First slide

Applications of determinant

Question

$If [\begin{array}{ll} 2 1 \\ 3 2 \end{array}] A [\begin{array}{cc} - 3 & 2 \\ 5 & - 3 \end{array}] = [\begin{array}{ll} 1 0 \\ 0 1 \end{array}], then the matrix A =$

Moderate

A

$[\begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix}]$
B

$[\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}]$
C

$[\begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix}]$
D

$[\begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix}]$

Solution

$\begin{array}{l} [\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}] A [\begin{array}{cc} - 3 & 2 \\ 5 & - 3 \end{array}] = [\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}] \\ \Rightarrow A = {[\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}]}^{- 1} [\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}] {[\begin{array}{cc} - 3 & 2 \\ 5 & - 3 \end{array}]}^{- 1} = [\begin{array}{cc} 2 & - 1 \\ - 3 & 2 \end{array}] [\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}] \frac{1}{(9 - 10)} [\begin{array}{ll} - 3 & - 2 \\ - 5 & - 3 \end{array}] \end{array}$
$= [\begin{matrix} 2 & - 1 \\ - 3 & 2 \end{matrix}] [\begin{matrix} 3 & 2 \\ 5 & 3 \end{matrix}] = [\begin{matrix} 6 - 5 & 4 - 3 \\ - 9 + 10 & - 6 + 6 \end{matrix}] = [\begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix}]$

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