Question

If \( \begin{bmatrix} 1 & 3 \\ 4 & 5 \end{bmatrix} \begin{bmatrix} x \\ 2 \end{bmatrix} = \begin{bmatrix} 5 \\ 6 \end{bmatrix} \), then value of \( x \) is :

• A. \( 1 \)
• B. \( 0 \)
• C. \( -1 \)
• D. \( 3 \)

Hint:

In a multiple-choice setting, you can also "plug and play." If you substitute the options into the matrix multiplication, only \( x = -1 \) satisfies both rows. For example, if \( x=1 \), row 1 would give \( 1+6=7 \), which does not equal 5. This saves time in exam conditions.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem presents a linear system in matrix form, typically represented as \( AX = B \), where \( A \) is the coefficient matrix, \( X \) is the vector of variables (here containing \( x \) and a constant 2), and \( B \) is the constant vector. Matrix equations are fundamental in solving simultaneous linear equations efficiently.
In this specific case, the product of a \( 2 \times 2 \) matrix and a \( 2 \times 1 \) vector results in another \( 2 \times 1 \) vector. The operation follows the "row-by-column" rule, where the components of the rows of the first matrix are multiplied by the components of the column of the second matrix and then summed. This results in two separate scalar equations that must both be satisfied for the same value of \( x \).

Step 2: Key Formula or Approach:
The multiplication of a \( 2 \times 2 \) matrix by a column vector is defined as:
\[ \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} = \begin{bmatrix} au + bv \\ cu + dv \end{bmatrix} \]
We will apply this to the left side of the given equation. Once we have the resulting vector, we set it equal to the vector on the right side. This leads to a system of two equations:
1) \( 1(x) + 3(2) = 5 \)
2) \( 4(x) + 5(2) = 6 \)
We only need to solve one of these to find \( x \), but using the second one serves as an excellent verification of our work.

Step 3: Detailed Explanation:
Step 3.1: Expanding the matrix multiplication.
The first row of the matrix is \( [1, 3] \) and the column vector is \( \begin{bmatrix} x \\ 2 \end{bmatrix} \).
Product for the first row: \( (1 \cdot x) + (3 \cdot 2) = x + 6 \).
The second row of the matrix is \( [4, 5] \).
Product for the second row: \( (4 \cdot x) + (5 \cdot 2) = 4x + 10 \).
So, the matrix equation becomes:
\[ \begin{bmatrix} x + 6 \\ 4x + 10 \end{bmatrix} = \begin{bmatrix} 5 \\ 6 \end{bmatrix} \]
Step 3.2: Setting up and solving the equations.
From the first row components:
\[ x + 6 = 5 \]
Subtracting 6 from both sides:
\[ x = 5 - 6 = -1 \]
Step 3.3: Verification with the second row components.
Now substitute \( x = -1 \) into the second equation:
\[ 4(-1) + 10 = -4 + 10 = 6 \]
Since \( 6 = 6 \), the value \( x = -1 \) is consistent across both equations. This confirms that the system is consistent and our calculation is correct.

Step 4: Final Answer:
After performing the matrix multiplication and solving the resulting linear equations, we determined that \( x = -1 \). Checking this value against all constraints confirms its accuracy. Thus, Option (3) is the correct answer.