Matrices and Determinants

21 soru

Soru 1Soru →

If the matrix P=(k3121k420)P = \begin{pmatrix} k & 3 & 1 \\ 2 & 1 & k \\ 4 & 2 & 0 \end{pmatrix} is singular, find the non-zero value of kk.

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Cevap: 6

Cevap

The non-zero value of kk is 6.
For matrix P to be singular, its determinant must be 0. Expanding along row 3 yields 4(3k - 1) - 2(k^2 - 2) = 12k - 4 - 2k^2 + 4 = -2k^2 + 12k = 0. Factoring gives -2k(k - 6) = 0, which yields k = 0 or k = 6. The non-zero value is 6.

Adım Adım Çözüm

1
Set the determinant of matrix P to 0
\det(P) = 0
A matrix is singular if and only if its determinant equals zero.
2
Evaluate the 3x3 determinant by expanding along the third row
4 \cdot (3k - 1) - 2 \cdot (k^2 - 2) + 0 = 0
Expanding along the third row takes advantage of the zero entry to simplify computation.
3
Expand and combine like terms
-2k^2 + 12k = 0
12k - 4 - 2k^2 + 4 reduces to -2k^2 + 12k.
4
Factor out common factors and solve for k
-2k(k - 6) = 0 \implies k = 0 \text{ or } k = 6
Applying the zero-product property.
5
Select the required root
k = 6
The question specifies the non-zero value of k.

Anahtar Kavram

Determinant of a 3x3 matrix and singular matrix condition
Tahmini Süre:2m 0s
Soru 2Soru →

If A=(2314)A = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix} and B=(510−2)B = \begin{pmatrix} 5 & 1 \\ 0 & -2 \end{pmatrix}, what is the product matrix ABAB?

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Cevap: (10−45−7)\begin{pmatrix} 10 & -4 \\ 5 & -7 \end{pmatrix}

Cevap

The matrix product ABAB is (10−45−7)\begin{pmatrix} 10 & -4 \\ 5 & -7 \end{pmatrix}.
Matrix multiplication is evaluated using the row-by-column rule. Evaluating each entry: top-left is 2(5)+3(0)=102(5)+3(0)=10, top-right is 2(1)+3(−2)=−42(1)+3(-2)=-4, bottom-left is 1(5)+4(0)=51(5)+4(0)=5, and bottom-right is 1(1)+4(−2)=−71(1)+4(-2)=-7. Thus, the product matrix is (10−45−7)\begin{pmatrix} 10 & -4 \\ 5 & -7 \end{pmatrix}.

Adım Adım Çözüm

1
Calculate the top-left element (Row 1 of A ⋅\cdot Column 1 of B)
2(5)+3(0)=10+0=102(5) + 3(0) = 10 + 0 = 10
Matrix multiplication takes the dot product of the corresponding row of the first matrix and column of the second matrix.
2
Calculate the top-right element (Row 1 of A ⋅\cdot Column 2 of B)
2(1)+3(−2)=2−6=−42(1) + 3(-2) = 2 - 6 = -4
Combine products of elements in row 1 of A and column 2 of B.
3
Calculate the bottom-left element (Row 2 of A ⋅\cdot Column 1 of B)
1(5)+4(0)=5+0=51(5) + 4(0) = 5 + 0 = 5
Combine products of elements in row 2 of A and column 1 of B.
4
Calculate the bottom-right element (Row 2 of A ⋅\cdot Column 2 of B)
1(1)+4(−2)=1−8=−71(1) + 4(-2) = 1 - 8 = -7
Combine products of elements in row 2 of A and column 2 of B.

Anahtar Kavram

Matrix Multiplication (Row-by-Column Rule)
Soru 3Soru →

What is the determinant of the 2×22 \times 2 matrix P=(5234)P = \begin{pmatrix} 5 & 2 \\ 3 & 4 \end{pmatrix}?

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Cevap: 14

Cevap

The determinant of matrix PP is 1414.
For any 2×22 \times 2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix}, the determinant is given by ad−bcad - bc. Substituting a=5a=5, b=2b=2, c=3c=3, and d=4d=4 yields (5×4)−(2×3)=20−6=14(5 \times 4) - (2 \times 3) = 20 - 6 = 14.

