A=⎛⎜⎝6−32104576⎞⎟⎠ and B=⎛⎜⎝1−9422−3313−10⎞⎟⎠, then what is the value of det(B−1(A−1B−1)−1A−1)? A. −1 C. 1 E. 5 B. 0 D. 3
Solution
Use the following matrix inversing theorem. (A−1B−1)−1=BAA−1⋅A=A⋅A−1=II⋅I=I and note that the determinant of identity matrix I always equals 1. Therefore, we have det(B−1(A−1B−1)−1A−1)=det(B−1(BA)A−1)=det((B−1B)(AA−1))=det(I⋅I)=det(I)=1 Thus, the determinant of B−1(A−1B−1)−1A−1 equals 1 (Answer C)
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Problem Number 2 Let M be a matrix such that M⋅(abcd)=(aba−cb−d). The determinant of matrix M equals ⋯⋅ A. −1 C. 1 E. 3 B. 0 D. 2
Solution
Given that M⋅(abcd)=(aba−cb−d) Therefore, we will have |M|⋅∣∣∣abcd∣∣∣=∣∣∣aba−cb−d∣∣∣|M|⋅(ad−bc)=a(b−d)−b(a−c)|M|⋅(ad−bc)=ab−ad−ab+bc|M|⋅(ad−bc)=−(ad−bc)|M|=−(ad−bc)ad−bc=−1 Thus, the determinant of matrix M equals −1 (Answer A)
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Problem Number 3 If given ∣∣
∣∣abcdefghi∣∣
∣∣=3, then ∣∣
∣∣2a+da4a+2d+g2b+eb4b+2e+h2c+fc4c+2f+i∣∣
∣∣=⋯⋅ A. −3 C. 0 E. 3 B. −2 D. 2
Solution
It is given that ∣∣
∣∣abcdefghi∣∣
∣∣=3 By swapping the first and second row entries, causing the determinant to be negative, we have ∣∣
∣∣defabcghi∣∣
∣∣=−3 Transpose the matrix, so we have ∣∣
∣∣dagebhfci∣∣
∣∣=−3 (Transposing a matrix doesn’t affect the determinant) After that, add correspondingly each entry in the first column to two times of each entry in the second column (it also won’t change the determinant). ∣∣
∣∣2a+dag2b+ebh2c+fci∣∣
∣∣=−3 Finally, add correspondingly each entry in the third column to two times of each entry in the first column (it also won’t change the determinant). ∣∣
∣∣2a+da4a+2d+g2b+eb4b+2e+h2c+fc4c+2f+i∣∣
∣∣=−3 Thus, the value of ∣∣
∣∣2a+da4a+2d+g2b+eb4b+2e+h2c+fc4c+2f+i∣∣
∣∣=−3 (Answer A)
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Problem Number 4 If matrix A=(37−1−2), then A27+A31+A40 equals ⋯⋅ A. (513−4) B. (−714−23) C. (7−1432) D. (1001) E. (−100−1)
Solution
Given A=(37−1−2) Try to spot the pattern by finding the power of matrix A. A2=(37−1−2)(37−1−2)=(27−1−3)A3=A2⋅A=(27−1−3)(37−1−2)=(−100−1)=−(1001)=−Iwith I denotes the identity matrix. After then, A27+A31+A40=A27(I+A4+A13)=(A3)9(I+A3A+(A3)4A)=(−I)9(I−IA+(−I)4A)=−I(I−A+A)=−I=(−100−1) Thus, A27+A31+A40=(−100−1) Note: Any power of a identity matrix always yields a same identity matrix. (Answer E)
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Problem Number 5 If a constant k satisfies the matrix equation (k110)(x−1y−1)=(0k), then the value of x+y=⋯⋅ A. (2+k)(1+k) B. (2−k)(1−k) C. (2+k)(1−k) D. (1−k)(1+k) E. (2−k)(1+k)
Solution
By applying matrix multiplication rule, we have (k110)(x−1y−1)=(0k)(k(x−1)+1(y−1)1(x−1)+0(y−1))=(0k)(k(x−1)+(y−1)x−1)=(0k) Hence, we have the following equation system. {k(x−1)+(y−1)=0(⋯1)x−1=k(⋯2) Substitute equation (2) into equation (1). k(k)+y−1=0⇔y=1−k2 Therefore, we have x+y=(k+1)+(1−k2)=(k+1)+(k+1)(1−k)=(k+1)(1+(1−k))=(k+1)(2−k) So, the value of x+y is (k+1)(2−k) or can be rewritten into (2−k)(1+k) (Answer E)
