Matrices P, Q, R and Singularity

The problem provides three matrices, $P$, $Q$, and $R$, and states that they are singular. A singular matrix is a square matrix whose determinant is equal to zero. We will use this property to find the values of the variables $a$, $b$, and $c$ present in the matrices.

Determining Variable 'a' from Matrix P

The first matrix is given as $P = \begin{bmatrix} 2 & 3a \\ 4 & 3 \end{bmatrix}$.

For $P$ to be singular, its determinant must be zero. The determinant of a 2x2 matrix $\begin{bmatrix} p & q \\ r & s \end{bmatrix}$ is calculated as $ps - qr$.

So, the determinant of $P$ is:

$\det(P) = (2 \times 3) - (3a \times 4)$

$\det(P) = 6 - 12a$

Since $P$ is singular, $\det(P) = 0$.

$6 - 12a = 0$

$12a = 6$

$a = \frac{6}{12} = \frac{1}{2}$

Thus, the value of $a$ is $\frac{1}{2}$.

Calculating Variable 'b' using Matrix Q

The second matrix is given as $Q = \begin{bmatrix} b & 5 \\ 2a & 6 \end{bmatrix}$.

For $Q$ to be singular, its determinant must be zero.

The determinant of $Q$ is:

$\det(Q) = (b \times 6) - (5 \times 2a)$

$\det(Q) = 6b - 10a$

Since $Q$ is singular, $\det(Q) = 0$.

$6b - 10a = 0$

We already found that $a = \frac{1}{2}$. Substituting this value:

$6b - 10 \times \frac{1}{2} = 0$

$6b - 5 = 0$

$6b = 5$

$b = \frac{5}{6}$

Thus, the value of $b$ is $\frac{5}{6}$.

Finding Variable 'c' from Matrix R

The third matrix is given as $R = \begin{bmatrix} a^2 + b^2 - c & 1 - c \\ c + 1 & c \end{bmatrix}$.

For $R$ to be singular, its determinant must be zero.

The determinant of $R$ is:

$\det(R) = (a^2 + b^2 - c) \times c - (1 - c) \times (c + 1)$

$\det(R) = (a^2 + b^2)c - c^2 - (1 \times c + 1 \times 1 - c \times c - c \times 1)$

$\det(R) = (a^2 + b^2)c - c^2 - (c + 1 - c^2 - c)$

$\det(R) = (a^2 + b^2)c - c^2 - (1 - c^2)$

$\det(R) = (a^2 + b^2)c - c^2 - 1 + c^2$

$\det(R) = (a^2 + b^2)c - 1$

Since $R$ is singular, $\det(R) = 0$.

$(a^2 + b^2)c - 1 = 0$

$c(a^2 + b^2) = 1$

Now we need to calculate $a^2 + b^2$ using the values $a = \frac{1}{2}$ and $b = \frac{5}{6}$.

$a^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$

$b^2 = \left(\frac{5}{6}\right)^2 = \frac{25}{36}$

$a^2 + b^2 = \frac{1}{4} + \frac{25}{36}$

To add these fractions, we find a common denominator, which is 36.

$a^2 + b^2 = \frac{1 \times 9}{4 \times 9} + \frac{25}{36} = \frac{9}{36} + \frac{25}{36} = \frac{9 + 25}{36} = \frac{34}{36} = \frac{17}{18}$

Substitute this sum back into the equation $c(a^2 + b^2) = 1$.

$c \times \frac{17}{18} = 1$

$c = \frac{18}{17}$

Thus, the value of $c$ is $\frac{18}{17}$.

Evaluating Expression $(2a + 6b + 17c)$

We need to find the value of the expression $(2a + 6b + 17c)$ using the determined values of $a$, $b$, and $c$.

We have $a = \frac{1}{2}$, $b = \frac{5}{6}$, and $c = \frac{18}{17}$.

Let's calculate each term:

• $2a = 2 \times \frac{1}{2} = 1$
• $6b = 6 \times \frac{5}{6} = 5$
• $17c = 17 \times \frac{18}{17} = 18$

Now, add these values together:

$2a + 6b + 17c = 1 + 5 + 18 = 24$

Therefore, the value of the expression $(2a + 6b + 17c)$ is 24.