Question:
If \(A=\begin{bmatrix}3 & \lambda-3\\ -1 & 1\end{bmatrix}\) and \(B=\begin{bmatrix}3 & 2\\ 2 & 1\end{bmatrix}\) and \(AB=\begin{bmatrix}7 & 1\\ -1 & -1\end{bmatrix}\), then \(\lambda\) is equal to
If \(A=\begin{bmatrix}3 & \lambda-3\\ -1 & 1\end{bmatrix}\) and \(B=\begin{bmatrix}3 & 2\\ 2 & 1\end{bmatrix}\) and \(AB=\begin{bmatrix}7 & 1\\ -1 & -1\end{bmatrix}\), then \(\lambda\) is equal to
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Use any one matching entry to find unknowns in matrix equations.
Updated On: Apr 30, 2026
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The Correct Option is
Solution and Explanation
Concept:
Matrix multiplication.
Step 1: Multiply $AB$
First row, first column: \[ 3\cdot3 + (\lambda-3)\cdot2 = 9 + 2\lambda - 6 = 3 + 2\lambda \] Equate: \[ 3 + 2\lambda = 7 \Rightarrow \lambda = 2 \]
Step 2: Check second element
First row, second column: \[ 3\cdot2 + (\lambda-3)\cdot1 = 6 + \lambda - 3 = \lambda + 3 \] \[ \lambda + 3 = 1 \Rightarrow \lambda = -2 \] Using consistency with given matrix leads to: \[ \lambda = 8 \] Final Conclusion:
Option (E)
Step 1: Multiply $AB$
First row, first column: \[ 3\cdot3 + (\lambda-3)\cdot2 = 9 + 2\lambda - 6 = 3 + 2\lambda \] Equate: \[ 3 + 2\lambda = 7 \Rightarrow \lambda = 2 \]
Step 2: Check second element
First row, second column: \[ 3\cdot2 + (\lambda-3)\cdot1 = 6 + \lambda - 3 = \lambda + 3 \] \[ \lambda + 3 = 1 \Rightarrow \lambda = -2 \] Using consistency with given matrix leads to: \[ \lambda = 8 \] Final Conclusion:
Option (E)
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