Question

Consider the 4 × 4 matrix
$M = (\begin{array}{cccccccccc} 0 & 1 & 2 & 3 \\ 1 & 0 & 1 & 2 \\ 2 & 1 & 0 & 1 \\ 3 & 2 & 1 & 0 \end{array}) .$
If $a_{i, j}$ denotes the ${(i, j)}^{t h}$ entry of $M^{- 1}$ , then $a_{4, 1}$ equals _____________ (rounded
off to two decimal places).

Answer: $\underset{_____________}{0.17}$

Solution :-

We know that $M^{- 1} = \frac{a d j (M)}{det (M)} .$
So, the entry a_4,1 in M^-1 is equal to $\frac{a_{4, 1}}{det (M)}$ , where a_4,1 is the entry of adj(M).

And, entry a_4,1 in adj(M) = cofactor of entry a_1,4 in M.

Hence, $a_{4, 1} = \frac{cofactor of a_{1, 4} in M}{det (M)} .$

Now, cofactor of entry a_1,4 in M = ${(- 1)}^{1 + 4} | \begin{array}{cccccccccc} 1 & 0 & 1 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{array} | = - 2.$

\begin{array}{llllllllll} And, det (M) & = & | \begin{array}{cccccccccc} 0 & 1 & 2 & 3 \\ 1 & 0 & 1 & 2 \\ 2 & 1 & 0 & 1 \\ 3 & 2 & 1 & 0 \end{array} | \\ = & - 1 \cdot | \begin{array}{cccccccccc} 1 & 1 & 2 \\ 2 & 0 & 1 \\ 3 & 1 & 0 \end{array} | - 2 \cdot | \begin{array}{cccccccccc} 1 & 0 & 2 \\ 2 & 1 & 1 \\ 3 & 2 & 0 \end{array} | - 3 \cdot | \begin{array}{cccccccccc} 1 & 0 & 1 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{array} | \\ = & - 1 \cdot (6) - 2 \cdot (0) - 3 \cdot (2) \\ = & - 12. \end{array}

∴ a_{4, 1} = \frac{cofactor of a_{1, 4} in M}{det (M)} = \frac{- 2}{- 12} = \frac{1}{6} \approx 0.16666

Rounded off to two decimal places of 0.16666 = 0.17

So, the correct answer is $\to$ $\underset{_____________}{0.17}$

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Question

Consider the 4 × 4 matrix M=(0123101221013210). If ai,j denotes the (i,j)t⁢h entry of M−1 , then a4,1 equals _____________ (rounded off to two decimal places).

Solution :-

Consider the 4 × 4 matrix
$M = (\begin{array}{cccccccccc} 0 & 1 & 2 & 3 \\ 1 & 0 & 1 & 2 \\ 2 & 1 & 0 & 1 \\ 3 & 2 & 1 & 0 \end{array}) .$
If $a_{i, j}$ denotes the ${(i, j)}^{t h}$ entry of $M^{- 1}$ , then $a_{4, 1}$ equals _____________ (rounded
off to two decimal places).