Question

Find AC when
$A \: = \: \begin{gathered}\left[\begin{array}{ccc}2 \: \: \: \: \: \: 1 \\ 0 \: - 2\end{array}\right]\end{gathered}$
$C\:=\:\begin{gathered}\left[\begin{array}{ccc} - 3 \: \: 2\\ - 1 \: \: \: 4\end{array}\right]$

Answer is

$\begin{gathered}\left[\begin{array}{ccc} \: - 7 \: \: \: \: \: 8 \\ 2 \: \: - 8\end{array}\right]$

Show step by step​

Answer 1

$\large\underline{\sf{Given- }}$

$\rm \: A = \begin{gathered}\sf\left[\begin{array}{cc}2&1\\0& - 2\end{array}\right]\end{gathered}$

and

$\rm \: C = \begin{gathered}\sf\left[\begin{array}{cc} - 3&2\\ - 1& 4\end{array}\right]\end{gathered}$

$\large\underline{\sf{To\:Find - }}$

$\rm \: AC$

$\large\underline{\sf{Solution-}}$

Given that,

$\rm \: A = \begin{gathered}\sf\left[\begin{array}{cc}2&1\\0& - 2\end{array}\right]\end{gathered}$

and

$\rm \: C = \begin{gathered}\sf\left[\begin{array}{cc} - 3&2\\ - 1& 4\end{array}\right]\end{gathered}$

Now, Consider

$\rm \: AC$

$\rm \: = \: \begin{gathered}\sf\left[\begin{array}{cc}2&1\\0& - 2\end{array}\right]\end{gathered} \times \begin{gathered}\sf\left[\begin{array}{cc} - 3&2\\ - 1& 4\end{array}\right]\end{gathered}$

$\rm \: = \: \begin{gathered}\sf\left[\begin{array}{cc}2 \times ( - 3) + 1 \times ( - 1)&2 \times 2 + 1 \times 4\\0 \times ( - 3) - 2 \times ( - 1)&0 \times 2 - 2 \times 4\end{array}\right]\end{gathered}$

$\rm \: = \: \begin{gathered}\sf\left[\begin{array}{cc} - 6 -1 &4 + 4\\ 0 + 2& 0 - 8\end{array}\right]\end{gathered}$

$\rm \: = \: \begin{gathered}\sf\left[\begin{array}{cc} - 7&8\\2& - 8\end{array}\right]\end{gathered}$

Hence,

$\rm\implies \: \: \boxed{\tt{ \: AC = \: \begin{gathered}\sf\left[\begin{array}{cc} - 7&8\\2& - 8\end{array}\right]\end{gathered}}}$

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ADDITIONAL INFORMATION

1. Matrix multiplication is defined when number of columns of pre multiplier is equal to number of rows of post multiplier.

2. Matrix multiplication may or may not be Commutative.

3. Matrix multiplication is Associative, i.e A(BC) = (AB)C

4. Matrix multiplication is Distributive, i.e A(B + C) = AB + AC

5. There exist a identity matrix I such that AI = IA = A

Answer 2

Multiply - Matrix

The matrix multiplication can be done only if the number of columns of the first matrix is equal to the number of rows of the second matrix.

Given matrices are,

$\longrightarrow A = \left[\begin{array}{cc}2 & 1 \\ 0 & - 2\end{array}\right]$

$\longrightarrow C = \left[\begin{array}{cc} -3 & 2\\ - 1 & 4\end{array}\right]$

We need to find out the value of AC, it means we have to multiply both the matrices.

Solution:

Multiply each row in the first matrix by each column in the second matrix, then simplify each element of the matrix by multiplying out all the expressions.

$AC = A \times C$

$\implies \left[\begin{array}{cc}2 & 1 \\ 0 & - 2\end{array}\right] \left[\begin{array}{cc} -3 & 2\\ - 1 & 4\end{array}\right]$

$\implies \left[\begin{array}{cc} 2(3) + 1(-1) & 2(2) + 1(4) \\ 0(-3) - 2(-1) & 0(2) - 2(4)\end{array}\right]$

$\implies \left[\begin{array}{cc} - 6 -1 &4 + 4\\ 0 + 2& 0 - 8\end{array}\right]$

$\implies \left[\begin{array}{cc} -7 & 8 \\ 2 & -8\end{array}\right]$

Therefore the required answer is:

$\boxed{AC = \left[\begin{array}{cc} -7 & 8 \\ 2 & -8\end{array}\right]}$