Matrices and Determinants

29 soru

Soru 1Soru

Given the 3×33 \times 3 matrix A=(x213121x0)A = \begin{pmatrix} x & 2 & 1 \\ 3 & 1 & 2 \\ 1 & x & 0 \end{pmatrix}, find the positive value of xx for which det(A)=2\det(A) = -2.

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Cevap: 2.5

Cevap

The positive value of xx is 2.5.
Expanding the determinant of matrix AA along the third row gives 1(41)x(2x3)=2x2+3x+31(4 - 1) - x(2x - 3) = -2x^2 + 3x + 3. Setting this equal to 2-2 yields 2x23x5=02x^2 - 3x - 5 = 0. Factoring gives (2x5)(x+1)=0(2x - 5)(x + 1) = 0, yielding solutions x=2.5x = 2.5 and x=1x = -1. Taking the positive value gives x=2.5x = 2.5.

Adım Adım Çözüm

1
Calculate the determinant of matrix AA in terms of xx
det(A)=2x2+3x+3\det(A) = -2x^2 + 3x + 3
Expanding along the third row simplifies computation because of the zero entry.
2
Set the determinant expression equal to 2-2 and rearrange terms
2x23x5=02x^2 - 3x - 5 = 0
Setting 2x2+3x+3=2-2x^2 + 3x + 3 = -2 forms a standard quadratic equation.
3
Factorize the quadratic equation to find the roots
x=2.5x = 2.5 or x=1x = -1
Factoring (2x5)(x+1)=0(2x - 5)(x + 1) = 0 yields two real solutions.
4
Filter for the positive value requested in the stem
x=2.5x = 2.5
The question specifically requires the positive value of xx.

Anahtar Kavram

Determinant of a 3x3 Matrix and Quadratic Equation Solving
Tahmini Süre:2m 30s
Soru 2Soru

If the matrix A=(3275)A = \begin{pmatrix} 3 & 2 \\ 7 & 5 \end{pmatrix}, which matrix represents the inverse A1A^{-1}?

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Cevap: (5273)\begin{pmatrix} 5 & -2 \\ -7 & 3 \end{pmatrix}

Cevap

(5273)\begin{pmatrix} 5 & -2 \\ -7 & 3 \end{pmatrix}
The correct inverse matrix is computed by evaluating the determinant det(A)=3(5)2(7)=1\det(A) = 3(5) - 2(7) = 1 and constructing the adjugate matrix by swapping the diagonal elements 33 and 55 while changing the signs of 22 and 77, resulting in (5273)\begin{pmatrix} 5 & -2 \\ -7 & 3 \end{pmatrix}.

Adım Adım Çözüm

1
Calculate the determinant of matrix AA
det(A)=(3)(5)(2)(7)=1514=1\det(A) = (3)(5) - (2)(7) = 15 - 14 = 1
The inverse requires dividing the adjugate matrix by the determinant of AA.
2
Find the adjugate of matrix AA
adj(A)=(5273)\text{adj}(A) = \begin{pmatrix} 5 & -2 \\ -7 & 3 \end{pmatrix}
For a 2x2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix}, swap the main diagonal elements (aa and dd) and change the signs of the off-diagonal elements (bb and cc).
3
Compute A1=1det(A)adj(A)A^{-1} = \frac{1}{\det(A)} \text{adj}(A)
A1=11(5273)=(5273)A^{-1} = \frac{1}{1} \begin{pmatrix} 5 & -2 \\ -7 & 3 \end{pmatrix} = \begin{pmatrix} 5 & -2 \\ -7 & 3 \end{pmatrix}
Multiply the adjugate matrix by the reciprocal of the determinant.

Anahtar Kavram

2x2 Matrix Inversion
Tahmini Süre:1m 0s
Soru 3Soru

Let A=(2113)A = \begin{pmatrix} 2 & 1 \\ -1 & 3 \end{pmatrix} and B=(x21y)B = \begin{pmatrix} x & 2 \\ 1 & y \end{pmatrix} be 2×22 \times 2 matrices with integer entries xx and yy. If the matrix product ABAB is singular and det(A+B)=0\det(A + B) = 0, what is the value of x+yx + y?

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Cevap: 3-3

Cevap

The value of x+yx + y is 3-3.
Since matrix ABAB is singular, det(AB)=det(A)det(B)=0\det(AB) = \det(A)\det(B) = 0. Evaluating det(A)=70\det(A) = 7 \neq 0, we find det(B)=xy2=0\det(B) = xy - 2 = 0, giving xy=2xy = 2. Computing A+B=(2+x303+y)A+B = \begin{pmatrix} 2+x & 3 \\ 0 & 3+y \end{pmatrix}, its determinant is (2+x)(3+y)=6+3x+2y+xy=0(2+x)(3+y) = 6 + 3x + 2y + xy = 0. Substituting xy=2xy = 2 yields 3x+2y=83x + 2y = -8. The integer pair satisfying both xy=2xy = 2 and 3x+2y=83x + 2y = -8 is x=2x = -2 and y=1y = -1. Therefore, x+y=3x + y = -3.

