Question

If $3A+4B^{T}=\begin{bmatrix}7&-10& 17\\ 0& 6& 31\end{bmatrix}$ and $2B-3A^{T}=\begin{bmatrix}-1& 18\\ 4&-6\\-5&-7\end{bmatrix}$ then $B=$

• A. $\begin{bmatrix}1& 3\\ -1& 0\\ -2&-4\end{bmatrix}$
• B. $\begin{bmatrix}1& 3\\ 1& 0\\ 2& 4\end{bmatrix}$
• C. $\begin{bmatrix}1& 3\\ -1& 0\\ 2& 4\end{bmatrix}$
• D. $\begin{bmatrix}-1&-3\\ 1& 0\\ 2& 4\end{bmatrix}$

Hint:

When solving matrix equations with transposes, transpose one equation to align the terms (A vs $A^{T}$) for easy elimination.

Answer 1

The Correct Option is C

Step 1: Concept
Use the properties of matrix transposes, specifically $(B^{T})^{T} = B$ and $(A+B)^{T} = A^{T} + B^{T}$.

Step 2: Meaning
We are given two linear equations involving matrices $A$ and $B$. To find $B$, we need to eliminate $A$. Taking the transpose of the second equation $2B-3A^{T}$ gives $2B^{T}-3A$.

Step 3: Analysis
From the first equation: $3A + 4B^{T} = M_{1}$. From the transposed second equation: $-3A + 2B^{T} = M_{2}^{T}$. Adding these two equations eliminates $A$, leaving $6B^{T} = M_{1} + M_{2}^{T}$.

Step 4: Conclusion
Calculating $M_{1} + M_{2}^{T}$ and dividing by 6 yields $B^{T}$, then transposing that result gives $B = \begin{bmatrix}1& 3\\ -1& 0\\ 2& 4\end{bmatrix}$. Final Answer: (C)