[Latihan UTBK – Matriks]

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Diketahui matriks B = (-1 1 2 1)
dan matriks A berukuran 2 × 2 yang mempunyai invers. Jika 2AB + B² = (-1 -4 8 5), maka nilai dari det (2A^-1) adalah….​

[Latihan UTBK - Matriks]

[Latihan UTBK – Matriks]

begin{array}{rcl}text{B}&=&begin{pmatrix}-1&2\1&1end{pmatrix}\~\text{B}^2&=&begin{pmatrix}-1&2\1&1end{pmatrix}begin{pmatrix}-1&2\1&1end{pmatrix}\~\&=&begin{pmatrix}(-1times -1)+(2times 1)&(-1times 2)+(2times 1)\(1times -1)+(1times 1)&(1times 2)+(1times 1)end{pmatrix}\~\&=&begin{pmatrix}3&0\0&3end{pmatrix}end{array}

Sehingga :

begin{array}{rcl}2text{A}begin{pmatrix}-1&2\1&1end{pmatrix}+begin{pmatrix}3&0\0&3end{pmatrix}&=&begin{pmatrix}-1&8\-4&5end{pmatrix}\~\2text{A}begin{pmatrix}-1&2\1&1end{pmatrix}&=&begin{pmatrix}-1&8\-4&5end{pmatrix}-begin{pmatrix}3&0\0&3end{pmatrix}\~\2text{A}begin{pmatrix}-1&2\1&1end{pmatrix}&=&begin{pmatrix}-1-3&8-0\-4-0&5-3end{pmatrix}\~\2text{A}begin{pmatrix}-1&2\1&1end{pmatrix}&=&begin{pmatrix}-4&8\-4&2end{pmatrix}\~\2text{A}&=&begin{pmatrix}-4&8\-4&2end{pmatrix}begin{pmatrix}-1&2\1&1end{pmatrix}^{-1}\~\2text{A}&=&begin{pmatrix}-4&8\-4&2end{pmatrix}frac{1}{(-1times 1)-(2times 1)}begin{pmatrix}1&-2\-1&-1end{pmatrix}\~\2text{A}&=&begin{pmatrix}-4&8\-4&2end{pmatrix}begin{pmatrix}-frac{1}{2}&frac{2}{3}\frac{1}{3}&frac{1}{3}end{pmatrix}\~\2text{A}&=&begin{pmatrix}left(-4times -frac{1}{2}right)+left(8times frac{1}{3}right)&left(-4times frac{2}{3}right)+left(8times frac{1}{3}right)\left(-4times -frac{1}{2}right)+left(2times frac{1}{3}right)&left(-4times frac{2}{3}right)+left(2times frac{1}{3}right)end{pmatrix}\~\2text{A}&=&begin{pmatrix}frac{14}{3}&0\-frac{4}{3}&-2end{pmatrix}\~\text{A}&=&frac{1}{2}begin{pmatrix}frac{14}{3}&0\-frac{4}{3}&-2end{pmatrix}\~\text{A}&=&begin{pmatrix}frac{7}{3}&0\-frac{2}{3}&-1end{pmatrix}\~\text{A}^{-1}&=&frac{1}{left(frac{7}{3}times -1right)-left(0times -frac{2}{3}right)}begin{pmatrix}-1&0\frac{2}{3}&frac{7}{3}end{pmatrix}\~\text{A}^{-1}&=&-frac{3}{7}begin{pmatrix}-1&0\frac{2}{3}&frac{7}{3}end{pmatrix}\~\text{A}^{-1}&=&begin{pmatrix}frac{3}{7}&0\-frac{2}{7}&-1end{pmatrix}\2text{A}^{-1}&=&2begin{pmatrix}frac{3}{7}&0\-frac{2}{7}&-1end{pmatrix}\2text{A}^{-1}&=&begin{pmatrix}frac{6}{7}&0\-frac{4}{7}&-2end{pmatrix}\~\det~left(2text{A}^{-1}right)&=&left(frac{6}{7}times -2right)-left(0times -frac{4}{7}right)end{array}

huge det~left(2text{A}^{-1}right)=-frac{12}{7}