Adım Adım Çözüm

1
Apply the determinant formula for a 2×22 \times 2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix}, which is det⁡(P)=ad−bc\det(P) = ad - bc.
\det(P) = (5)(4) - (2)(3)
The determinant of a 2×22 \times 2 matrix is defined as the product of the main diagonal elements minus the product of the off-diagonal elements.
2
Evaluate the arithmetic expression.
20 - 6 = 14
Perform multiplication followed by subtraction to get the final determinant value.

Anahtar Kavram

Determinant of a 2x2 Matrix
Tahmini Süre:45s
Soru 4Soru →

Given the matrices A=(4x−13)A = \begin{pmatrix} 4 & x \\ -1 & 3 \end{pmatrix} and B=(2134)B = \begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix}, if the determinant of the product matrix ABAB is 9595, what is the value of xx?

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Cevap: 7

Cevap

The value of xx is 77.
Using the property det⁡(AB)=det⁡(A)det⁡(B)\det(AB) = \det(A)\det(B), we find det⁡(B)=(2)(4)−(1)(3)=5\det(B) = (2)(4) - (1)(3) = 5 and det⁡(A)=(4)(3)−(x)(−1)=12+x\det(A) = (4)(3) - (x)(-1) = 12 + x. Setting 5(12+x)=955(12 + x) = 95 yields 60+5x=9560 + 5x = 95, which solves to x=7x = 7.

Adım Adım Çözüm

1
Calculate the determinant of matrix BB
det⁡(B)=(2)(4)−(1)(3)=5\det(B) = (2)(4) - (1)(3) = 5
The determinant of a 2×22 \times 2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix} is ad−bcad - bc.
2
Express the determinant of matrix AA in terms of xx
det⁡(A)=(4)(3)−(x)(−1)=12+x\det(A) = (4)(3) - (x)(-1) = 12 + x
Applying the determinant formula to matrix AA gives 12−(−x)=12+x12 - (-x) = 12 + x.
3
Apply the determinant product property det⁡(AB)=det⁡(A)⋅det⁡(B)\det(AB) = \det(A) \cdot \det(B)
5(12+x)=955(12 + x) = 95
The determinant of the product of two square matrices is equal to the product of their individual determinants.
4
Solve the linear equation for xx
60+5x=95  ⟹  5x=35  ⟹  x=760 + 5x = 95 \implies 5x = 35 \implies x = 7
Subtract 6060 from both sides and divide by 55 to isolate xx.

Anahtar Kavram

Determinant of Matrix Product Property
Soru 5Soru →

Given the matrices A=(2143)A = \begin{pmatrix} 2 & 1 \\ 4 & 3 \end{pmatrix} and B=(1−125)B = \begin{pmatrix} 1 & -1 \\ 2 & 5 \end{pmatrix}, find the determinant of the matrix C=2A−BC = 2A - B.

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Cevap: -15

Cevap

The determinant of the matrix C=2A−BC = 2A - B is −15-15.
Scalar multiplication gives 2A=(4286)2A = \begin{pmatrix} 4 & 2 \\ 8 & 6 \end{pmatrix}. Subtracting BB yields C=(3361)C = \begin{pmatrix} 3 & 3 \\ 6 & 1 \end{pmatrix}. Evaluating the determinant gives det⁡(C)=(3)(1)−(3)(6)=3−18=−15\det(C) = (3)(1) - (3)(6) = 3 - 18 = -15.

Adım Adım Çözüm

1
Multiply matrix AA by scalar 22
2A=(4286)2A = \begin{pmatrix} 4 & 2 \\ 8 & 6 \end{pmatrix}
Scalar multiplication requires multiplying each entry of matrix AA by 22.
2
Subtract matrix BB from matrix 2A2A entry-wise to find matrix CC
C=(4−12−(−1)8−26−5)=(3361)C = \begin{pmatrix} 4 - 1 & 2 - (-1) \\ 8 - 2 & 6 - 5 \end{pmatrix} = \begin{pmatrix} 3 & 3 \\ 6 & 1 \end{pmatrix}
Subtract corresponding entries of matrix BB from 2A2A.
3
Calculate the determinant of matrix CC
det⁡(C)=(3)(1)−(3)(6)=3−18=−15\det(C) = (3)(1) - (3)(6) = 3 - 18 = -15
The determinant of a 2×22 \times 2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix} is ad−bcad - bc.

Anahtar Kavram

Matrix Operations and Determinants
Soru 6Soru →

If matrix AA has dimension 2×32 \times 3 and matrix BB has dimension 3×43 \times 4, what is the dimension of the matrix product ABAB?

