Question

If P, Q and R are three singular matrices given by \(P = \begin{bmatrix} 2 & 3a \\ 4 & 3 \end{bmatrix}\), \(Q = \begin{bmatrix} b & 5 \\ 2a & 6 \end{bmatrix}\) and \(R = \begin{bmatrix} a^2 + b^2 - c & 1-c \\ c+1 & c \end{bmatrix}\), then the value of \((2a + 6b + 17c)\) is

• A. 30
• B. 18
• C. 34
• D. 24

Hint:

The most important piece of information for this question is the term "singular matrix". Always remember that the determinant of a singular matrix is zero. This is the key to setting up the equations needed to solve for the variables.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
A matrix is singular if its determinant is equal to zero. We are given three singular matrices P, Q, and R. We will use this property to find the values of the variables a, b, and c.
Step 2: Key Formula or Approach:
For a 2x2 matrix \(M = \begin{bmatrix} p & q
r & s \end{bmatrix}\), the determinant is given by \(|M| = ps - qr\).
If M is singular, then \(|M| = 0\), which implies \(ps - qr = 0\).
Step 3: Detailed Explanation:
For matrix P:
Given \(P = \begin{bmatrix} 2 & 3a
4 & 3 \end{bmatrix}\) is singular, so \(|P| = 0\).
\[ (2)(3) - (3a)(4) = 0 \] \[ 6 - 12a = 0 \] \[ 12a = 6 \] \[ a = \frac{6}{12} = \frac{1}{2} \] For matrix Q:
Given \(Q = \begin{bmatrix} b & 5
2a & 6 \end{bmatrix}\) is singular, so \(|Q| = 0\).
\[ (b)(6) - (5)(2a) = 0 \] \[ 6b - 10a = 0 \] Substitute the value of \(a = 1/2\):
\[ 6b - 10\left(\frac{1}{2}\right) = 0 \] \[ 6b - 5 = 0 \] \[ 6b = 5 \] \[ b = \frac{5}{6} \] For matrix R:
Given \(R = \begin{bmatrix} a^2 + b^2 - c & 1-c
c+1 & c \end{bmatrix}\) is singular, so \(|R| = 0\).
\[ (a^2 + b^2 - c)(c) - (1-c)(c+1) = 0 \] \[ c(a^2 + b^2) - c^2 - (1 - c^2) = 0 \] \[ c(a^2 + b^2) - c^2 - 1 + c^2 = 0 \] \[ c(a^2 + b^2) - 1 = 0 \] \[ c(a^2 + b^2) = 1 \] Substitute the values of a and b:
\[ a^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \] \[ b^2 = \left(\frac{5}{6}\right)^2 = \frac{25}{36} \] \[ c\left(\frac{1}{4} + \frac{25}{36}\right) = 1 \] \[ c\left(\frac{9}{36} + \frac{25}{36}\right) = 1 \] \[ c\left(\frac{34}{36}\right) = 1 \] \[ c\left(\frac{17}{18}\right) = 1 \] \[ c = \frac{18}{17} \] Step 4: Final Answer:
We need to find the value of \((2a + 6b + 17c)\).
\[ 2a = 2\left(\frac{1}{2}\right) = 1 \] \[ 6b = 6\left(\frac{5}{6}\right) = 5 \] \[ 17c = 17\left(\frac{18}{17}\right) = 18 \] \[ 2a + 6b + 17c = 1 + 5 + 18 = 24 \] The value is 24.