Math  /  Algebra

Question[14762862]undefinedR1+R2R2[1470420]\left[\begin{array}{cc:c} 1 & -4 & 7 \\ -6 & 28 & -62 \end{array}\right] \xrightarrow{\square \cdot R_{1}+R_{2} \rightarrow R_{2}}\left[\begin{array}{cc:c} 1 & -4 & 7 \\ 0 & 4 & -20 \end{array}\right]
Step 3:

Studdy Solution
Verify =6\square = 6 with the other equations:
Second element: 6(4)+28=24+28=46 \cdot (-4) + 28 = -24 + 28 = 4
Third element: 67+(62)=4262=206 \cdot 7 + (-62) = 42 - 62 = -20
Both verifications are correct, confirming =6\square = 6.
The value of \square is:
6 \boxed{6}

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