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Cevap: 2×42 \times 4

Cevap

The dimension of the matrix product ABAB is 2×42 \times 4.
For two matrices to be multiplied, the number of columns in the first matrix must equal the number of rows in the second matrix. When a matrix of size m×nm \times n is multiplied by a matrix of size n×pn \times p, the resulting product matrix has size m×pm \times p. For A2×3A_{2 \times 3} and B3×4B_{3 \times 4}, the outer dimensions give 2×42 \times 4.

Adım Adım Çözüm

1
Identify the dimensions of the given matrices AA and BB
Matrix AA is of order 2×32 \times 3 (m=2,n=3m = 2, n = 3) and Matrix BB is of order 3×43 \times 4 (n=3,p=4n = 3, p = 4).
Matrix multiplication requirement requires the number of columns of the first matrix to match the number of rows of the second matrix.
2
Apply the matrix multiplication dimension rule
The product matrix ABAB has mm rows and pp columns, resulting in dimension 2×42 \times 4.
When multiplying an m×nm \times n matrix by an n×pn \times p matrix, the resulting matrix has dimensions m×pm \times p.

Anahtar Kavram

Matrix Multiplication Dimension Compatibility and Resulting Order
Tahmini Süre:45s
Soru 7Soru →

If the determinant of the matrix M=(x325)M = \begin{pmatrix} x & 3 \\ 2 & 5 \end{pmatrix} is 1414, find the value of xx.

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Cevap: 4

Cevap

The value of xx is 44.
For matrix M=(x325)M = \begin{pmatrix} x & 3 \\ 2 & 5 \end{pmatrix}, the determinant is calculated as (x)(5)−(3)(2)=5x−6(x)(5) - (3)(2) = 5x - 6. Equating this to 1414 gives 5x−6=145x - 6 = 14, which simplifies to 5x=205x = 20, yielding x=4x = 4.

Adım Adım Çözüm

1
Apply the 2×22 \times 2 determinant formula det⁡=ad−bc\det = ad - bc
\det(M) = (x \times 5) - (3 \times 2) = 5x - 6
The determinant of a 2×22 \times 2 matrix is the product of the main diagonal minus the product of the anti-diagonal.
2
Set the determinant equal to the given value 1414
5x - 6 = 14
The problem states that the determinant is equal to 14.
3
Solve the linear equation for xx
5x = 20 \implies x = 4
Adding 6 to both sides gives 5x=205x = 20, and dividing by 5 yields x=4x = 4.

Anahtar Kavram

Determinant of a 2x2 Matrix
Soru 8Soru →

What is the determinant of the matrix P=(5327)P = \begin{pmatrix} 5 & 3 \\ 2 & 7 \end{pmatrix}?

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Cevap: 29

Cevap

29
For a 2×22 \times 2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix}, the determinant is given by ad−bcad - bc. Substituting the values a=5a = 5, b=3b = 3, c=2c = 2, and d=7d = 7 yields (5×7)−(3×2)=35−6=29(5 \times 7) - (3 \times 2) = 35 - 6 = 29.

Adım Adım Çözüm

1
Identify the values of a,b,c,a, b, c, and dd from the given 2×22 \times 2 matrix P=(5327)P = \begin{pmatrix} 5 & 3 \\ 2 & 7 \end{pmatrix}.
a=5,b=3,c=2,d=7a = 5, b = 3, c = 2, d = 7
To set up the values for the 2×22 \times 2 determinant formula.
2
Multiply the elements of the main diagonal and the secondary diagonal.
Main diagonal product: 5×7=355 \times 7 = 35; Secondary diagonal product: 3×2=63 \times 2 = 6.
The determinant of a 2×22 \times 2 matrix is the difference between the main diagonal product and the secondary diagonal product.
3
Subtract the product of the secondary diagonal from the product of the main diagonal.
Determinant = 35−6=2935 - 6 = 29.
Completes the formula det⁡(P)=ad−bc\det(P) = ad - bc.

Anahtar Kavram

Determinant of a 2x2 Matrix
Soru 9Soru →

Given that the matrix A=(12k0k32−11)A = \begin{pmatrix} 1 & 2 & k \\ 0 & k & 3 \\ 2 & -1 & 1 \end{pmatrix} is singular, what is the positive value of kk?

