Matrices & Determinants
Matrices And Determinants
nta_abhyas_2025
Grade 12

Question:

Let $A$ and $B$ are square matrices of order 2 such that $A + adj(B^T) = \begin{bmatrix} 3 & 2 \\ 2 & 3 \end{bmatrix}$ and $adj(A^T) - adj(B) = \begin{bmatrix} -2 & -1 \\ -1 & -2 \end{bmatrix}$, then $A + 2A^T + 3A^T$ is
0
1
2
3

Step-by-Step Solution

Key Concept: Solve the system of matrix equations by combining them to isolate $A$, then verify by substitution.
From the four equations, we can set up a system. Using $2A - X + Y^T = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$, we get $2A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} + X - Y^T$. Computing: $X - Y^T = \begin{bmatrix} 3 & 2 \\ 2 & 3 \end{bmatrix} - \begin{bmatrix} -2 & 1 \\ -1 & -2 \end{bmatrix} = \begin{bmatrix} 5 & 1 \\ 3 & 5 \end{bmatrix}$. Therefore $2A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} + \begin{bmatrix} 5 & 1 \\ 3 & 5 \end{bmatrix} = \begin{bmatrix} 6 & 2 \\ 4 & 6 \end{bmatrix}$, but direct evaluation gives $A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$.
Correct Answer: 1

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