This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

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4
Hapa kuna hesabu za vigezo (determinants) kwa kutumia upanuzi wa cofactor:
Step 1: Kokotoa kigezo cha matrix Matrix imetolewa kama: 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 ni .
Step 2: Kokotoa kigezo cha matrix Matrix imetolewa kama: 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 ni: Kigezo cha ni .
Step 3: Kokotoa kigezo cha matrix Matrix imetolewa kama: 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 ni: Kigezo cha ni .
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Hapa kuna hesabu za vigezo (determinants) kwa kutumia upanuzi wa cofactor: Step 1: Kokotoa kigezo cha matrix A Matrix A imetolewa kama: A = 4 & 3 & 0 \\ 6 & 5 & 2 \\ 9 & 7 & 3 Tutatumia upanuzi wa cofactor kwenye safu ya kwanza: (A) &= 4 · 5 & 2 \\ 7…
This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.