First slide

Applications of determinant

Question

If $A = [\begin{array}{lll} 0 1 2 \\ 1 2 3 \\ 3 x 1 \end{array}]$ and $A^{- 1} = [\begin{array}{ccc} 1 / 2 & - 1 / 2 & 1 / 2 \\ - 4 & 3 & y \\ 5 / 2 & - 3 / 2 & 1 / 2 \end{array}]$ , then

Moderate

A

$x = 1, y = - 1$
B

$x = - 1, y = 1$
C

$x = 2, y = - 1 / 2$
D

$x = 1 / 2, y = 1 / 2$

Solution

We have
$\begin{array}{l} [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] = A A^{- 1} = [\begin{array}{lll} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & x & 1 \end{array}] [\begin{array}{ccc} 1 / 2 & - 1 / 2 & 1 / 2 \\ - 4 & 3 & y \\ 5 / 2 & - 3 / 2 & 1 / 2 \end{array}] \\ = [\begin{array}{ccc} 1 & 0 & y + 1 \\ 0 & 1 & 2 (y + 1) \\ 4 (1 - x) & 3 (x - 1) & 2 + x y \end{array}] \end{array}$
$\begin{array}{l} \Rightarrow 1 - x = 0, x - 1 = 0, \\ y + 1 = 0, y + 1 = 0, 2 + x y = 1 \\ ∴ x = 1, y = - 1 . \end{array}$

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