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 objective is to solve the matrix equation \(AX = B\). This is achieved by performing the matrix multiplication on the left side and then equating the elements of the resulting matrix with the elements of the matrix on the right side, thereby forming a system of linear equations.

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

Step 3: Execution:
Given the equation \(AX = B\), substitute the matrices:
\[ \begin{bmatrix} 3 & 7
4 & -2 \end{bmatrix} \begin{bmatrix} \alpha
-2 \end{bmatrix} = \begin{bmatrix} 7
32 \end{bmatrix} \]
Conduct 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 the corresponding elements yields a system of two linear equations:
1) \(3\alpha - 14 = 7\)
2) \(4\alpha + 4 = 32\)
Solve the first equation for \(\alpha\):
\[ 3\alpha = 7 + 14 \]
\[ 3\alpha = 21 \]
\[ \alpha = \frac{21}{3} = 7 \]
Verify with the second equation:
\[ 4(7) + 4 = 28 + 4 = 32 \]
The value \(\alpha = 7\) is consistent with both equations.

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