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Cevap: 3

Cevap

The positive value of kk is 33.
A square matrix is singular when its determinant is zero. Expanding det⁡(A)\det(A) along the first column yields 1⋅(k+3)+2⋅(6−k2)=−2k2+k+151 \cdot (k + 3) + 2 \cdot (6 - k^2) = -2k^2 + k + 15. Equating this to zero gives the quadratic equation 2k2−k−15=02k^2 - k - 15 = 0, which factors as (2k+5)(k−3)=0(2k + 5)(k - 3) = 0. The roots are k=−2.5k = -2.5 and k=3k = 3. The positive value is 33.

Adım Adım Çözüm

1
Calculate the determinant of matrix A
\det(A) = -2k^2 + k + 15
Expanding along the first column simplifies the calculation because of the zero entry.
2
Set the determinant to zero for singularity
2k^2 - k - 15 = 0
A matrix is singular if and only if its determinant equals zero.
3
Solve the quadratic equation for k
k = 3 or k = -2.5
Factoring 2k2−k−15=(2k+5)(k−3)=02k^2 - k - 15 = (2k + 5)(k - 3) = 0 yields two roots.
4
Choose the positive solution
k = 3
The question explicitly requires the positive value of kk.

Anahtar Kavram

Singular matrix definition and 3x3 determinant evaluation
Soru 10Soru →

Given that xx is a positive real number and the determinant of the matrix A=(x102x3014)A = \begin{pmatrix} x & 1 & 0 \\ 2 & x & 3 \\ 0 & 1 & 4 \end{pmatrix} is 22, find the value of xx.

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Cevap: 2

Cevap

The positive value of xx is 22.
Expanding the determinant along the first row yields det⁡(A)=x(4x−3)−1(8−0)=4x2−3x−8\det(A) = x(4x - 3) - 1(8 - 0) = 4x^2 - 3x - 8. Equating this to 22 gives 4x2−3x−10=04x^2 - 3x - 10 = 0. Factoring the quadratic gives (4x+5)(x−2)=0(4x + 5)(x - 2) = 0, which yields x=2x = 2 or x=−1.25x = -1.25. Because xx must be positive, the correct value is 22.

Adım Adım Çözüm

1
Expand the 3×33 \times 3 determinant along the first row
\det(A) = x(4x - 3) - 1(8 - 0) + 0 = 4x^2 - 3x - 8
Expanding along the first row leverages the zero entry to simplify calculation of the determinant.
2
Equate the determinant expression to the given determinant value
4x^2 - 3x - 10 = 0
Setting the calculated determinant equal to 22 creates a quadratic equation in terms of xx.
3
Factor the quadratic equation to find the candidate values for xx
(4x + 5)(x - 2) = 0 \implies x = 2 \text{ or } x = -1.25
Factoring determines all algebraic solutions that satisfy the determinant equation.
4
Apply the positivity constraint given in the problem statement
x = 2
The question restricts xx to positive real numbers, discarding the negative root.

Anahtar Kavram

Determinant of a 3x3 matrix and quadratic equations
Tahmini Süre:2m 0s
Soru 11Soru →

What is the determinant of the matrix A=(6−321)A = \begin{pmatrix} 6 & -3 \\ 2 & 1 \end{pmatrix}?

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Cevap: 12

Cevap

12
For any 2×22 \times 2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix}, the determinant is given by ad−bcad - bc. Substituting a=6a = 6, b=−3b = -3, c=2c = 2, and d=1d = 1 yields (6)(1)−(−3)(2)=6−(−6)=12(6)(1) - (-3)(2) = 6 - (-6) = 12.

Adım Adım Çözüm

1
Identify the elements of the 2x2 matrix A=(abcd)A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}
a=6a = 6, b=−3b = -3, c=2c = 2, d=1d = 1
Standard entries for determinant formula det⁡(A)=ad−bc\det(A) = ad - bc.
2
Compute the product of the main diagonal elements (adad) and off-diagonal elements (bcbc)
ad=(6)(1)=6ad = (6)(1) = 6 and bc=(−3)(2)=−6bc = (-3)(2) = -6
These form the components of the 2x2 determinant formula.
3
Subtract bcbc from adad
det⁡(A)=6−(−6)=6+6=12\det(A) = 6 - (-6) = 6 + 6 = 12
Subtracting a negative quantity is equivalent to addition.