Adım Adım Çözüm

1
Calculate the determinant of matrix AA.
det(A)=(2)(3)(1)(1)=6+1=7\det(A) = (2)(3) - (1)(-1) = 6 + 1 = 7.
Knowing det(A)\det(A) helps simplify the condition det(AB)=0\det(AB) = 0 using determinant properties.
2
Use the singularity of ABAB to find a relationship between xx and yy.
det(AB)=det(A)det(B)=7(xy2)=0    xy=2\det(AB) = \det(A)\det(B) = 7(xy - 2) = 0 \implies xy = 2.
The determinant of a product of matrices equals the product of their determinants.
3
Form the matrix A+BA + B and compute its determinant.
A+B=(2+x303+y)    det(A+B)=(2+x)(3+y)(3)(0)=(2+x)(3+y)=6+3x+2y+xyA + B = \begin{pmatrix} 2+x & 3 \\ 0 & 3+y \end{pmatrix} \implies \det(A+B) = (2+x)(3+y) - (3)(0) = (2+x)(3+y) = 6 + 3x + 2y + xy.
Setting this determinant to zero gives a second equation involving xx and yy.
4
Substitute xy=2xy = 2 into det(A+B)=0\det(A+B) = 0 and solve for integer values of xx and yy.
6+3x+2y+2=0    3x+2y=86 + 3x + 2y + 2 = 0 \implies 3x + 2y = -8. Testing integer pairs (x,y)(x,y) for xy=2xy = 2 gives x=2x = -2 and y=1y = -1.
3(2)+2(1)=83(-2) + 2(-1) = -8 is satisfied only by x=2,y=1x = -2, y = -1 among integer pairs.
5
Compute x+yx + y.
x+y=2+(1)=3x + y = -2 + (-1) = -3.
This answers the question directly.

Anahtar Kavram

Determinants of matrix products and sums, matrix singularity, and integer solutions to matrix equations.
Soru 4Soru

If the matrix P=(x386)P = \begin{pmatrix} x & 3 \\ 8 & 6 \end{pmatrix} is a singular matrix, what is the value of xx?

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Cevap: 4

Cevap

The value of xx is 44.
For a 2×22 \times 2 matrix to be singular, its determinant must be zero. For matrix PP, the determinant is (x6)(38)=6x24(x \cdot 6) - (3 \cdot 8) = 6x - 24. Setting 6x24=06x - 24 = 0 gives 6x=246x = 24, which simplifies to x=4x = 4.

Adım Adım Çözüm

1
State the condition for a matrix to be singular.
A matrix is singular if its determinant is equal to zero: det(P)=0\det(P) = 0.
By definition, a singular matrix has a determinant equal to zero.
2
Calculate the determinant of matrix PP.
\det(P) = (x \times 6) - (3 \times 8) = 6x - 24.
The determinant of a 2×22 \times 2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix} is calculated using the formula adbcad - bc.
3
Solve the linear equation 6x24=06x - 24 = 0 for xx.
6x = 24 \implies x = 4.
Adding 2424 to both sides and dividing by 66 isolates xx.

Anahtar Kavram

Singular matrix condition and 2×22 \times 2 determinant calculation
Soru 5Soru

Given the matrices A=(k231)A = \begin{pmatrix} k & 2 \\ 3 & 1 \end{pmatrix} and B=(1120)B = \begin{pmatrix} 1 & -1 \\ 2 & 0 \end{pmatrix}, what is the smaller positive integer value of kk for which det(A2+B)=4\det(A^2 + B) = 4?

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Cevap: 5

Cevap

The smaller positive integer value of kk is 5.
Evaluating A2A^2 via matrix multiplication gives (k2+62k+23k+37)\begin{pmatrix} k^2 + 6 & 2k + 2 \\ 3k + 3 & 7 \end{pmatrix}. Adding BB gives A2+B=(k2+72k+13k+57)A^2 + B = \begin{pmatrix} k^2 + 7 & 2k + 1 \\ 3k + 5 & 7 \end{pmatrix}. Computing det(A2+B)=(k2+7)(7)(2k+1)(3k+5)=k213k+44\det(A^2 + B) = (k^2 + 7)(7) - (2k + 1)(3k + 5) = k^2 - 13k + 44. Setting this to 4 results in k213k+40=0k^2 - 13k + 40 = 0, which yields roots k=5k = 5 and k=8k = 8. The smaller value is 5.

