First slide

Applications of determinant

Question

The matrix A satisfying $A [\begin{array}{ll} 1 5 \\ 0 1 \end{array}] = [\begin{array}{cc} 3 & - 1 \\ 6 & 0 \end{array}]$ is

Easy

A

$[\begin{array}{cc} 3 & 2 \\ 6 & - 3 \end{array}]$
B

$[\begin{array}{cc} 3 & - 16 \\ 6 & - 30 \end{array}]$
C

$[\begin{array}{cc} 3 & - 16 \\ 6 & 30 \end{array}]$
D

$[\begin{array}{cc} 3 & - 3 \\ 6 & 2 \end{array}]$

Solution

We know that if $A C = B$ , then $A = B C^{- 1}$ ,
$\begin{array}{l} ∴ A = [\begin{array}{cc} 3 & - 1 \\ 6 & 0 \end{array}] {[\begin{array}{ll} 1 & 5 \\ 0 & 1 \end{array}]}^{- 1} = [\begin{array}{cc} 3 & - 1 \\ 6 & 0 \end{array}] [\begin{array}{cc} 1 & - 5 \\ 0 & 1 \end{array}] \\ = [\begin{array}{cc} 3 & - 16 \\ 6 & - 30 \end{array}] \end{array}$

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