Anahtar Kavram

Determinant of a 2x2 Matrix
Tahmini Süre:45s
Soru 12Soru →

Given the 3×33 \times 3 matrix M=(k213k−12141)M = \begin{pmatrix} k & 2 & 1 \\ 3 & k-1 & 2 \\ 1 & 4 & 1 \end{pmatrix}, where k>5k > 5. If det⁡(M2)=100\det(M^2) = 100 and det⁡(M)<0\det(M) < 0, what is the value of kk?

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Cevap: 7

Cevap

7
Using the matrix determinant power identity det⁡(M2)=(det⁡(M))2=100\det(M^2) = (\det(M))^2 = 100 and the condition det⁡(M)<0\det(M) < 0, we find det⁡(M)=−10\det(M) = -10. Expanding det⁡(M)\det(M) along the first row yields det⁡(M)=k(k−9)−2(1)+1(13−k)=k2−10k+11\det(M) = k(k-9) - 2(1) + 1(13-k) = k^2 - 10k + 11. Setting this equal to −10-10 gives the quadratic equation k2−10k+21=0k^2 - 10k + 21 = 0, which factors as (k−3)(k−7)=0(k-3)(k-7) = 0. Given k>5k > 5, the unique solution is k=7k = 7.

Adım Adım Çözüm

1
Apply determinant properties for matrix powers
det⁡(M)=−10\det(M) = -10
Because det⁡(M2)=(det⁡(M))2=100\det(M^2) = (\det(M))^2 = 100 and it is given that det⁡(M)<0\det(M) < 0, taking the negative square root gives det⁡(M)=−10\det(M) = -10.
2
Evaluate the determinant of matrix MM using first-row expansion
det⁡(M)=k2−10k+11\det(M) = k^2 - 10k + 11
Expanding along row 1 gives k((k−1)(1)−8)−2(3(1)−2(1))+1(3(4)−(k−1)(1))=k2−9k−2+13−k=k2−10k+11k((k-1)(1) - 8) - 2(3(1) - 2(1)) + 1(3(4) - (k-1)(1)) = k^2 - 9k - 2 + 13 - k = k^2 - 10k + 11.
3
Set up and simplify the quadratic equation for kk
k2−10k+21=0k^2 - 10k + 21 = 0
Equating k2−10k+11k^2 - 10k + 11 to −10-10 yields k2−10k+21=0k^2 - 10k + 21 = 0.
4
Solve the quadratic equation and enforce the inequality condition k>5k > 5
k=7k = 7
Factoring (k−3)(k−7)=0(k-3)(k-7) = 0 gives k=3k = 3 or k=7k = 7. Applying the restriction k>5k > 5 selects k=7k = 7.

Anahtar Kavram

Determinant Properties of Matrix Powers and 3x3 Matrix Expansion
Soru 13Soru →

Given the matrices P=(2314)P = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix} and Q=(50−21)Q = \begin{pmatrix} 5 & 0 \\ -2 & 1 \end{pmatrix}, what is the product matrix PQPQ?

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Cevap: (43−34)\begin{pmatrix} 4 & 3 \\ -3 & 4 \end{pmatrix}

Cevap

The product matrix PQPQ is (43−34)\begin{pmatrix} 4 & 3 \\ -3 & 4 \end{pmatrix}.
To find PQPQ, multiply each row of matrix PP by each column of matrix QQ. The entry in row 1, column 1 is 2(5)+3(−2)=42(5) + 3(-2) = 4. The entry in row 1, column 2 is 2(0)+3(1)=32(0) + 3(1) = 3. The entry in row 2, column 1 is 1(5)+4(−2)=−31(5) + 4(-2) = -3. The entry in row 2, column 2 is 1(0)+4(1)=41(0) + 4(1) = 4. Putting these together yields (43−34)\begin{pmatrix} 4 & 3 \\ -3 & 4 \end{pmatrix}.

Adım Adım Çözüm

1
Compute the entries of the top row of PQPQ using the first row of PP and both columns of QQ.
First entry: (2)(5)+(3)(−2)=10−6=4(2)(5) + (3)(-2) = 10 - 6 = 4. Second entry: (2)(0)+(3)(1)=0+3=3(2)(0) + (3)(1) = 0 + 3 = 3.
Matrix entry (1,1)(1,1) is the dot product of Row 1 of PP and Column 1 of QQ; entry (1,2)(1,2) is the dot product of Row 1 of PP and Column 2 of QQ.
2
Compute the entries of the bottom row of PQPQ using the second row of PP and both columns of QQ.
Third entry: (1)(5)+(4)(−2)=5−8=−3(1)(5) + (4)(-2) = 5 - 8 = -3. Fourth entry: (1)(0)+(4)(1)=0+4=4(1)(0) + (4)(1) = 0 + 4 = 4.
Matrix entry (2,1)(2,1) is the dot product of Row 2 of PP and Column 1 of QQ; entry (2,2)(2,2) is the dot product of Row 2 of PP and Column 2 of QQ.
3
Assemble the computed entries into a 2×22 \times 2 matrix.
PQ=(43−34)PQ = \begin{pmatrix} 4 & 3 \\ -3 & 4 \end{pmatrix}.
Combining row results gives the final product matrix.

