Answer to Question #87349 in Linear Algebra for Daniel

Question

⎛ 2 0 1 ⎞
9.Given that A = , ⎜ k 2 3 ⎟ what is the value of k, if A is said to be a singular matrix?
⎝ 2 1 4 ⎠

a.8
b.10
c.7
d.-6





⎛1 2 3⎞
10.let A= ⎜4 5 0⎟ then the cofactor of matrix A is the matrix
⎝2 1 4⎠,

a.

⎛ 20 −16 −6 ⎞
A = ⎜−5 −2 3 ⎟
⎝−13 12 −41 ⎠

b.
⎛ 20 −16 −6 ⎞
A = ⎜−5 −2 3 ⎟
⎝−13 12 −3 ⎠

c.
⎛ 20 −16 −6 ⎞
A = ⎜−5 −2 3 ⎟
⎝−15 12 −3 ⎠

d.
⎛ 20 −12 −6 ⎞
A = ⎜−5 −2 3 ⎟
⎝−13 12 −41 ⎠

Answer 1

9. Solution: The matrix A is said to be a singular, if

$$det(A)=0.$$

Hence 

$$0=det(A)=\begin{vmatrix} 2 & 0 & 1 \\ k & 2 & 3 \\ 2 & 1 & 4 \end{vmatrix} =2\cdot 2\cdot 4 +2 \cdot 0 \cdot 3+k \cdot 1 \cdot 1 -$$ $$-2 \cdot 2 \cdot 1- k \cdot 0 \cdot 4 -2 \cdot 1 \cdot 3=k+6,$$

i.e. $k=-6$.

Answer: d. -6

10. Solution: First find the cofactor of each element of matrix A.

$$A_{11}={(-1)}^{1+1} \begin{vmatrix} 5 & 0 \\ 1 & 4 \end{vmatrix} =20; \ A_{12}={(-1)}^{1+2} \begin{vmatrix} 4 & 0 \\ 2 & 4 \end{vmatrix}=-16;$$ $$A_{13}={(-1)}^{1+3} \begin{vmatrix} 4 & 5 \\ 2 & 1 \end{vmatrix} =-6;$$

$$A_{21}={(-1)}^{2+1} \begin{vmatrix} 2 & 3 \\ 1 & 4 \end{vmatrix} =-5; \ A_{22}={(-1)}^{2+2}\begin{vmatrix} 1 & 3 \\ 2 & 4 \end{vmatrix} =-2;$$

$$A_{23}={(-1)}^{2+3}\begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix} =3;$$

$$A_{31}={(-1)}^{3+1} \begin{vmatrix} 2 & 3 \\ 5 & 0 \end{vmatrix} =-15; \ A_{32}={(-1)}^{3+2} \begin{vmatrix} 1 & 3 \\ 4 & 0 \end{vmatrix} =12;$$ $$A_{33}={(-1)}^{3+3}\begin{vmatrix} 1 & 2 \\ 4 & 5 \end{vmatrix} =-3.$$

Therefore, the cofactor matrix $A^C$ of the matrix $A$ is equal to

$$A^C=\begin{pmatrix} 20 & -16 & -6 \\ -5 & -2 & 3 \\ -15 & 12 & -3 \end{pmatrix}.$$

Answer: c. $\begin{pmatrix} 20 & -16 & -6 \\ -5 & -2 & 3 \\ -15 & 12 & -3 \end{pmatrix}$ .