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halljamya77 halljamya77
  • 06-11-2026
  • Mathematics
Answered

$\left[\begin{array}{ccc}

1 & 3 & 1 \\

-2 & 1 & 0

\end{array}\right] \times \left[\begin{array}{cc}

2 & -2 \\

3 & 5 \\

4 & 1

\end{array}\right]$


A$\left[\begin{array}{cc}12 & 7 \\ 7 & 10\end{array}\right]$


B$\left[\begin{array}{cc}11 & 6 \\ 7 & 9\end{array}\right]$


C$\left[\begin{array}{cc}15 & 14 \\ -1 & 9\end{array}\right]$


D$\left[\begin{array}{cc}15 & 18 \\ 7 & 1\end{array}\right]$

Verified Answer :

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