Adım Adım Çözüm

1
Compute matrix A2A^2 using matrix multiplication (row by column).
A2=(k231)(k231)=(k2+62k+23k+37)A^2 = \begin{pmatrix} k & 2 \\ 3 & 1 \end{pmatrix} \begin{pmatrix} k & 2 \\ 3 & 1 \end{pmatrix} = \begin{pmatrix} k^2 + 6 & 2k + 2 \\ 3k + 3 & 7 \end{pmatrix}
Matrix multiplication requires taking the dot product of rows of the first matrix with columns of the second matrix.
2
Add matrix BB to A2A^2.
A2+B=(k2+62k+23k+37)+(1120)=(k2+72k+13k+57)A^2 + B = \begin{pmatrix} k^2 + 6 & 2k + 2 \\ 3k + 3 & 7 \end{pmatrix} + \begin{pmatrix} 1 & -1 \\ 2 & 0 \end{pmatrix} = \begin{pmatrix} k^2 + 7 & 2k + 1 \\ 3k + 5 & 7 \end{pmatrix}
Matrix addition is performed by adding corresponding entries.
3
Calculate the determinant of A2+BA^2 + B and set it equal to 4.
det(A2+B)=(k2+7)(7)(2k+1)(3k+5)=7k2+49(6k2+13k+5)=k213k+44=4\det(A^2 + B) = (k^2 + 7)(7) - (2k + 1)(3k + 5) = 7k^2 + 49 - (6k^2 + 13k + 5) = k^2 - 13k + 44 = 4
The determinant of a 2×22 \times 2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix} is adbcad - bc.
4
Solve the quadratic equation for kk.
k213k+40=0    (k5)(k8)=0    k=5 or k=8k^2 - 13k + 40 = 0 \implies (k - 5)(k - 8) = 0 \implies k = 5 \text{ or } k = 8
Factoring the quadratic equation gives the two possible integer roots.

Anahtar Kavram

Matrix multiplication, matrix addition, determinant of a 2x2 matrix, and quadratic equation solving.
Soru 6Soru

Two 2×22 \times 2 matrices are given as A=(2143)A = \begin{pmatrix} 2 & -1 \\ 4 & 3 \end{pmatrix} and B=(102k)B = \begin{pmatrix} 1 & 0 \\ 2 & k \end{pmatrix}. If the determinant of the product matrix ABAB is equal to 3030, what is the value of kk?

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Cevap: 3

Cevap

The value of k is 3.
The determinant of a product of square matrices equals the product of their individual determinants: det(AB)=det(A)det(B)\det(AB) = \det(A) \cdot \det(B). Evaluating det(A)\det(A) yields (2)(3)(1)(4)=10(2)(3) - (-1)(4) = 10, and det(B)\det(B) yields (1)(k)(0)(2)=k(1)(k) - (0)(2) = k. Substituting these into det(AB)=30\det(AB) = 30 gives 10k=3010k = 30, leading directly to k=3k = 3.

Adım Adım Çözüm

1
Evaluate the determinant of matrix A
\det(A) = 10
Using the 2×22 \times 2 determinant formula det(abcd)=adbc\det\begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad - bc, we calculate det(A)=(2)(3)(1)(4)=6+4=10\det(A) = (2)(3) - (-1)(4) = 6 + 4 = 10.
2
Evaluate the determinant of matrix B
\det(B) = k
Calculating the determinant of matrix BB yields det(B)=(1)(k)(0)(2)=k\det(B) = (1)(k) - (0)(2) = k.
3
Use the product property of determinants to solve for k
k = 3
Since det(AB)=det(A)det(B)\det(AB) = \det(A) \cdot \det(B), we have 10k=3010k = 30. Dividing both sides by 10 yields k=3k = 3.

Anahtar Kavram

Determinant of Matrix Product
Tahmini Süre:1m 30s
Soru 7Soru

Given the matrices A=(231m)A = \begin{pmatrix} 2 & 3 \\ 1 & m \end{pmatrix} and B=(42)B = \begin{pmatrix} 4 \\ -2 \end{pmatrix}, if the matrix product AB=(26)AB = \begin{pmatrix} 2 \\ 6 \end{pmatrix}, what is the value of mm?

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Cevap: 1-1

Cevap

The value of mm is 1-1.
Multiplying the 2×22 \times 2 matrix by the 2×12 \times 1 column vector yields a 2×12 \times 1 result where the second row entry is 1(4)+m(2)=42m1(4) + m(-2) = 4 - 2m. Setting this equal to the second row element 66 gives 42m=64 - 2m = 6, which simplifies to m=1m = -1.

Adım Adım Çözüm

1
Compute the matrix product ABAB
AB=(2(4)+3(2)1(4)+m(2))=(8642m)=(242m)AB = \begin{pmatrix} 2(4) + 3(-2) \\ 1(4) + m(-2) \end{pmatrix} = \begin{pmatrix} 8 - 6 \\ 4 - 2m \end{pmatrix} = \begin{pmatrix} 2 \\ 4 - 2m \end{pmatrix}
Matrix multiplication requires taking the dot product of each row of the first matrix with the column vector.
2
Equate the resulting vector to the given vector (26)\begin{pmatrix} 2 \\ 6 \end{pmatrix}
42m=64 - 2m = 6
Corresponding elements of equal matrices must be equal.
3
Solve the linear equation for mm
2m=642m=2m=1-2m = 6 - 4 \Rightarrow -2m = 2 \Rightarrow m = -1
Isolating the variable mm by subtracting 44 from both sides and dividing by 2-2.

Anahtar Kavram

Matrix Multiplication and Equality of Matrices
Soru 8Soru

Given that the determinant of the 3×33 \times 3 matrix M=(3102x1042)M = \begin{pmatrix} 3 & 1 & 0 \\ 2 & x & -1 \\ 0 & 4 & 2 \end{pmatrix} is equal to 2020, calculate the value of xx.

