Matrices

[sharp idol]

Hii, I don't understand how to find the variables

[woven crow]

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[random obsidian]

do you know how a scalar multiplies with h a matrix?

[sharp idol]

um no..

[random obsidian]

i'm assuming you do know, then we simply just get the following:
$$\begin{bmatrix}2a&6a\a&4a\end{bmatrix}=\begin{bmatrix}x&27\y&z\end{bmatrix}$$

[cursive otter]

wolfqz

[random obsidian]

oh, do you not understand what i did in the above step?

[sharp idol]

you added a variable in the first matrix? just a guess

[random obsidian]

there was an a outside the matrix

i just uhh multiplied them

[sharp idol]

ohh

[random obsidian]

from here we can find the value of a, which gives the values of x,y,z

because when two matrices are equal, then each of their correspondinng terms are equal so 6a=27

and ofc z=4a and x=2a

[sharp idol]

ohh i think i understand now

lemme try solving it