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Explain Solution for RD Sharma Maths Class 12 Chapter 6 Adjoint and inverse of a Matrix Exercise very short answer type Question 27 maths textbook solution

Answers (1)

VSQ : 27

Answer : $A^{-1}=\frac{1}{19}A$

Given : $A.=\left[\begin{array}{cc} 2 & 3 \\ 5 & -2 \end{array}\right]$

Hint : to find $A^{-1^{}}$ ,find adj A & $\left | A \right |$

Solution : $A.=\left[\begin{array}{cc} 2 & 3 \\ 5 & -2 \end{array}\right]$

$\left | A \right |=-4-15$

$=-19$

$\left | A \right |\neq 0$

So, $A^{-1}$ exists

$adjA=\left[\begin{array}{cc} -2 & -5 \\ -3 & 2 \end{array}\right]^{T}$

$=\left[\begin{array}{cc} -2 & -3 \\ -5 & -2 \end{array}\right]$

Now $A^{-1}=\frac{adjA}{\left | A \right |}$

$A^{-1}=-\frac{1}{19}$ $\left[\begin{array}{cc} -2 & -3 \\ -5 & -2 \end{array}\right]$

$A^{-1}=\frac{1}{19}\left[\begin{array}{cc} 2 & 3 \\ 5 & -2 \end{array}\right]$

$A^{-1}=\frac{1}{19}A$

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