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: Understanding the Concept:
The problem involves solving a matrix equation \(AX = B\). This can be solved by performing the matrix multiplication on the left-hand side and then equating the corresponding elements of the resulting matrix with the matrix on the right-hand side. This will yield a system of linear equations.
Step 2: Key Formula or Approach:
The multiplication of a 2x2 matrix with a 2x1 matrix is 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: Detailed Explanation:
We are given the equation \(AX = B\). Substituting the given 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:
\[ \begin{bmatrix} (3)(\alpha) + (7)(-2)
(4)(\alpha) + (-2)(-2) \end{bmatrix} = \begin{bmatrix} 7
32 \end{bmatrix} \] \[ \begin{bmatrix} 3\alpha - 14
4\alpha + 4 \end{bmatrix} = \begin{bmatrix} 7
32 \end{bmatrix} \] By equating the corresponding elements, we get a system of two linear equations:
1) \(3\alpha - 14 = 7\)
2) \(4\alpha + 4 = 32\)
Let's solve the first equation for \(\alpha\):
\[ 3\alpha = 7 + 14 \] \[ 3\alpha = 21 \] \[ \alpha = \frac{21}{3} = 7 \] We can verify this result using the second equation:
\[ 4(7) + 4 = 28 + 4 = 32 \] The value \(\alpha = 7\) satisfies both equations.
Step 4: Final Answer:
The value of \(\alpha\) is 7.