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Cevap: 2

Cevap

The value of xx is 22.
Expanding the matrix MM along its top row yields det(M)=3(2x+4)1(4)=6x+8\det(M) = 3(2x + 4) - 1(4) = 6x + 8. Setting this expression equal to 2020 gives 6x+8=206x + 8 = 20, which simplifies to 6x=126x = 12, yielding x=2x = 2.

Adım Adım Çözüm

1
Expand the 3x3 matrix along the first row
\det(M) = 3(2x - (-4)) - 1(4 - 0) + 0
Cofactor expansion along a row containing a zero simplifies the computation of a 3x3 determinant.
2
Simplify the algebraic expression for the determinant
\det(M) = 6x + 8
Distribute the coefficients and combine like constant terms.
3
Solve the linear equation for x
x = 2
Subtract 8 from 20 to get 12, then divide by 6.

Anahtar Kavram

Determinant of a 3x3 Matrix via Cofactor Expansion
Soru 9Soru

Given that the matrix A=(x25x3)A = \begin{pmatrix} x & 2 \\ 5 & x - 3 \end{pmatrix} is a singular matrix, what is the positive value of xx?

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Cevap: 55

Cevap

The positive value of xx is 55.
For a matrix to be singular, its determinant must be zero. Calculating the determinant of AA yields det(A)=x(x3)(2)(5)=x23x10\det(A) = x(x - 3) - (2)(5) = x^2 - 3x - 10. Setting this to zero and factoring gives (x5)(x+2)=0(x - 5)(x + 2) = 0, leading to x=5x = 5 or x=2x = -2. The positive solution is 55.

Adım Adım Çözüm

1
Set the determinant of the matrix equal to zero.
det(A)=(x)(x3)(2)(5)=0\det(A) = (x)(x - 3) - (2)(5) = 0
By definition, a matrix is singular if and only if its determinant is zero.
2
Expand and simplify the algebraic equation.
x23x10=0x^2 - 3x - 10 = 0
Expanding x(x3)x(x - 3) gives x23xx^2 - 3x, and subtracting 1010 forms a standard quadratic equation.
3
Factor the quadratic equation to solve for xx.
(x5)(x+2)=0    x=5 or x=2(x - 5)(x + 2) = 0 \implies x = 5 \text{ or } x = -2
The factors of 10-10 that sum to 3-3 are 5-5 and +2+2.
4
Select the positive value requested by the question.
x=5x = 5
The question specifically asks for the positive value of xx.

Anahtar Kavram

Singular Matrices and Determinants
Soru 10Soru

If the matrix M=(2x436)M = \begin{pmatrix} 2x & 4 \\ 3 & 6 \end{pmatrix} is a singular matrix, what is the value of xx?

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Cevap: 11

Cevap

The value of xx is 11.
A matrix is singular when its determinant equals zero. Evaluating the determinant of (2x436)\begin{pmatrix} 2x & 4 \\ 3 & 6 \end{pmatrix} gives (2x)(6)(4)(3)=12x12(2x)(6) - (4)(3) = 12x - 12. Setting 12x12=012x - 12 = 0 leads directly to x=1x = 1.

Adım Adım Çözüm

1
Recall the condition for a matrix to be singular.
A matrix is singular if and only if its determinant is equal to zero, so det(M)=0\det(M) = 0.
Singular matrices have no inverse because their determinant is zero.
2
Calculate the determinant of matrix M=(2x436)M = \begin{pmatrix} 2x & 4 \\ 3 & 6 \end{pmatrix}.
\det(M) = (2x)(6) - (4)(3) = 12x - 12.
The determinant of a 2×22 \times 2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix} is adbcad - bc.
3
Set the determinant equal to zero and solve for xx.
12x12=0    12x=12    x=112x - 12 = 0 \implies 12x = 12 \implies x = 1.
Solving the linear equation yields the required parameter value.

Anahtar Kavram

Condition for Singular Matrix and 2x2 Determinant Evaluation
Tahmini Süre:45s
Soru 11Soru

For the matrix M=(k312k0152)M = \begin{pmatrix} k & 3 & 1 \\ 2 & k & 0 \\ 1 & 5 & 2 \end{pmatrix}, the determinant of MM is equal to 44. What is the positive value of kk?

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Cevap: 2

Cevap

The positive value of kk is 22.
Expanding the determinant along the second row yields det(M)=2(65)+k(2k1)=2k2k2\det(M) = -2(6-5) + k(2k-1) = 2k^2 - k - 2. Setting this equal to 44 yields 2k2k6=02k^2 - k - 6 = 0, which factors into (2k+3)(k2)=0(2k+3)(k-2)=0. The positive value is 22.

Adım Adım Çözüm

1
Evaluate the determinant of MM in terms of kk using row 2 cofactor expansion
\det(M) = 2k^2 - k - 2
Expanding along the second row gives 2(65)+k(2k1)0=2+2k2k-2(6 - 5) + k(2k - 1) - 0 = -2 + 2k^2 - k.
2
Set the determinant equal to the given value 4 and rearrange into standard quadratic form
2k^2 - k - 6 = 0
Subtracting 4 from both sides gives 2k2k6=02k^2 - k - 6 = 0.
3
Solve the quadratic equation by factorization
k = -1.5 \text{ or } k = 2
Factoring (2k+3)(k2)=0(2k + 3)(k - 2) = 0 gives roots k=1.5k = -1.5 and k=2k = 2.
4
Select the positive root
k = 2
The question specifically asks for the positive value of kk.

