+aa+b∣=2(a+b+c)(a2+b2+c2βˆ’abβˆ’bcβˆ’ca){\begin{vmatrix} a+b & b+c & c+a \\ c+a & a+b & b+c \\ b+c & c+a & a+b \\ \end{vmatrix} = 2(a+b+c)(a^2+b^2+c^2 - ab - bc - ca)}

This might seem a bit intimidating at first glance, with all the a's, b's, and c's flying around. But hey, that's what makes math fun, right? The goal is to manipulate this determinant using some clever tricks – specifically, column operations – to simplify it and reveal the beautiful underlying structure. We'll follow the hints provided, which are our roadmap to success. So, let’s get into the nitty-gritty.

Understanding the Goal

Our ultimate goal is to transform the left-hand side (LHS) of the equation (the determinant) into the right-hand side (RHS). The RHS is a product of two factors: 2(a+b+c){2(a+b+c)} and (a2+b2+c2βˆ’abβˆ’bcβˆ’ca){(a^2+b^2+c^2 - ab - bc - ca)}. This tells us what we should be aiming for during our simplification process. We need to somehow make these factors appear as we perform our column operations. This is a bit like a treasure hunt, where we're looking for these specific "treasures" within the determinant.

Decoding the Hints

The hints are the keys to unlocking this problem. They suggest a sequence of column operations. Let's break them down:

  1. C1β†’C1βˆ’C2βˆ’C3{C_1 \to C_1 - C_2 - C_3}: This means we'll replace the first column C1{C_1} with the result of subtracting the second column C2{C_2} and the third column C3{C_3} from it.
  2. C2β†’C2βˆ’C1{C_2 \to C_2 - C_1}: Next, we'll replace the second column C2{C_2} with the result of subtracting the new first column C1{C_1} from it.
  3. C3β†’C3βˆ’C1{C_3 \to C_3 - C_1}: Finally, we'll replace the third column C3{C_3} with the result of subtracting the new first column C1{C_1} from it.

These operations are designed to introduce zeros and create opportunities for factoring, bringing us closer to our target on the RHS. Remember that column operations, unlike row operations, do not change the determinant's value if done correctly. So, we're free to use these operations to reshape the determinant without altering its fundamental value. Now, with the plan in place, let's start calculating!

Applying the Column Operations: The Transformation

Alright, let's get our hands dirty and start applying these column operations, one step at a time. This is where the magic really happens.

Step 1: C1β†’C1βˆ’C2βˆ’C3{C_1 \to C_1 - C_2 - C_3}

Let's apply the first column operation. We'll replace the first column with C1βˆ’C2βˆ’C3{C_1 - C_2 - C_3}. This means:

  • New C1{C_1} element 1: (a+b)βˆ’(b+c)βˆ’(c+a)=a+bβˆ’bβˆ’cβˆ’cβˆ’a=βˆ’2c{(a+b) - (b+c) - (c+a) = a+b-b-c-c-a = -2c}
  • New C1{C_1} element 2: (c+a)βˆ’(a+b)βˆ’(b+c)=c+aβˆ’aβˆ’bβˆ’bβˆ’c=βˆ’2b{(c+a) - (a+b) - (b+c) = c+a-a-b-b-c = -2b}
  • New C1{C_1} element 3: (b+c)βˆ’(c+a)βˆ’(a+b)=b+cβˆ’cβˆ’aβˆ’aβˆ’b=βˆ’2a{(b+c) - (c+a) - (a+b) = b+c-c-a-a-b = -2a}

So, after this operation, our determinant looks like this:

βˆ£βˆ’2cb+cc+aβˆ’2ba+bb+cβˆ’2ac+aa+b∣{\begin{vmatrix} -2c & b+c & c+a \\ -2b & a+b & b+c \\ -2a & c+a & a+b \\ \end{vmatrix}}

Notice that the first column is now βˆ’2c,βˆ’2b,βˆ’2a{-2c, -2b, -2a}. We're already seeing some simplification happen. Let's move on to the next step!