Anahtar Kavram

Matrix Multiplication (Row-by-Column Dot Product)
Tahmini Süre:1m 15s
Soru 14Soru →

Given the matrices A=(x312)A = \begin{pmatrix} x & 3 \\ 1 & 2 \end{pmatrix} and B=(2134)B = \begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix}, what is the value of xx such that the determinant of the product matrix ABAB is 2525?

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Cevap: 4

Cevap

The value of xx is 4.
The determinant of a product of matrices equals the product of their determinants, det⁡(AB)=det⁡(A)det⁡(B)\det(AB) = \det(A)\det(B). Since det⁡(B)=5\det(B) = 5 and det⁡(A)=2x−3\det(A) = 2x - 3, setting 5(2x−3)=255(2x - 3) = 25 yields 2x−3=52x - 3 = 5, giving x=4x = 4.

Adım Adım Çözüm

1
Calculate the determinant of matrix B
det⁡(B)=(2)(4)−(1)(3)=8−3=5\det(B) = (2)(4) - (1)(3) = 8 - 3 = 5
The determinant of a 2x2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix} is given by ad−bcad - bc.
2
Express the determinant of matrix A in terms of x
det⁡(A)=(x)(2)−(3)(1)=2x−3\det(A) = (x)(2) - (3)(1) = 2x - 3
Apply the 2x2 determinant formula to matrix A.
3
Use the determinant product rule det(AB) = det(A) * det(B)
det⁡(AB)=(2x−3)⋅5=10x−15\det(AB) = (2x - 3) \cdot 5 = 10x - 15
The determinant of the product of two square matrices equals the product of their individual determinants.
4
Set det(AB) equal to 25 and solve for x
10x−15=25  ⟹  10x=40  ⟹  x=410x - 15 = 25 \implies 10x = 40 \implies x = 4
Equate the determinant expression to the given value 25 to isolate x.

Anahtar Kavram

Determinant of a Matrix Product
Soru 15Soru →

Given two matrices A=(x223)A = \begin{pmatrix} x & 2 \\ 2 & 3 \end{pmatrix} and B=(x1−11)B = \begin{pmatrix} x & 1 \\ -1 & 1 \end{pmatrix}, where x>0x > 0. If the determinant of the product matrix ABAB is 2020, find the value of xx.

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Cevap: 3

Cevap

The positive value of xx is 3.
Using the property det⁡(AB)=det⁡(A)⋅det⁡(B)\det(AB) = \det(A) \cdot \det(B), we compute det⁡(A)=3x−4\det(A) = 3x - 4 and det⁡(B)=x+1\det(B) = x + 1. Equating their product to 20 gives (3x−4)(x+1)=20(3x - 4)(x + 1) = 20, which simplifies to the quadratic 3x2−x−24=03x^2 - x - 24 = 0. Factorizing yields (3x+8)(x−3)=0(3x + 8)(x - 3) = 0. Since xx must be a positive number, the correct value is x=3x = 3.

Adım Adım Çözüm

1
Calculate the determinant of matrix A
\det(A) = 3x - 4
The determinant of a 2x2 matrix \begin{pmatrix} a & b \\ c & d \end{pmatrix} is ad - bc.
2
Calculate the determinant of matrix B
\det(B) = x + 1
Applying ad - bc gives (x)(1) - (1)(-1) = x + 1.
3
Use the determinant product property det(AB) = det(A) * det(B)
(3x - 4)(x + 1) = 20
The determinant of the product of two square matrices equals the product of their individual determinants.
4
Form and solve the quadratic equation
3x^2 - x - 24 = 0, which factorizes into (3x + 8)(x - 3) = 0
Expanding (3x - 4)(x + 1) gives 3x^2 - x - 4. Subtracting 20 yields 3x^2 - x - 24 = 0.
5
Determine the positive solution for x
x = 3
Solving the factors gives x = -8/3 or x = 3. Since x must be positive (x > 0), x = 3.

