Question

If \(\begin{bmatrix} a+2&3b+2c\\c+3&7d+6 \end{bmatrix}=\begin{bmatrix} 2&-3\\3c&-8 \end{bmatrix}\), then the values of a,b,c and d are.

• A. a=-2, b=0, c=-3, d=-2
• B. a=0, b=-2, c=-2, d=\(\frac{3}{2}\)
• C. a=-1, b=-3, c=\(\frac{-3}{2}\), d=2
• D. a=0, b=-2, c=\(\frac{3}{2}\), d=-2

Answer 1

The Correct Option is D

To find the values of \(a\), \(b\), \(c\), and \(d\), compare the elements of the matrices: \(\begin{bmatrix} a+2 & 3b+2c \\ c+3 & 7d+6 \end{bmatrix} = \begin{bmatrix} 2 & -3 \\ 3c & -8 \end{bmatrix}\)

Each element of the matrices must be equal:

1. \(a+2 = 2\)

2. \(3b+2c = -3\)

3. \(c+3 = 3c\)

4. \(7d+6 = -8\)

Solving these equations:

From equation 1: \(a+2=2 \rightarrow a=0\)

From equation 3: \(c+3=3c \rightarrow 3c-c=3 \rightarrow 2c=3 \rightarrow c=\frac{3}{2}\)

From equation 4: \(7d+6=-8 \rightarrow 7d=-14 \rightarrow d=-2\)

Substituting \(c = \frac{3}{2}\) into equation 2: \[3b+2\left(\frac{3}{2}\right)=-3\]

\(3b+3=-3 \rightarrow 3b=-6 \rightarrow b=-2\)

Therefore, the values are \(a=0\), \(b=-2\), \(c=\frac{3}{2}\), \(d=-2\).