Question

If \(A = \begin{bmatrix} 3 & 7 \\ 4 & -2 \end{bmatrix}\), \(X = \begin{bmatrix} \alpha \\ -2 \end{bmatrix}\), \(B = \begin{bmatrix} 7 \\ 32 \end{bmatrix}\) and \(AX = B\), then the value of the \(\alpha\) is

• A. 7
• B. 4/3
• C. 1
• D. 5

Hint:

When solving matrix equations like \(AX = B\), you can either set up and solve the system of linear equations (as shown here) or find the inverse of A and compute \(X = A^{-1}B\). For simple systems, direct multiplication and comparison is often faster.

Answer 1

The Correct Option is A

Step 1: Conceptualization:
The task is to solve the matrix equation \(AX = B\). This is achieved by performing the matrix multiplication on the left side and then equating the resultant matrix elements with the elements of the matrix on the right side. This process generates a system of linear equations.

Step 2: Method:
The multiplication of a 2x2 matrix by a 2x1 matrix is defined as follows:
\[ \begin{bmatrix} a & b
c & d \end{bmatrix} \begin{bmatrix} x
y \end{bmatrix} = \begin{bmatrix} ax+by
cx+dy \end{bmatrix} \]

Step 3: Execution:
The given equation is \(AX = B\). Substituting the provided matrices:
\[ \begin{bmatrix} 3 & 7
4 & -2 \end{bmatrix} \begin{bmatrix} \alpha
-2 \end{bmatrix} = \begin{bmatrix} 7
32 \end{bmatrix} \]
Perform the matrix multiplication on the left side:
\[ \begin{bmatrix} (3)(\alpha) + (7)(-2)
(4)(\alpha) + (-2)(-2) \end{bmatrix} = \begin{bmatrix} 7
32 \end{bmatrix} \]
\[ \begin{bmatrix} 3\alpha - 14
4\alpha + 4 \end{bmatrix} = \begin{bmatrix} 7
32 \end{bmatrix} \]
Equating corresponding elements yields two linear equations:
1) \(3\alpha - 14 = 7\)
2) \(4\alpha + 4 = 32\)
Solving the first equation for \(\alpha\):
\[ 3\alpha = 7 + 14 \]
\[ 3\alpha = 21 \]
\[ \alpha = \frac{21}{3} = 7 \]
Verification using the second equation:
\[ 4(7) + 4 = 28 + 4 = 32 \]
The value \(\alpha = 7\) satisfies both equations.

Step 4: Conclusion:
The determined value for \(\alpha\) is 7.