Anahtar Kavram

Determinant of Matrix Product and 2x2 Determinants
Tahmini Süre:2m 0s
Soru 16Soru →

Given the matrices A=(4213)A = \begin{pmatrix} 4 & 2 \\ 1 & 3 \end{pmatrix} and B=(1−12k)B = \begin{pmatrix} 1 & -1 \\ 2 & k \end{pmatrix}, find the value of kk such that the determinant of the product matrix ABAB is equal to 50.

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Cevap: 3

Cevap

The value of kk is 3.
By applying the determinant product rule det⁡(AB)=det⁡(A)det⁡(B)\det(AB) = \det(A)\det(B), we find det⁡(A)=(4)(3)−(2)(1)=10\det(A) = (4)(3) - (2)(1) = 10. Given det⁡(AB)=50\det(AB) = 50, it follows that det⁡(B)=50/10=5\det(B) = 50 / 10 = 5. Since det⁡(B)=(1)(k)−(−1)(2)=k+2\det(B) = (1)(k) - (-1)(2) = k + 2, setting k+2=5k + 2 = 5 gives k=3k = 3.

Adım Adım Çözüm

1
Calculate the determinant of matrix A
\det(A) = (4 \times 3) - (2 \times 1) = 12 - 2 = 10
The determinant of a 2x2 matrix \begin{pmatrix} a & b \\ c & d \end{pmatrix} is ad - bc.
2
Apply the product property of determinants
\det(B) = \frac{\det(AB)}{\det(A)} = \frac{50}{10} = 5
For any two square matrices of the same dimension, \det(AB) = \det(A) \cdot \det(B).
3
Express the determinant of matrix B in terms of k and solve
\det(B) = (1)(k) - (-1)(2) = k + 2 = 5 \implies k = 3
Equating the calculated determinant formula for B to its numerical value of 5.

Anahtar Kavram

Determinant Product Property and 2x2 Matrix Determinant
Tahmini Süre:1m 30s
Soru 17Soru →

Which of the following is the inverse of the matrix M=(5273)M = \begin{pmatrix} 5 & 2 \\ 7 & 3 \end{pmatrix}?

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Cevap: (3−2−75)\begin{pmatrix} 3 & -2 \\ -7 & 5 \end{pmatrix}

Cevap

(3−2−75)\begin{pmatrix} 3 & -2 \\ -7 & 5 \end{pmatrix}
For any non-singular 2x2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix}, the inverse is 1ad−bc(d−b−ca)\frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}. Here, ad−bc=(5)(3)−(2)(7)=1ad - bc = (5)(3) - (2)(7) = 1, so the inverse matrix is (3−2−75)\begin{pmatrix} 3 & -2 \\ -7 & 5 \end{pmatrix}.

Adım Adım Çözüm

1
Calculate the determinant of matrix M
\det(M) = (5)(3) - (2)(7) = 15 - 14 = 1
For a 2x2 matrix M=(abcd)M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}, the determinant is given by det⁡(M)=ad−bc\det(M) = ad - bc.
2
Find the adjugate matrix of M
\text{adj}(M) = \begin{pmatrix} 3 & -2 \\ -7 & 5 \end{pmatrix}
The adjugate of a 2x2 matrix is obtained by swapping the main diagonal elements (aa and dd) and negating the off-diagonal elements (bb and cc).
3
Apply the matrix inverse formula
M^{-1} = \frac{1}{\det(M)} \text{adj}(M) = \frac{1}{1}\begin{pmatrix} 3 & -2 \\ -7 & 5 \end{pmatrix} = \begin{pmatrix} 3 & -2 \\ -7 & 5 \end{pmatrix}
Multiplying the adjugate matrix by the reciprocal of the determinant gives the exact inverse matrix.

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Inverse of a 2x2 Matrix
Soru 18Soru →

Given the matrix A=(3142)A = \begin{pmatrix} 3 & 1 \\ 4 & 2 \end{pmatrix}, what is the determinant of the matrix 3A3A?

