Exercise 1.8.71 True or False? If you say the statement may be false, you must give an explicit example - with numbers! If you say the statement is always true, you must give clear explanation.

Matrix $A$ imetolewa kama:

$$A = \begin{bmatrix} 4 & 3 & 0 \\ 6 & 5 & 2 \\ 9 & 7 & 3 \end{bmatrix}$$

Tutatumia upanuzi wa cofactor kwenye safu ya kwanza:
\begin{align*} \det(A) &= 4 \cdot \det \begin{bmatrix} 5 & 2 \ 7 & 3 \end{bmatrix} - 3 \cdot \det \begin{bmatrix} 6 & 2 \ 9 & 3 \end{bmatrix} + 0 \cdot \det \begin{bmatrix} 6 & 5 \ 9 & 7 \end{bmatrix} \ &= 4((5)(3) - (2)(7)) - 3((6)(3) - (2)(9)) + 0 \ &= 4(15 - 14) - 3(18 - 18) + 0 \ &= 4(1) - 3(0) + 0 \ &= 4 \end{align*}
Kigezo cha $A$ ni $\boxed{\text{4}}$.

Matrix $B$ imetolewa kama:

$$B = \begin{bmatrix} 6 & 0 & 0 & 5 \\ 1 & 7 & 2 & -5 \\ 2 & 0 & 0 & 0 \\ 8 & 3 & 1 & 8 \end{bmatrix}$$

Tutatumia upanuzi wa cofactor kwenye safu ya tatu kwa sababu ina namba nyingi za sifuri:
\begin{align*} \det(B) &= 2 \cdot (-1)^{3+1} \det \begin{bmatrix} 0 & 0 & 5 \ 7 & 2 & -5 \ 3 & 1 & 8 \end{bmatrix} + 0 + 0 + 0 \ &= 2 \cdot \det \begin{bmatrix} 0 & 0 & 5 \ 7 & 2 & -5 \ 3 & 1 & 8 \end{bmatrix} \end{align*}
Sasa, kokotoa kigezo cha matrix ndogo ya 3x3 kwa kutumia upanuzi wa cofactor kwenye safu ya kwanza:
\begin{align*} \det \begin{bmatrix} 0 & 0 & 5 \ 7 & 2 & -5 \ 3 & 1 & 8 \end{bmatrix} &= 0 - 0 + 5 \cdot (-1)^{1+3} \det \begin{bmatrix} 7 & 2 \ 3 & 1 \end{bmatrix} \ &= 5((7)(1) - (2)(3)) \ &= 5(7 - 6) \ &= 5(1) \ &= 5 \end{align*}
Kwa hiyo, kigezo cha $B$ ni:

$$\det(B) = 2 \cdot 5 = 10$$

Kigezo cha $B$ ni $\boxed{\text{10}}$.

Matrix $C$ imetolewa kama:

$$C = \begin{bmatrix} 4 & 0 & 0 & 0 & 0 \\ 7 & -1 & 0 & 0 & 0 \\ 2 & 6 & 3 & 0 & 0 \\ 5 & -8 & -4 & -3 & 0 \\ 5 & 0 & 0 & 0 & 2 \end{bmatrix}$$

Tutatumia upanuzi wa cofactor kwenye safu ya kwanza:
\begin{align*} \det(C) &= 4 \cdot (-1)^{1+1} \det \begin{bmatrix} -1 & 0 & 0 & 0 \ 6 & 3 & 0 & 0 \ -8 & -4 & -3 & 0 \ 0 & 0 & 0 & 2 \end{bmatrix} + 0 + 0 + 0 + 0 \ &= 4 \cdot \det \begin{bmatrix} -1 & 0 & 0 & 0 \ 6 & 3 & 0 & 0 \ -8 & -4 & -3 & 0 \ 0 & 0 & 0 & 2 \end{bmatrix} \end{align*}
Sasa, kokotoa kigezo cha matrix ndogo ya 4x4 kwa kutumia upanuzi wa cofactor kwenye safu ya kwanza:
\begin{align*} \det \begin{bmatrix} -1 & 0 & 0 & 0 \ 6 & 3 & 0 & 0 \ -8 & -4 & -3 & 0 \ 0 & 0 & 0 & 2 \end{bmatrix} &= -1 \cdot (-1)^{1+1} \det \begin{bmatrix} 3 & 0 & 0 \ -4 & -3 & 0 \ 0 & 0 & 2 \end{bmatrix} + 0 + 0 + 0 \ &= -1 \cdot \det \begin{bmatrix} 3 & 0 & 0 \ -4 & -3 & 0 \ 0 & 0 & 2 \end{bmatrix} \end{align*}
Kisha, kokotoa kigezo cha matrix ndogo ya 3x3 kwa kutumia upanuzi wa cofactor kwenye safu ya kwanza:
\begin{align*} \det \begin{bmatrix} 3 & 0 & 0 \ -4 & -3 & 0 \ 0 & 0 & 2 \end{bmatrix} &= 3 \cdot (-1)^{1+1} \det \begin{bmatrix} -3 & 0 \ 0 & 2 \end{bmatrix} + 0 + 0 \ &= 3((-3)(2) - (0)(0)) \ &= 3(-6 - 0) \ &= 3(-6) \ &= -18 \end{align*}
Kwa hiyo, kigezo cha $C$ ni:

$$\det(C) = 4 \cdot (-1) \cdot (-18) = 4 \cdot 18 = 72$$

Kigezo cha $C$ ni $\boxed{\text{72}}$.