Let . If then A

Engineering

Mathematics

Multiplication of Matrices

Symmetric and Skew Symmetric Matrices

Question

Let $[\begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array}] [\begin{array}{cc} 1 & 2 \\ 0 & 1 \end{array}] [\begin{array}{cc} 1 & 3 \\ 0 & 1 \end{array}] . [\begin{array}{cc} 1 & n - 1 \\ 0 & 1 \end{array}] = [\begin{array}{cc} 1 & 78 \\ 0 & 1 \end{array}]$ . If $A = [\begin{array}{cc} 1 & n \\ 0 & 1 \end{array}]$ then A^–1 = ?

$[\begin{array}{cc} 1 & 12 \\ 0 & 1 \end{array}]$

$[\begin{array}{cc} 1 & - 13 \\ 0 & 1 \end{array}]$

$[\begin{array}{cc} 1 & - 12 \\ 0 & 1 \end{array}]$

$[\begin{array}{cc} 1 & 0 \\ - 13 & 1 \end{array}]$

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$[\begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array}] [\begin{array}{cc} 1 & 2 \\ 0 & 1 \end{array}] [\begin{array}{cc} 1 & 3 \\ 0 & 1 \end{array}] . [\begin{array}{cc} 1 & n - 1 \\ 0 & 1 \end{array}] = [\begin{array}{cc} 1 & 78 \\ 0 & 1 \end{array}]$
$\Rightarrow \frac{n (n - 1)}{2} = 78 \Rightarrow n = 13$

$A = [\begin{array}{cc} 1 & 13 \\ 0 & 1 \end{array}]$
so $A^{- 1} = [\begin{array}{cc} 1 & - 13 \\ 0 & 1 \end{array}]$ .

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