Step 2: C2β†’C2βˆ’C1{C_2 \to C_2 - C_1}

Now, let's perform the second column operation: C2β†’C2βˆ’C1{C_2 \to C_2 - C_1}. This means we subtract the new C1{C_1} (which we just calculated) from C2{C_2}:

  • New C2{C_2} element 1: (b+c)βˆ’(βˆ’2c)=b+c+2c=b+3c{(b+c) - (-2c) = b+c+2c = b+3c}
  • New C2{C_2} element 2: (a+b)βˆ’(βˆ’2b)=a+b+2b=a+3b{(a+b) - (-2b) = a+b+2b = a+3b}
  • New C2{C_2} element 3: (c+a)βˆ’(βˆ’2a)=c+a+2a=c+3a{(c+a) - (-2a) = c+a+2a = c+3a}

Our determinant is now:

βˆ£βˆ’2cb+3cc+aβˆ’2ba+3bb+cβˆ’2ac+3aa+b∣{\begin{vmatrix} -2c & b+3c & c+a \\ -2b & a+3b & b+c \\ -2a & c+3a & a+b \\ \end{vmatrix}}

Getting closer, guys! The second column is starting to show some interesting patterns. Let's proceed to the final step and see what we get.

Step 3: C3β†’C3βˆ’C1{C_3 \to C_3 - C_1}

Here comes the final column operation: C3β†’C3βˆ’C1{C_3 \to C_3 - C_1}. Subtract the new C1{C_1} from C3{C_3}:

  • New C3{C_3} element 1: (c+a)βˆ’(βˆ’2c)=c+a+2c=3c+a{(c+a) - (-2c) = c+a+2c = 3c+a}
  • New C3{C_3} element 2: (b+c)βˆ’(βˆ’2b)=b+c+2b=3b+c{(b+c) - (-2b) = b+c+2b = 3b+c}
  • New C3{C_3} element 3: (a+b)βˆ’(βˆ’2a)=a+b+2a=3a+b{(a+b) - (-2a) = a+b+2a = 3a+b}

After this final transformation, our determinant becomes:

βˆ£βˆ’2cb+3c3c+aβˆ’2ba+3b3b+cβˆ’2ac+3a3a+b∣{\begin{vmatrix} -2c & b+3c & 3c+a \\ -2b & a+3b & 3b+c \\ -2a & c+3a & 3a+b \\ \end{vmatrix}}

This looks quite different from where we started. But don't worry, we are far from the end. We will do some more operations to get the final solution!

Unveiling the Solution: The Final Steps

We have performed all the column operations, but we still do not have the expected solution. Let's do some more operation to reach the final answer.

Factor Out -2

Notice that there is a βˆ’2{-2} in the first column, we can factor it out.

βˆ£βˆ’2(c)b+3c3c+aβˆ’2(b)a+3b3b+cβˆ’2(a)c+3a3a+b∣{\begin{vmatrix} -2(c) & b+3c & 3c+a \\ -2(b) & a+3b & 3b+c \\ -2(a) & c+3a & 3a+b \\ \end{vmatrix}}

=βˆ’2∣cb+3c3c+aba+3b3b+cac+3a3a+b∣{= -2\begin{vmatrix} c & b+3c & 3c+a \\ b & a+3b & 3b+c \\ a & c+3a & 3a+b \\ \end{vmatrix}}

Now we can perform C1β†’C1+C2+C3{C1 \to C1+C2+C3}

=βˆ’2∣c+b+3c+3c+ab+3c3c+ab+a+3b+3b+ca+3b3b+ca+c+3a+3a+bc+3a3a+b∣{= -2\begin{vmatrix} c+b+3c+3c+a & b+3c & 3c+a \\ b+a+3b+3b+c & a+3b & 3b+c \\ a+c+3a+3a+b & c+3a & 3a+b \\ \end{vmatrix}}

=βˆ’2∣a+b+7cb+3c3c+aa+7b+ca+3b3b+c7a+b+cc+3a3a+b∣{= -2\begin{vmatrix} a+b+7c & b+3c & 3c+a \\ a+7b+c & a+3b & 3b+c \\ 7a+b+c & c+3a & 3a+b \\ \end{vmatrix}}

This doesn't seem to be working. So we will take a different approach!

Alternative Approach - Summing All Rows and Applying Operations

This approach will start by summing all the rows, which can help reveal the (a+b+c){(a+b+c)} factor that we are looking for. Summing the rows in a determinant doesn't change its value, because essentially, we are not changing the relationships between the columns or rows, only the way we are expressing them. The beauty of this is that the result is easier to deal with.

