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

24

The correct answer is Option (4) → 24

Given three singular matrices.

Matrix $P=\begin{pmatrix}2 & 3a\\ 4 & 3\end{pmatrix}$

$\det P = 2\cdot 3 - 4\cdot 3a = 6 - 12a = 0$

$6 = 12a$

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

Matrix $Q=\begin{pmatrix}b & 5\\ 2a & 6\end{pmatrix}$

$\det Q = b\cdot 6 - 5\cdot 2a = 6b - 10a = 0$

Substitute $a=\frac{1}{2}$:

$6b - 10\left(\frac{1}{2}\right)=0$

$6b - 5=0$

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

Matrix $R=\begin{pmatrix}a^{2}+b^{2}-c & 1-c\\ c+1 & c\end{pmatrix}$

$\det R = (a^{2}+b^{2}-c)\cdot c - (1-c)(c+1)=0$

Set determinant zero:

$ca^{2}+cb^{2}-1=0$

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

$c=\frac{1}{a^{2}+b^{2}}$

Compute $a^{2}+b^{2}$:

$a=\frac{1}{2}$ gives $a^{2}=\frac{1}{4}$

$b=\frac{5}{6}$ gives $b^{2}=\frac{25}{36}$

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

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

Compute $2a+6b+17c$:

$2a = 1$

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

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

Sum:

$1 + 5 + 18 = 24$

Final answer: $24$