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ICE Princess25 [194]
2 years ago
11

------»Kindly Don't espam!!⟵(o_O)

itle=" \: \: \: " alt=" \: \: \: " align="absmiddle" class="latex-formula">
​

Mathematics
2 answers:
muminat2 years ago
7 0

Answer:

No, this matrix is singular

Step-by-step explanation:

-1*-2*3=6

2*1*-2=-4

1*2*-1=-2

So, 6+(-4)+(-2)=0       -SINGULAR-

Inessa05 [86]2 years ago
4 0

Answer:

Step-by-step explanation:

If determinant ≠ 0, then it is a non-singular matrix

<h2>Determinant =  a₁(b₂c₃ -b₃c₂) - b₁(a₂c₃-a₃c₂)+c₁(a₂b₃-a₃b₂)</h2><h2>= 1(3-4) - 2(6 - 2) + (-1)(-4+ 1)</h2><h2>= 1*(-1) - 2*4 + (-1)*(-1)</h2><h2>= -1  - 8 + 1</h2><h2>= -8 </h2><h2>So, it is a non singular matrix</h2>
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