Anahtar Kavram

Evaluating a 3x3 matrix determinant using cofactor expansion and solving the resulting quadratic equation for an unknown parameter.
Soru 12Soru

Given the matrices A=(2x13)A = \begin{pmatrix} 2 & x \\ -1 & 3 \end{pmatrix} and B=(142y)B = \begin{pmatrix} 1 & 4 \\ 2 & y \end{pmatrix}, if AB=BAAB = BA, what is the value of x+yx + y?

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Cevap: 3-3

Cevap

The value of x+yx + y is 3-3.
Evaluating the matrix products ABAB and BABA using standard matrix multiplication rules yields AB=(2+2x8+xy54+3y)AB = \begin{pmatrix} 2+2x & 8+xy \\ 5 & -4+3y \end{pmatrix} and BA=(2x+124y2x+3y)BA = \begin{pmatrix} -2 & x+12 \\ 4-y & 2x+3y \end{pmatrix}. Equating entry-by-entry gives 2+2x=2    x=22+2x = -2 \implies x = -2 and 5=4y    y=15 = 4-y \implies y = -1. Adding these together gives x+y=3x + y = -3.

Adım Adım Çözüm

1
Compute the product matrix ABAB
AB=(2(1)+x(2)2(4)+x(y)1(1)+3(2)1(4)+3(y))=(2+2x8+xy54+3y)AB = \begin{pmatrix} 2(1) + x(2) & 2(4) + x(y) \\ -1(1) + 3(2) & -1(4) + 3(y) \end{pmatrix} = \begin{pmatrix} 2 + 2x & 8 + xy \\ 5 & -4 + 3y \end{pmatrix}
Matrix multiplication requires taking the dot product of rows from the first matrix and columns from the second matrix.
2
Compute the product matrix BABA
BA=(1(2)+4(1)1(x)+4(3)2(2)+y(1)2(x)+y(3))=(2x+124y2x+3y)BA = \begin{pmatrix} 1(2) + 4(-1) & 1(x) + 4(3) \\ 2(2) + y(-1) & 2(x) + y(3) \end{pmatrix} = \begin{pmatrix} -2 & x + 12 \\ 4 - y & 2x + 3y \end{pmatrix}
Evaluate BABA by multiplying rows of BB by columns of AA.
3
Equate corresponding entries of ABAB and BABA since AB=BAAB = BA
From row 1, col 1: 2+2x=2    2x=4    x=22 + 2x = -2 \implies 2x = -4 \implies x = -2.
From row 2, col 1: 5=4y    y=15 = 4 - y \implies y = -1.
Two matrices are equal if and only if all corresponding entries are equal.
4
Verify consistency on remaining entries and calculate x+yx + y
Row 1, col 2 check: 8+(2)(1)=108 + (-2)(-1) = 10 and 2+12=10-2 + 12 = 10.
Row 2, col 2 check: 4+3(1)=7-4 + 3(-1) = -7 and 2(2)+3(1)=72(-2) + 3(-1) = -7.
Sum: x+y=2+(1)=3x + y = -2 + (-1) = -3.
Verifying consistency ensures the system of equations has a valid unique solution.

Anahtar Kavram

Matrix Multiplication Commutativity and Matrix Equality
Tahmini Süre:2m 30s
Soru 13Soru

If the matrix A=(1022k1314)A = \begin{pmatrix} 1 & 0 & 2 \\ 2 & k & 1 \\ 3 & 1 & 4 \end{pmatrix} is singular, what is the value of kk?

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Cevap: 32\frac{3}{2}

Cevap

32\frac{3}{2}
A matrix is singular if its determinant equals zero. Expanding the determinant of matrix AA along its first row gives 1(4k1)0+2(23k)=2k+31(4k - 1) - 0 + 2(2 - 3k) = -2k + 3. Setting 2k+3=0-2k + 3 = 0 yields k=32k = \frac{3}{2}.

Adım Adım Çözüm

1
Apply the condition for a singular matrix
A matrix is singular when its determinant equals zero, so det(A)=0\det(A) = 0.
By definition, square matrices with zero determinant are singular.
2
Expand det(A)\det(A) along the first row
det(A)=1(4k1)0(83)+2(23k)=4k1+46k=2k+3\det(A) = 1(4k - 1) - 0(8 - 3) + 2(2 - 3k) = 4k - 1 + 4 - 6k = -2k + 3.
Laplace expansion along the first row simplifies computation due to the zero entry.
3
Solve for kk
2k+3=0    2k=3    k=32-2k + 3 = 0 \implies 2k = 3 \implies k = \frac{3}{2}.
Isolating the variable gives the required value of kk.

Anahtar Kavram

Singular Matrix Condition and 3x3 Determinant Expansion
Soru 14Soru

Find the product matrix ABAB given that A=(1203)A = \begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix} and B=(4512)B = \begin{pmatrix} 4 & 5 \\ 1 & 2 \end{pmatrix}.