So let's proceed

Summing all the rows, we get R1βˆ’>R1+R2+R3{R_1 -> R_1 + R_2 + R_3}

∣a+b+c+a+b+cb+c+a+b+c+ac+a+b+c+a+bc+aa+bb+cb+cc+aa+b∣{\begin{vmatrix} a+b+c+a+b+c & b+c+a+b+c+a & c+a+b+c+a+b \\ c+a & a+b & b+c \\ b+c & c+a & a+b \\ \end{vmatrix}}

=∣2a+2b+2c2a+2b+2c2a+2b+2cc+aa+bb+cb+cc+aa+b∣{= \begin{vmatrix} 2a+2b+2c & 2a+2b+2c & 2a+2b+2c \\ c+a & a+b & b+c \\ b+c & c+a & a+b \\ \end{vmatrix}}

=∣2(a+b+c)2(a+b+c)2(a+b+c)c+aa+bb+cb+cc+aa+b∣{= \begin{vmatrix} 2(a+b+c) & 2(a+b+c) & 2(a+b+c) \\ c+a & a+b & b+c \\ b+c & c+a & a+b \\ \end{vmatrix}}

We can factor out 2(a+b+c){2(a+b+c)} from the first row:

2(a+b+c)∣111c+aa+bb+cb+cc+aa+b∣{2(a+b+c)\begin{vmatrix} 1 & 1 & 1 \\ c+a & a+b & b+c \\ b+c & c+a & a+b \\ \end{vmatrix}}

Now, apply the following column operations:

  • C2β†’C2βˆ’C1{C_2 \to C_2 - C_1}
  • C3β†’C3βˆ’C1{C_3 \to C_3 - C_1}

=2(a+b+c)∣100c+aa+bβˆ’cβˆ’ab+cβˆ’cβˆ’ab+cc+aβˆ’bβˆ’ca+bβˆ’bβˆ’c∣{= 2(a+b+c)\begin{vmatrix} 1 & 0 & 0 \\ c+a & a+b-c-a & b+c-c-a \\ b+c & c+a-b-c & a+b-b-c \\ \end{vmatrix}}

=2(a+b+c)∣100c+abβˆ’cbβˆ’ab+caβˆ’baβˆ’c∣{= 2(a+b+c)\begin{vmatrix} 1 & 0 & 0 \\ c+a & b-c & b-a \\ b+c & a-b & a-c \\ \end{vmatrix}}

Now, expand the determinant along the first row. The determinant simplifies to:

2(a+b+c)[1βˆ—((bβˆ’c)(aβˆ’c)βˆ’(bβˆ’a)(aβˆ’b))]=2(a+b+c)[abβˆ’acβˆ’bc+c2βˆ’(abβˆ’b2βˆ’a2+ab)]{2(a+b+c)[1 * ((b-c)(a-c) - (b-a)(a-b))] = 2(a+b+c)[ab - ac - bc + c^2 - (ab - b^2 - a^2 + ab)]}

=2(a+b+c)[abβˆ’acβˆ’bc+c2βˆ’ab+b2+a2βˆ’ab]{= 2(a+b+c)[ab - ac - bc + c^2 - ab + b^2 + a^2 - ab]}

=2(a+b+c)[a2+b2+c2βˆ’abβˆ’bcβˆ’ac]{= 2(a+b+c)[a^2 + b^2 + c^2 - ab - bc - ac]}

And there we have it! We have successfully shown that:

∣a+bb+cc+ac+aa+bb+cb+cc+aa+b∣=2(a+b+c)(a2+b2+c2βˆ’abβˆ’bcβˆ’ca){\begin{vmatrix} a+b & b+c & c+a \\ c+a & a+b & b+c \\ b+c & c+a & a+b \\ \end{vmatrix} = 2(a+b+c)(a^2+b^2+c^2 - ab - bc - ca)}

Conclusion: Determinant Mastery Achieved!

Woohoo! We've made it through the determinant. This problem is now conquered. Remember, the key to solving determinant problems is to practice and remember key operations. The more you work with these concepts, the more natural they will become. Keep up the great work, and don't be afraid to take on any mathematical challenge that comes your way! Until next time, keep exploring the awesome world of math!