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Home  >>  CBSE XII  >>  Math  >>  Determinants
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True or False: $(aA)^{-1}= \Large{ \frac{1}{a}}\normalsize A^{-1}$, where $a$ is a any real number and $A$ is a square matrix.

$\begin{array}{1 1} True \\ False \end{array}$

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Toolbox:
  • $A^{-1}=\frac{1}{|A|}(adj A)$
Let us consider $A=\begin{bmatrix}4 & 2\\3 & 2\end{bmatrix}$
aA=$\begin{bmatrix}4a & 2a\\3 a& 2a\end{bmatrix}$
To determine $(aA)^{-1}$ let us find |aA|.
$|aA|=8a^2-6a^2=2a^2$
adj (aA) can be obtained by interchanging the elements of $a_{11}$ and $a_{22}$ and changing the symbols of $a_{12}$ and $a_{21}$
Therefore $adj(aA)=\begin{bmatrix}2a & -2a\\-3a & 4a\end{bmatrix}$
$(aA)^{-1}=\frac{1}{2a^2}\begin{bmatrix}2a & -2a\\-3a & 4a\end{bmatrix}$
Taking a as the common factor
$(aA)^{-1}=\frac{a}{2a^2}\begin{bmatrix}2 & -2\\-3 & 4\end{bmatrix}$
$\quad=\frac{1}{2a}\begin{bmatrix}2 & -2\\3 & 4\end{bmatrix}=\frac{1}{a}\bigg(\frac{1}{2}\begin{bmatrix}2 & -2\\3 & 4\end{bmatrix}\bigg)$
Now $A^{-1}=\frac{1}{2}\begin{bmatrix}2 &-2\\3 & 4\end{bmatrix}$
Therefore $(aA)^{-1}=\frac{1}{a}A^{-1}$
Hence the statement is True.
answered Mar 27, 2013 by sreemathi.v
 

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