Q84 / 180 · COMEDK 2024 (Evening Shift) · Mathematics
$\text { If } 3 A+4 B^t=\left(\begin{array}{ccc}
7 & -10 & 17 \\
0 & 6 & 31
\end{array}\right) \text { and } 2 B-3 A^t=\left(\begin{array}{cc}
-1 & 18 \\
4 & -6 \\
-5 & -7
\end{array}\right) \text { then }(5 B)^t=$
- A$\left(\begin{array}{ccc} 5 & 5 & 10 \\ 15 & 0 & 20 \end{array}\right)$
- Correct answerB$\left(\begin{array}{ccc} 5 & -5 & 10 \\ 15 & 0 & 20 \end{array}\right)$
- C$\left(\begin{array}{ccc} 5 & -5 & -10 \\ 15 & 0 & -20 \end{array}\right)$
- D$\left(\begin{array}{ccc} -5 & 5 & 10 \\ -15 & 0 & 20 \end{array}\right)$
Solution
Given equations:
1) $3A + 4B^t = \begin{pmatrix} 7 & -10 & 17 \\ 0 & 6 & 31 \end{pmatrix}$
2) $2B - 3A^t = \begin{pmatrix} -1 & 18 \\ 4 & -6 \\ -5 & -7 \end{pmatrix}$
Taking the transpose of equation (2):
$(2B - 3A^t)^t = \begin{pmatrix} -1 & 18 \\ 4 & -6 \\ -5 & -7 \end{pmatrix}^t$
$2B^t - 3A = \begin{pmatrix} -1 & 4 & -5 \\ 18 & -6 & -7 \end{pmatrix}$
Now we have a system of equations in $A$ and $B^t$:
i) $3A + 4B^t = \begin{pmatrix} 7 & -10 & 17 \\ 0 & 6 & 31 \end{pmatrix}$
ii) $-3A + 2B^t = \begin{pmatrix} -1 & 4 & -5 \\ 18 & -6 & -7 \end{pmatrix}$
Adding (i) and (ii):
$6B^t = \begin{pmatrix} 7-1 & -10+4 & 17-5 \\ 0+18 & 6-6 & 31-7 \end{pmatrix} = \begin{pmatrix} 6 & -6 & 12 \\ 18 & 0 & 24 \end{pmatrix}$
$B^t = \begin{pmatrix} 1 & -1 & 2 \\ 3 & 0 & 4 \end{pmatrix}$
We need $(5B)^t = 5B^t$:
$5B^t = 5 \begin{pmatrix} 1 & -1 & 2 \\ 3 & 0 & 4 \end{pmatrix} = \begin{pmatrix} 5 & -5 & 10 \\ 15 & 0 & 20 \end{pmatrix}$
Answer: $\left(\begin{array}{ccc} 5 & -5 & 10 \\ 15 & 0 & 20 \end{array}\right)$
1) $3A + 4B^t = \begin{pmatrix} 7 & -10 & 17 \\ 0 & 6 & 31 \end{pmatrix}$
2) $2B - 3A^t = \begin{pmatrix} -1 & 18 \\ 4 & -6 \\ -5 & -7 \end{pmatrix}$
Taking the transpose of equation (2):
$(2B - 3A^t)^t = \begin{pmatrix} -1 & 18 \\ 4 & -6 \\ -5 & -7 \end{pmatrix}^t$
$2B^t - 3A = \begin{pmatrix} -1 & 4 & -5 \\ 18 & -6 & -7 \end{pmatrix}$
Now we have a system of equations in $A$ and $B^t$:
i) $3A + 4B^t = \begin{pmatrix} 7 & -10 & 17 \\ 0 & 6 & 31 \end{pmatrix}$
ii) $-3A + 2B^t = \begin{pmatrix} -1 & 4 & -5 \\ 18 & -6 & -7 \end{pmatrix}$
Adding (i) and (ii):
$6B^t = \begin{pmatrix} 7-1 & -10+4 & 17-5 \\ 0+18 & 6-6 & 31-7 \end{pmatrix} = \begin{pmatrix} 6 & -6 & 12 \\ 18 & 0 & 24 \end{pmatrix}$
$B^t = \begin{pmatrix} 1 & -1 & 2 \\ 3 & 0 & 4 \end{pmatrix}$
We need $(5B)^t = 5B^t$:
$5B^t = 5 \begin{pmatrix} 1 & -1 & 2 \\ 3 & 0 & 4 \end{pmatrix} = \begin{pmatrix} 5 & -5 & 10 \\ 15 & 0 & 20 \end{pmatrix}$
Answer: $\left(\begin{array}{ccc} 5 & -5 & 10 \\ 15 & 0 & 20 \end{array}\right)$
Hence, the correct answer is option: 'B'.