Animated Solution for Mathematics - Matrices and Determinants: If A=[2−112], B=[1101], C=ABAT and X=ATC2A, then detX is equal to :
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Visualized Solution
Introduction to Matrices A and B
Given Matrix A=[2−112]
Given Matrix B=[1101]
Objective: Find det(X) where X=ATC2A
Determinant Formula for 2×2
For a matrix M=[acbd]
The determinant is: det(M)=ad−bc
Setting up det(A)
Substitute values from Matrix A:
det(A)=(2)(2)−(1)(−1)
Calculating det(A)
Main diagonal product: (2)(2)=2
Off-diagonal product: (1)(−1)=−1
det(A)=2−(−1)=3
Calculating det(B)
For Matrix B:
det(B)=(1)(1)−(0)(1)
det(B)=1−0=1
Properties of Determinants
Product Rule: det(MN)=det(M)⋅det(N)
Transpose Rule: det(MT)=det(M)
Power Rule: det(Mk)=(detM)k
Setting up det(C)
Given C=ABAT
Apply properties: det(C)=det(A)⋅det(B)⋅det(AT)
Calculating det(C)
Substitute known values:
det(A)=3, det(B)=1, det(AT)=3
det(C)=3⋅1⋅3=9
Setting up det(X)
Given X=ATC2A
Apply properties: det(X)=det(AT)⋅det(C2)⋅det(A)
Substituting Values for X
Using Power Rule: det(C2)=(detC)2=92
Substitute all values:
det(X)=3⋅92⋅3
Final Computation
det(X)=3⋅81⋅3
det(X)=9⋅81
det(X)=729
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The Sigma Insight: Properties of Determinants
Solution Diagram
Analyzing the Setup
Imagine you are staring at a complex matrix equation. You see A, you see B, and then you see C=ABAT and X=ATC2A.
Your first instinct might be to start multiplying, to dive into the rows and columns, and to perform a massive amount of arithmetic. Stop. Take a breath.
In the world of JEE Advanced, the most elegant path is rarely the one that requires the most writing. Today, we are going to solve for det(X) without ever performing a full matrix multiplication.
The Determinant Toolkit
First, let us look at our building blocks. We have:
A=[2−112],B=[1101]
To find the determinant of a 2×2 matrix M=[acbd], we use the classic formula det(M)=ad−bc.
For matrix A, this is (2)(2)−(1)(−1), which simplifies to 2−(−1)=3. For matrix B, it is even simpler: (1)(1)−(0)(1)=1.
We have our values: det(A)=3 and det(B)=1.
The Magic of Properties
Now, we enter the realm of matrix properties. This is where the JEE examiners test your conceptual depth.
We know that det(MN)=det(M)⋅det(N). We also know that det(AT)=det(A).
With these tools, we can dismantle the expression for C=ABAT. Instead of calculating C, we calculate its determinant directly:
det(C)=det(A)⋅det(B)⋅det(AT)
Substituting our known values, we get det(C)=3⋅1⋅3=9. See how the complexity just melted away?
The Grand Finale
Finally, we tackle X=ATC2A. Applying the same property, we get:
det(X)=det(AT)⋅det(C2)⋅det(A)
We know det(C2)=(detC)2, so det(C2)=92=81.
Now, we just plug everything in: det(X)=3⋅81⋅3. That is 9⋅81, which equals 729.
We have arrived at the answer, not through brute force, but through the elegant application of mathematical laws. Remember, in mathematics, the most powerful tool is often the one that lets you see the structure before you start the calculation.