2019-04-01T10:38:28-04:00
3. A necessary and sufficient condition for a matrix (square) A to be invertible is that
a.A is not equal zero
b.|A|≠0
c.|A|>0
d.A<0
4. Given that x + 2y + 3z =1, 3x + 2y + z = 4, x + 3y + 2z = 0. What is x,y and z?
a.(7/4,-3/4,1/4)
b.(5/4,-2/4,1/4)
c.(1,-3/4,1/4)
d.(1,-2,3)
1
2019-04-09T10:39:20-0400
3b - it is a known theorem
4a
x = ∣ 1 2 3 4 2 1 0 3 2 ∣ ∣ 1 2 3 3 2 1 1 3 2 ∣ = 4 + 36 − 16 − 3 4 + 2 + 27 − 6 − 12 − 3 = 7 4 x=\frac{\begin{vmatrix}
1 & 2 & 3 \\
4 & 2 & 1\\
0&3&2
\end{vmatrix}}{\begin{vmatrix}
1 & 2 & 3 \\
3 & 2 & 1\\
1&3&2
\end{vmatrix}}=\frac{4+36-16-3}{4+2+27-6-12-3}=\frac{7}{4} x = ∣ ∣ 1 3 1 2 2 3 3 1 2 ∣ ∣ ∣ ∣ 1 4 0 2 2 3 3 1 2 ∣ ∣ = 4 + 2 + 27 − 6 − 12 − 3 4 + 36 − 16 − 3 = 4 7
y = ∣ 1 1 3 3 4 1 1 0 2 ∣ ∣ 1 2 3 3 2 1 1 3 2 ∣ = 8 + 1 − 12 − 6 4 + 2 + 27 − 6 − 12 − 3 = − 3 4 y=\frac{\begin{vmatrix}
1 & 1 & 3 \\
3 & 4 & 1\\
1&0&2
\end{vmatrix}}{\begin{vmatrix}
1 & 2 & 3 \\
3 & 2 & 1\\
1&3&2
\end{vmatrix}}=\frac{8+1-12-6}{4+2+27-6-12-3}=-\frac{3}{4} y = ∣ ∣ 1 3 1 2 2 3 3 1 2 ∣ ∣ ∣ ∣ 1 3 1 1 4 0 3 1 2 ∣ ∣ = 4 + 2 + 27 − 6 − 12 − 3 8 + 1 − 12 − 6 = − 4 3
z = ∣ 1 2 1 3 2 4 1 3 0 ∣ ∣ 1 2 3 3 2 1 1 3 2 ∣ = 8 + 9 − 2 − 12 4 + 2 + 27 − 6 − 12 − 3 = 1 4 z=\frac{\begin{vmatrix}
1 & 2 & 1 \\
3 & 2 & 4\\
1&3&0
\end{vmatrix}}{\begin{vmatrix}
1 & 2 & 3 \\
3 & 2 & 1\\
1&3&2
\end{vmatrix}}=\frac{8+9-2-12}{4+2+27-6-12-3}=\frac{1}{4} z = ∣ ∣ 1 3 1 2 2 3 3 1 2 ∣ ∣ ∣ ∣ 1 3 1 2 2 3 1 4 0 ∣ ∣ = 4 + 2 + 27 − 6 − 12 − 3 8 + 9 − 2 − 12 = 4 1
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