Question

If the transpose of matrix A is matrix B, where \(A=\begin{bmatrix} 2a & 1 & 3b \\ 1 & 2 & 4c \\ 5 & 6 & 0 \end{bmatrix}\)and \(B=\begin{bmatrix} 4 & 1 & 5 \\ 1 & 2 & 6 \\ 9 & 3 & 0 \end{bmatrix}\)then the value of 3a + 2b + 4c is :

• A. \(\frac{23}{4}\)
• B. 9
• C. 15
• D. 17

Answer 1

The Correct Option is C

The problem states that the transpose of matrix A is matrix B, where:

\(A=\begin{bmatrix} 2a & 1 & 3b \\ 1 & 2 & 4c \\ 5 & 6 & 0 \end{bmatrix}\)

and

\(B=\begin{bmatrix} 4 & 1 & 5 \\ 1 & 2 & 6 \\ 9 & 3 & 0 \end{bmatrix}\)

Since B is the transpose of A, we have:

• \(2a = 4\), \(1 = 1\), \(3b = 9\)
• \(1 = 1\), \(2 = 2\), \(4c = 3\)
• \(5 = 5\), \(6 = 6\), \(0 = 0\)

Finding the unknowns:

• From \(2a = 4\), \(a = \frac{4}{2} = 2\)
• From \(3b = 9\), \(b = \frac{9}{3} = 3\)
• From \(4c = 3\), \(c = \frac{3}{4}\)

Now, calculate \(3a + 2b + 4c\):

• \(3a = 3 \times 2 = 6\)
• \(2b = 2 \times 3 = 6\)
• \(4c = 4 \times \frac{3}{4} = 3\)
• Total: \(6 + 6 + 3 = 15\)

Thus, the value of \(3a + 2b + 4c\) is: 15