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Cevap: (6936)\begin{pmatrix} 6 & 9 \\ 3 & 6 \end{pmatrix}

Cevap

(6936)\begin{pmatrix} 6 & 9 \\ 3 & 6 \end{pmatrix}
The matrix product ABAB is found by performing row-by-column multiplication. Computing row 1 gives (1)(4)+(2)(1)=6(1)(4)+(2)(1) = 6 and (1)(5)+(2)(2)=9(1)(5)+(2)(2) = 9. Computing row 2 gives (0)(4)+(3)(1)=3(0)(4)+(3)(1) = 3 and (0)(5)+(3)(2)=6(0)(5)+(3)(2) = 6. Thus, the resulting matrix is (6936)\begin{pmatrix} 6 & 9 \\ 3 & 6 \end{pmatrix}.

Adım Adım Çözüm

1
Set up row-by-column matrix multiplication formula for 2×22 \times 2 matrices.
The entry at row ii, column jj of product ABAB is the dot product of row ii of AA and column jj of BB.
Matrix multiplication requires combining rows of the left matrix with columns of the right matrix.
2
Calculate row 1 entries of the product matrix.
First element: (1)(4)+(2)(1)=4+2=6(1)(4) + (2)(1) = 4 + 2 = 6. Second element: (1)(5)+(2)(2)=5+4=9(1)(5) + (2)(2) = 5 + 4 = 9.
Multiply elements of row 1 of AA by corresponding elements of columns 1 and 2 of BB and sum them.
3
Calculate row 2 entries of the product matrix.
Third element: (0)(4)+(3)(1)=0+3=3(0)(4) + (3)(1) = 0 + 3 = 3. Fourth element: (0)(5)+(3)(2)=0+6=6(0)(5) + (3)(2) = 0 + 6 = 6.
Multiply elements of row 2 of AA by corresponding elements of columns 1 and 2 of BB and sum them.
4
Assemble the product matrix.
AB=(6936)AB = \begin{pmatrix} 6 & 9 \\ 3 & 6 \end{pmatrix}.
Combine all computed row and column entries into a single 2×22 \times 2 matrix.

Anahtar Kavram

Matrix Multiplication (2×22 \times 2 Matrix Product)
Soru 15Soru

Given the matrices A=(2134)A = \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix} and B=(1x25)B = \begin{pmatrix} 1 & x \\ -2 & 5 \end{pmatrix}, if the determinant of the product matrix ABAB is equal to 121121, find the value of xx.

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Cevap: 3

Cevap

The value of xx is 33.
By applying the property det(AB)=det(A)det(B)\det(AB) = \det(A) \cdot \det(B), we compute det(A)=(2)(4)(1)(3)=11\det(A) = (2)(4) - (-1)(3) = 11 and det(B)=(1)(5)(x)(2)=5+2x\det(B) = (1)(5) - (x)(-2) = 5 + 2x. Substituting into the equation gives 11(5+2x)=12111(5 + 2x) = 121, which yields 5+2x=115 + 2x = 11 and leads to x=3x = 3.

Adım Adım Çözüm

1
Find the determinant of matrix AA
det(A)=(2)(4)(1)(3)=11\det(A) = (2)(4) - (-1)(3) = 11
The determinant of a 2×22 \times 2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix} is calculated as adbcad - bc.
2
Express the determinant of matrix BB in terms of xx
det(B)=(1)(5)(x)(2)=5+2x\det(B) = (1)(5) - (x)(-2) = 5 + 2x
Apply the 2×22 \times 2 determinant formula to matrix BB.
3
Apply the determinant product rule
det(AB)=det(A)det(B)=11(5+2x)=121\det(AB) = \det(A) \cdot \det(B) = 11(5 + 2x) = 121
For any square matrices AA and BB, det(AB)=det(A)det(B)\det(AB) = \det(A) \det(B).
4
Solve the resulting linear equation for xx
x=3x = 3
Dividing 121121 by 1111 gives 5+2x=115 + 2x = 11, which simplifies to 2x=62x = 6 and x=3x = 3.

Anahtar Kavram

Determinant of a Matrix Product
Tahmini Süre:1m 30s
Soru 16Soru

Given the matrices A=(2x13)A = \begin{pmatrix} 2 & x \\ -1 & 3 \end{pmatrix} and B=(1042)B = \begin{pmatrix} 1 & 0 \\ 4 & 2 \end{pmatrix}, if det(A1BT)=15\det(A^{-1} B^T) = \frac{1}{5}, what is the value of xx?

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Cevap: 4

Cevap

The value of xx is 4.
Using the properties det(BT)=det(B)=2\det(B^T) = \det(B) = 2 and det(A1)=1det(A)=16+x\det(A^{-1}) = \frac{1}{\det(A)} = \frac{1}{6+x}, the product rule gives det(A1BT)=26+x\det(A^{-1}B^T) = \frac{2}{6+x}. Equating this to 15\frac{1}{5} gives 6+x=106+x = 10, so x=4x = 4.

