The element in the first row and third column of the inverse of the matrix $⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & - 3 0 & 1 & 2 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦$ is.

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Solution

The correct option is D $7$
Let $A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & - 3 0 & 1 & 2 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦$
$| A | = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 2 & - 3 0 & 1 & 2 0 & 0 & 1 \end{matrix} ∣ ∣ ∣ ∣$
$= ∣ ∣ ∣ \begin{matrix} 1 & 2 0 & 1 \end{matrix} ∣ ∣ ∣ = 1$
and $a d j (A) = ⎡ ⎢ ⎣ \begin{matrix} 1 & - 2 & 7 0 & 1 & - 2 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦$
Hence, $A^{- 1} = \frac{1}{| A |} a d j (A)$
$= ⎡ ⎢ ⎣ \begin{matrix} 1 & - 2 & 7 0 & 1 & - 2 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦$
So, required element
$A_{13} = 7$

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The element in the first row and third column of the inverse of the matrix ⎡⎢⎣12−3012001⎤⎥⎦ is.

The correct option is D 7Let A=⎡⎢⎣12−3012001⎤⎥⎦|A|=∣∣ ∣∣12−3012001∣∣ ∣∣=∣∣∣1201∣∣∣=1and adj(A)=⎡⎢⎣1−2701−2001⎤⎥⎦Hence, A−1=1|A|adj(A)=⎡⎢⎣1−2701−2001⎤⎥⎦So, required elementA13=7

The element in the first row and third column of the inverse of the matrix $⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & - 3 0 & 1 & 2 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦$ is.