Need Solution for R.D.Sharma Maths Class 12 Chapter 7 Solution of Simultaneous Linear Equation Exercise Very short answer Question 6 Maths Textbook Solution.

$\rightarrow 2$

Given

$\rightarrow A=\left[\begin{array}{ll} 2 & 4 \\ 4 & 3 \end{array}\right], \quad X=\left[\begin{array}{l} n \\ 1 \end{array}\right], B=\left[\begin{array}{c} 8 \\ 11 \end{array}\right], A X=B$

Explanation

$\rightarrow AX=B$

$\left[\begin{array}{ll} 2 & 4 \\ 4 & 3 \end{array}\right]\left[\begin{array}{l} n \\ 1 \end{array}\right]=\left[\begin{array}{c} 8 \\ 11 \end{array}\right]$

\begin{aligned} &{\left[\begin{array}{c} 2 \times n+4 \times 1 \\ 4 \times n+3 \times 1 \end{array}\right]=\left[\begin{array}{c} 8 \\ 11 \end{array}\right]} \\\\ &{\left[\begin{array}{l} 2 n+4 \\ 4 n+3 \end{array}\right]=\left[\begin{array}{c} 8 \\ 11 \end{array}\right]} \end{aligned}

Comparing both sides we get

\begin{aligned} &2 n+4=8 \\ &2 n=8-4 \\ &2 n=4 \end{aligned}

\begin{aligned} &n=\frac{4}{2} \\ &n=2 \end{aligned}

\begin{aligned} &4 n+3=11 \\ &4 n=11-3 \\ &4 n=8 \end{aligned}

\begin{aligned} &n=8 / 4 \\ &n=2 \end{aligned}

Hence $n=2$