Question:
If \(A\) and \(B\) are \(n\times n\) square matrices such that
\[
(2A+B)^2+(A-3B)^2=5A^2-2AB+10B^2,
\]
then \(ABAB=\)
If \(A\) and \(B\) are \(n\times n\) square matrices such that
\[
(2A+B)^2+(A-3B)^2=5A^2-2AB+10B^2,
\]
then \(ABAB=\)
Show Hint
Whenever matrix expressions are expanded and compared, check whether the comparison implies \(AB=BA\). Once matrices commute, many algebraic simplifications become possible.
Updated On: Jun 18, 2026
- \[ \frac{1}{2}\Big[(A-B)^2+(A+B)^2\Big] \]
- \(4AB\)
- \[ \frac{1}{2}\Big[(A+B)^2-(A-B)^2\Big] \]
- \(A^2B^2\)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Expand the given matrix expression.
\[ (2A+B)^2 = 4A^2+2AB+2BA+B^2 \] and \[ (A-3B)^2 = A^2-3AB-3BA+9B^2. \] Adding, \[ (2A+B)^2+(A-3B)^2 = 5A^2-AB-BA+10B^2. \]
Step 2: Compare with the given equation.
Given, \[ 5A^2-AB-BA+10B^2 = 5A^2-2AB+10B^2. \] Cancelling common terms, \[ -AB-BA=-2AB. \] Therefore, \[ AB+BA=2AB. \] Hence, \[ BA=AB. \] Thus \(A\) and \(B\) commute.
Step 3: Evaluate \(ABAB\).
Since \[ AB=BA, \] we have \[ ABAB=A(BA)B. \] Replacing \(BA\) by \(AB\), \[ ABAB=A(AB)B. \] Therefore, \[ ABAB=A^2B^2. \]
Step 4: Final conclusion.
Hence, \[ \boxed{ABAB=A^2B^2} \]
\[ (2A+B)^2 = 4A^2+2AB+2BA+B^2 \] and \[ (A-3B)^2 = A^2-3AB-3BA+9B^2. \] Adding, \[ (2A+B)^2+(A-3B)^2 = 5A^2-AB-BA+10B^2. \]
Step 2: Compare with the given equation.
Given, \[ 5A^2-AB-BA+10B^2 = 5A^2-2AB+10B^2. \] Cancelling common terms, \[ -AB-BA=-2AB. \] Therefore, \[ AB+BA=2AB. \] Hence, \[ BA=AB. \] Thus \(A\) and \(B\) commute.
Step 3: Evaluate \(ABAB\).
Since \[ AB=BA, \] we have \[ ABAB=A(BA)B. \] Replacing \(BA\) by \(AB\), \[ ABAB=A(AB)B. \] Therefore, \[ ABAB=A^2B^2. \]
Step 4: Final conclusion.
Hence, \[ \boxed{ABAB=A^2B^2} \]
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