Question
if $A=\left[\begin{array}{ll}2 & 1 \\ 1 & 3\end{array}\right]$ and $B=\left[\begin{array}{c}3 \\ -11\end{array}\right]$ Find the matrix $X$ such that $AX = B$
✓
Answer
Let the order of the matrix $X$ be a $x b.$
$AX = B$
$\left[\begin{array}{ll}2 & 1 \\ 1 & 3\end{array}\right]_{2 \times 2} \times X_{a \times b}=\left[\begin{array}{c}3 \\ -11\end{array}\right]_{2 \times 1}$
Clearly the order of the matrix $X$ is $2 \times 1$
Let $X=[(x),(y)]$
$\left[\begin{array}{ll}2 & 1 \\ 1 & 3\end{array}\right] \times\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}3 \\ -11\end{array}\right]$
Comparing the two matrices we get
$2x = y = 3 ...(1)$
$x = 3y = -11 ...(2)$
Multiplying $(1)$ with $3$ we get
$6x + 3y = 9 ...(3)$
Substracting $(2)$ from $(3)$ we get
$5x = 20$
$x = 4$
From $(1)$ we have
$y = 3 - 2x = 3 - 8 = -5$
$\therefore X=\left[\begin{array}{c}4 \\ -5\end{array}\right]$
$AX = B$
$\left[\begin{array}{ll}2 & 1 \\ 1 & 3\end{array}\right]_{2 \times 2} \times X_{a \times b}=\left[\begin{array}{c}3 \\ -11\end{array}\right]_{2 \times 1}$
Clearly the order of the matrix $X$ is $2 \times 1$
Let $X=[(x),(y)]$
$\left[\begin{array}{ll}2 & 1 \\ 1 & 3\end{array}\right] \times\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}3 \\ -11\end{array}\right]$
Comparing the two matrices we get
$2x = y = 3 ...(1)$
$x = 3y = -11 ...(2)$
Multiplying $(1)$ with $3$ we get
$6x + 3y = 9 ...(3)$
Substracting $(2)$ from $(3)$ we get
$5x = 20$
$x = 4$
From $(1)$ we have
$y = 3 - 2x = 3 - 8 = -5$
$\therefore X=\left[\begin{array}{c}4 \\ -5\end{array}\right]$
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