Adım Adım Çözüm

1
Calculate the determinants of matrices AA and BB.
\det(A) = (2)(3) - (-1)(x) = 6 + x and and \det(B) = (1)(2) - (0)(4) = 2$.
The determinant of a 2x2 matrix is computed using adbcad - bc.
2
Apply determinant rules for inverse and transpose matrices.
\det(B^T) = \det(B) = 2 and and \det(A^{-1}) = \frac{1}{\det(A)} = \frac{1}{6+x}$.
The determinant of a transpose equals the original determinant, and the determinant of an inverse is its reciprocal.
3
Express det(A1BT)\det(A^{-1} B^T) using the product rule of determinants.
\det(A^{-1} B^T) = \det(A^{-1}) \cdot \det(B^T) = \frac{1}{6+x} \cdot 2 = \frac{2}{6+x}$.
For square matrices, det(MN)=det(M)det(N)\det(MN) = \det(M)\det(N).
4
Equate the expression to 15\frac{1}{5} and solve for xx.
\frac{2}{6+x} = \frac{1}{5} \implies 10 = 6 + x \implies x = 4$.
Cross-multiplying and simplifying gives x=4x = 4.

Anahtar Kavram

Properties of Determinants (Product, Inverse, and Transpose)
Soru 17Soru

If P=(3124)P = \begin{pmatrix} 3 & 1 \\ 2 & 4 \end{pmatrix} and Q=(1205)Q = \begin{pmatrix} 1 & -2 \\ 0 & 5 \end{pmatrix}, what is the product matrix PQPQ?

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Cevap: (31216)\begin{pmatrix} 3 & -1 \\ 2 & 16 \end{pmatrix}

Cevap

The product matrix PQPQ is (31216)\begin{pmatrix} 3 & -1 \\ 2 & 16 \end{pmatrix}.
The matrix containing elements 3, -1 in the top row and 2, 16 in the bottom row correctly applies the row-by-column multiplication rule for matrices PP and QQ.

Adım Adım Çözüm

1
Set up the row-by-column matrix multiplication PQPQ
PQ=(3124)(1205)PQ = \begin{pmatrix} 3 & 1 \\ 2 & 4 \end{pmatrix} \begin{pmatrix} 1 & -2 \\ 0 & 5 \end{pmatrix}
Matrix multiplication requires computing dot products of rows of the first matrix with columns of the second matrix.
2
Calculate the first row elements of the product matrix
Row 1, Column 1: 3(1)+1(0)=3+0=33(1) + 1(0) = 3 + 0 = 3; Row 1, Column 2: 3(2)+1(5)=6+5=13(-2) + 1(5) = -6 + 5 = -1
Multiply elements of Row 1 of PP by corresponding elements of Columns 1 and 2 of QQ and sum them.
3
Calculate the second row elements of the product matrix
Row 2, Column 1: 2(1)+4(0)=2+0=22(1) + 4(0) = 2 + 0 = 2; Row 2, Column 2: 2(2)+4(5)=4+20=162(-2) + 4(5) = -4 + 20 = 16
Multiply elements of Row 2 of PP by corresponding elements of Columns 1 and 2 of QQ and sum them.
4
Assemble the resulting 2×22 \times 2 matrix
PQ=(31216)PQ = \begin{pmatrix} 3 & -1 \\ 2 & 16 \end{pmatrix}
Combine all computed entry values into matrix form.

Anahtar Kavram

Matrix Multiplication (Row-by-Column Rule)
Tahmini Süre:1m 30s
Soru 18Soru

Given the matrices A=(x142)A = \begin{pmatrix} x & 1 \\ 4 & 2 \end{pmatrix} and B=(231x)B = \begin{pmatrix} 2 & 3 \\ 1 & x \end{pmatrix}, what is the sum of all real values of xx for which the matrix C=ABBTC = AB - B^T is singular?

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Cevap: 52\frac{5}{2}

Cevap

The sum of all real values of xx is 52\frac{5}{2}.
Evaluating ABAB yields (2x+14x102x+12)\begin{pmatrix} 2x + 1 & 4x \\ 10 & 2x + 12 \end{pmatrix} and subtracting BT=(213x)B^T = \begin{pmatrix} 2 & 1 \\ 3 & x \end{pmatrix} gives matrix C=(2x14x17x+12)C = \begin{pmatrix} 2x - 1 & 4x - 1 \\ 7 & x + 12 \end{pmatrix}. Setting det(C)=(2x1)(x+12)7(4x1)=0\det(C) = (2x - 1)(x + 12) - 7(4x - 1) = 0 results in the quadratic equation 2x25x5=02x^2 - 5x - 5 = 0. Since the discriminant Δ=65>0\Delta = 65 > 0, both roots are real, and their sum is given by ba=52-\frac{b}{a} = \frac{5}{2}.