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Problem Number 6 If x:y=5:4, then the value of x and y that satisfy (2101)⎛⎜⎝xy453025⎞⎟⎠(510)=1.360 is ⋯⋅ A. x=1 and y=45 B. x=45 and y=1 C. x=5 and y=4 D. x=−10 and y=−8 E. x=10 and y=8
Solution
Problem Number 7 It is given that A=(2135) holds a certain relationship to matrix B=(−531−2). If matrix C=(321−5) and matrix D have the same relationship, then the value of C+D=⋯⋅ A. (8338) D. (3883) B. (833−8) E. (−8−3−38) C. (38−83)
Soal Nomor 1 (Soal UN SMA Semuan Jurusan) Diketahui matriks A = ( 3 y 5 − 1 ) , B = ( x 5 − 3 6 ) A = ( 3 y 5 − 1 ) , B = ( x 5 − 3 6 ) , dan C = ( − 3 − 1 y 9 ) C = ( − 3 − 1 y 9 ) . Jika A + B − C = ( 8 5 x − x − 4 ) A + B − C = ( 8 5 x − x − 4 ) , maka nilai x + 2 x y + y x + 2 x y + y adalah ⋯ ⋅ ⋯ ⋅ A. 8 8 C. 18 18 E. 22 22 B. 12 12 D. 20 20 Pembahasan Diketahui: A = ( 3 y 5 − 1 ) B = ( x 5 − 3 6 ) C = ( − 3 − 1 y 9 ) A = ( 3 y 5 − 1 ) B = ( x 5 − 3 6 ) C = ( − 3 − 1 y 9 ) Dengan demikian, A + B − C = ( 8 5 x − x − 4 ) ( 3 y 5 − 1 ) + ( x 5 − 3 6 ) − ( − 3 − 1 y 9 ) = ( 8 5 x − x − 4 ) ( 3 + x − ( − 3 ) y + 5 − ( − 1 ) 5 + ( − 3 ) − y − 1 + 6 − 9 ) = ( 8 5 x − x − 4...
Soal Nomor 1 Seorang ibu ingin mempunyai 2 2 orang anak. Kemungkinan kelahiran anak laki-laki dan perempuan diasumsikan sama. Peluang kedua anaknya perempuan adalah ⋯ ⋅ ⋯ ⋅ A. 1 4 1 4 C. 3 4 3 4 B. 1 2 1 2 D. 1 1 Pembahasan Peluang kelahiran anak laki-laki sama dengan peluang kelahiran anak perempuan, yaitu 1 2 1 2 . Peluang kedua anaknya perempuan ( 2 2 kejadian) adalah 1 2 perempuan × 1 2 perempuan = 1 4 1 2 ⏟ perempuan × 1 2 ⏟ perempuan = 1 4 (Jawaban A) [collapse] Soal Nomor 2 Dalam kantong terdapat tiga bola berwarna merah diberi nomor 1 – 3 1 – 3 , lima bola berwarna kuning diberi nomor 4 – 8 4 – 8 , dan empat bola berwarna hijau diberi nomor 9 – 12 9 – 12 . Tiga bola di...
Soal Nomor 1 Farly mempunyai kelereng merah, biru, dan hijau. Perbandingan antara banyak kelereng merah dan biru adalah 3 : 4 3 : 4 . Jumlah kelereng merah dan hijau adalah 27 27 . Jika dua kali banyak kelereng biru ditambah banyak kelereng hijau sama dengan 37 37 , maka banyak kelereng merah, biru, dan hijau berturut-turut yang dimiliki Farly adalah ⋯ ⋅ ⋯ ⋅ A. 12 , 16 12 , 16 , dan 20 20 B. 12 , 16 12 , 16 , dan 18 18 C. 12 , 16 12 , 16 , dan 15 15 D. 6 , 8 6 , 8 , dan 21 21 E. 6 , 8 6 , 8 , dan 15 15 Pembahasan Misalkan x , y , z x , y , z berturut-turut menyatakan banyaknya kelereng merah, biru, dan hijau. Perbandingan antara banyak kelereng merah ( x ) ( x ) dan biru ( y ) ( y ) adalah 3 : 4 3 : 4 . Secara matematis, ditulis x y = 3 4 ⇔ 4 x − 3 y = 0 x y = 3 4 ⇔ 4 x − 3 y = 0 Jumlah kelereng merah ( x ) ( x ) dan hijau ( z ) ( z ) ...
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