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Cevap: 1818

Cevap

The determinant of the matrix 3A3A is 1818.
For any 2×22 \times 2 matrix, scaling the matrix by a factor kk scales its determinant by k2k^2. Since det⁡(A)=(3)(2)−(1)(4)=2\det(A) = (3)(2) - (1)(4) = 2, multiplying the matrix by 33 yields det⁡(3A)=32×2=9×2=18\det(3A) = 3^2 \times 2 = 9 \times 2 = 18. Alternatively, computing 3A=(93126)3A = \begin{pmatrix} 9 & 3 \\ 12 & 6 \end{pmatrix} directly gives det⁡(3A)=(9)(6)−(3)(12)=54−36=18\det(3A) = (9)(6) - (3)(12) = 54 - 36 = 18.

Adım Adım Çözüm

1
Calculate the determinant of matrix AA.
\det(A) = (3)(2) - (1)(4) = 6 - 4 = 2.
The determinant of a 2×22 \times 2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix} is given by ad−bcad - bc.
2
Apply the determinant scalar multiplication property for an n×nn \times n matrix.
\det(3A) = 3^2 \det(A) = 9 \times 2 = 18.
For any n×nn \times n square matrix AA and scalar kk, the identity det⁡(kA)=kndet⁡(A)\det(kA) = k^n \det(A) holds. Here, k=3k=3 and n=2n=2.

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Scalar Multiplication Property of Determinants
Tahmini Süre:1m 15s
Soru 19Soru →

Given the matrix A=(y53y+2)A = \begin{pmatrix} y & 5 \\ 3 & y+2 \end{pmatrix}, if the determinant of AA is 99 and y>0y > 0, calculate the value of yy.

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Cevap: 4

Cevap

The value of yy is 4.
Expanding the determinant of AA yields det⁡(A)=y(y+2)−15=y2+2y−15\det(A) = y(y+2) - 15 = y^2 + 2y - 15. Equating this to 9 gives y2+2y−24=0y^2 + 2y - 24 = 0, which factors as (y+6)(y−4)=0(y+6)(y-4) = 0. The roots are y=−6y = -6 and y=4y = 4. Given that y>0y > 0, the required value is 4.

Adım Adım Çözüm

1
Find the expression for the determinant of matrix AA
det⁡(A)=y(y+2)−(5)(3)=y2+2y−15\det(A) = y(y+2) - (5)(3) = y^2 + 2y - 15
The determinant of a 2×22 \times 2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix} is given by ad−bcad - bc.
2
Form and solve the quadratic equation
y2+2y−15=9  ⟹  y2+2y−24=0  ⟹  (y+6)(y−4)=0y^2 + 2y - 15 = 9 \implies y^2 + 2y - 24 = 0 \implies (y + 6)(y - 4) = 0
Set the determinant expression equal to the given determinant value of 9 and rearrange into standard quadratic form.
3
Apply the given domain restriction y>0y > 0
y=4y = 4
The root y=−6y = -6 is discarded because yy must be strictly positive.

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Determinant of a 2x2 matrix and solving non-linear determinant equations
Soru 20Soru →

Given the matrices M=(2−134)M = \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix} and N=(1023)N = \begin{pmatrix} 1 & 0 \\ 2 & 3 \end{pmatrix}, what is the determinant of the matrix product MNMN?

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Cevap: 33

Cevap

33
Using the property that the determinant of a product of matrices equals the product of their individual determinants, det⁡(M)=(2)(4)−(−1)(3)=11\det(M) = (2)(4) - (-1)(3) = 11 and det⁡(N)=(1)(3)−(0)(2)=3\det(N) = (1)(3) - (0)(2) = 3. Multiplying these gives det⁡(MN)=11×3=33\det(MN) = 11 \times 3 = 33.

Adım Adım Çözüm

1
Calculate the determinant of matrix MM
\det(M) = (2)(4) - (-1)(3) = 8 + 3 = 11
The determinant of a 2×22 \times 2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix} is given by ad−bcad - bc.
2
Calculate the determinant of matrix NN
\det(N) = (1)(3) - (0)(2) = 3 - 0 = 3
Apply the determinant formula ad−bcad - bc to matrix NN.
3
Apply the determinant product property to find det⁡(MN)\det(MN)
\det(MN) = \det(M) \times \det(N) = 11 \times 3 = 33
For any two square matrices of the same order, det⁡(MN)=det⁡(M)⋅det⁡(N)\det(MN) = \det(M) \cdot \det(N).

Anahtar Kavram

Determinant Property of Matrix Products
Tahmini Süre:1m 30s
Sayfa 1 / 2Sonraki
Matrices and Determinants Alıştırma Soruları — JAMB UTME | Examkin