Adım Adım Çözüm

1
Compute the matrix product ABAB
AB=(x142)(231x)=(2x+14x102x+12)AB = \begin{pmatrix} x & 1 \\ 4 & 2 \end{pmatrix} \begin{pmatrix} 2 & 3 \\ 1 & x \end{pmatrix} = \begin{pmatrix} 2x + 1 & 4x \\ 10 & 2x + 12 \end{pmatrix}
Matrix multiplication requires taking the dot product of rows of the first matrix with columns of the second matrix.
2
Find the transpose of matrix BB, denoted BTB^T
BT=(213x)B^T = \begin{pmatrix} 2 & 1 \\ 3 & x \end{pmatrix}
The transpose of a matrix is formed by swapping its rows and columns.
3
Compute matrix C=ABBTC = AB - B^T
C=(2x+14x102x+12)(213x)=(2x14x17x+12)C = \begin{pmatrix} 2x + 1 & 4x \\ 10 & 2x + 12 \end{pmatrix} - \begin{pmatrix} 2 & 1 \\ 3 & x \end{pmatrix} = \begin{pmatrix} 2x - 1 & 4x - 1 \\ 7 & x + 12 \end{pmatrix}
Perform element-wise subtraction of matrix BTB^T from matrix ABAB.
4
Set the determinant of matrix CC to zero for singularity
det(C)=(2x1)(x+12)7(4x1)=2x2+23x12(28x7)=2x25x5=0\det(C) = (2x - 1)(x + 12) - 7(4x - 1) = 2x^2 + 23x - 12 - (28x - 7) = 2x^2 - 5x - 5 = 0
A matrix is singular if and only if its determinant is zero.
5
Find the sum of the real roots of the quadratic equation 2x25x5=02x^2 - 5x - 5 = 0
Discriminant Δ=(5)24(2)(5)=65>0\Delta = (-5)^2 - 4(2)(-5) = 65 > 0. By Vieta's formula, x1+x2=ba=52=52x_1 + x_2 = -\frac{b}{a} = -\frac{-5}{2} = \frac{5}{2}.
Since the discriminant is positive, two distinct real roots exist, and their sum is given by ba-\frac{b}{a}.

Anahtar Kavram

Singular matrix condition, matrix multiplication, transpose operations, and Vieta's formulas.
Soru 19Soru

Given the 2×22 \times 2 matrix M=(7243)M = \begin{pmatrix} 7 & -2 \\ 4 & 3 \end{pmatrix}, what is the value of det(M)\det(M)?

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Cevap: 29

Cevap

The determinant of matrix MM is 29.
The determinant of a 2×22 \times 2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix} is calculated as adbcad - bc. For the matrix M=(7243)M = \begin{pmatrix} 7 & -2 \\ 4 & 3 \end{pmatrix}, we compute (7)(3)(2)(4)=21(8)=29(7)(3) - (-2)(4) = 21 - (-8) = 29.

Adım Adım Çözüm

1
Identify the entries a,b,c,da, b, c, d from the matrix M=(7243)M = \begin{pmatrix} 7 & -2 \\ 4 & 3 \end{pmatrix}.
a=7a = 7, b=2b = -2, c=4c = 4, and d=3d = 3.
These entries correspond to the standard 2×22 \times 2 matrix representation (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix}.
2
Compute the determinant using the formula det(M)=adbc\det(M) = ad - bc.
det(M)=(7)(3)(2)(4)=21+8=29\det(M) = (7)(3) - (-2)(4) = 21 + 8 = 29.
Multiplying the main diagonal entries and subtracting the product of the off-diagonal entries yields the determinant.

Anahtar Kavram

Determinant of a 2x2 Matrix
Soru 20Soru

Given the matrices A=(x21302141)A = \begin{pmatrix} x & 2 & 1 \\ 3 & 0 & 2 \\ 1 & 4 & 1 \end{pmatrix} and B=(2153)B = \begin{pmatrix} 2 & -1 \\ 5 & 3 \end{pmatrix}, find the value of xx if det(A)+det(B)=5\det(A) + \det(B) = 5.

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Cevap: 2

Cevap

The value of x is 2.
Evaluating \det(B) gives 11, and cofactor expansion of \det(A) along the top row yields 10 - 8x. Summing these determinants gives (10 - 8x) + 11 = 5, which simplifies to 8x = 16, resulting in x = 2.

Adım Adım Çözüm

1
Calculate the determinant of the 2x2 matrix B
\det(B) = 11
For a 2x2 matrix \begin{pmatrix} a & b \\ c & d \end{pmatrix}, the determinant is ad - bc. Thus, \det(B) = (2)(3) - (-1)(5) = 6 + 5 = 11.
2
Calculate the determinant of the 3x3 matrix A in terms of x
\det(A) = 10 - 8x
Expanding along the first row: \det(A) = x\begin{vmatrix} 0 & 2 \\ 4 & 1 \end{vmatrix} - 2\begin{vmatrix} 3 & 2 \\ 1 & 1 \end{vmatrix} + 1\begin{vmatrix} 3 & 0 \\ 1 & 4 \end{vmatrix} = x(0 - 8) - 2(3 - 2) + 1(12 - 0) = -8x - 2 + 12 = 10 - 8x.
3
Set up and solve the linear equation \det(A) + \det(B) = 5
x = 2
Substituting the computed determinants gives (10 - 8x) + 11 = 5, which simplifies to 21 - 8x = 5. Rearranging yields 8x = 16, so x = 2.

Anahtar Kavram

Determinants of 2x2 and 3x3 matrices and linear equation solving
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Matrices and Determinants Alıştırma Soruları — JAMB UTME | Examkin