To verify that the given matrices [tex]\( A \)[/tex] do not have an inverse, we need to check if the determinant of each matrix is zero. A matrix does not have an inverse if and only if its determinant is zero.
### Exercise 9
For matrix [tex]\( A_9 \)[/tex]:
[tex]\[ A_9 = \begin{bmatrix} 0 & 0 & 0 \\ 1 & 2 & 1 \\ 3 & 2 & 1 \end{bmatrix} \][/tex]
The determinant of a [tex]\( 3 \times 3 \)[/tex] matrix [tex]\( A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \)[/tex] is given by:
[tex]\[ \text{det}(A) = a(ei − fh) − b(di − fg) + c(dh − eg) \][/tex]
For [tex]\( A_9 \)[/tex]:
[tex]\[ a = 0, b = 0, c = 0 \][/tex]
[tex]\[ d = 1, e = 2, f = 1 \][/tex]
[tex]\[ g = 3, h = 2, i = 1 \][/tex]
Let's compute the determinant:
[tex]\[
\text{det}(A_9) = 0(2 \cdot 1 − 1 \cdot 2) − 0(1 \cdot 1 − 1 \cdot 3) + 0(1 \cdot 2 − 2 \cdot 3) = 0
\][/tex]
Since the determinant of [tex]\( A_9 \)[/tex] is [tex]\( 0 \)[/tex], matrix [tex]\( A_9 \)[/tex] does not have an inverse.
### Exercise 10
For matrix [tex]\( A_10 \)[/tex]:
[tex]\[ A_{10} = \begin{bmatrix} 0 & 4 & 2 \\ 0 & 1 & 7 \\ 0 & 3 & 9 \end{bmatrix} \][/tex]
Similarly, let's compute the determinant for [tex]\( A_{10} \)[/tex]:
[tex]\[ a = 0, b = 4, c = 2 \][/tex]
[tex]\[ d = 0, e = 1, f = 7 \][/tex]
[tex]\[ g = 0, h = 3, i = 9 \][/tex]
[tex]\[
\text{det}(A_{10}) = 0(1 \cdot 9 − 7 \cdot 3) − 4(0 \cdot 9 − 7 \cdot 0) + 2(0 \cdot 3 − 1 \cdot 0) = 0
\][/tex]
Since the determinant of [tex]\( A_{10} \)[/tex] is [tex]\( 0 \)[/tex], matrix [tex]\( A_{10} \)[/tex] does not have an inverse.
### Conclusion
For both given matrices [tex]\( A_9 \)[/tex] and [tex]\( A_{10} \)[/tex], the determinants are zero ([tex]\( \det(A_9) = 0 \)[/tex] and [tex]\( \det(A_{10}) = 0 \)[/tex]). Therefore, neither [tex]\( A_9 \)[/tex] nor [tex]\( A_{10} \)[/tex] have an inverse.
