Hard Further Maths Matrices: arithmetic and determinants Questions

Challenging, exam-style Further Maths Matrices: arithmetic and determinants questions with worked solutions. Stretch yourself on the hardest matrices, inverse, 2x2, multiplication problems.

matricesinverse2x2multiplication3x3matrix-power
Further Maths34 questionsStep-by-step solutions
Question 1
8 markschallenging
Consider the simultaneous equations x+y+z=4x+y+z=4, 2x−y+z=82x-y+z=8 and x+2y−z=−3x+2y-z=-3. Which of the following is the solution?
Show worked solution

Worked solution

  1. Write the system as a matrix equation

    (1112−1112−1)(xyz)=(48−3)\begin{pmatrix}1 & 1 & 1 \\ 2 & -1 & 1 \\ 1 & 2 & -1\end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix}=\begin{pmatrix}4 \\ 8 \\ -3\end{pmatrix}

    Separate coefficients, unknowns and constants.

  2. Find the inverse of the coefficient matrix

    A−1=17(−1323−215−1−3)\mathbf{A}^{-1}=\frac{1}{7}\begin{pmatrix}-1 & 3 & 2 \\ 3 & -2 & 1 \\ 5 & -1 & -3\end{pmatrix}

    The determinant is non-zero so the inverse exists.

  3. Compute A−1b\mathbf{A}^{-1}\mathbf{b}

    x=(−17372737−271757−17−37)(48−3)=(2−13)\mathbf{x}=\begin{pmatrix}-\frac{1}{7} & \frac{3}{7} & \frac{2}{7} \\ \frac{3}{7} & -\frac{2}{7} & \frac{1}{7} \\ \frac{5}{7} & -\frac{1}{7} & -\frac{3}{7}\end{pmatrix}\begin{pmatrix}4 \\ 8 \\ -3\end{pmatrix}=\begin{pmatrix}2 \\ -1 \\ 3\end{pmatrix}

    This gives the unique solution.

  4. Write the system in matrix form

    (1112−1112−1)(xyz)=(48−3)\begin{pmatrix}1 & 1 & 1 \\ 2 & -1 & 1 \\ 1 & 2 & -1\end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix}=\begin{pmatrix}4 \\ 8 \\ -3\end{pmatrix}

    Collect the coefficients into a matrix and the unknowns into a column vector.

  5. Compute the determinant of the coefficient matrix

    det⁡A=7\det\mathbf{A}=7

    A non-zero determinant guarantees a unique solution.

  6. Find the inverse of the coefficient matrix

    A−1=(−17372737−271757−17−37)\mathbf{A}^{-1}=\begin{pmatrix}-\frac{1}{7} & \frac{3}{7} & \frac{2}{7} \\ \frac{3}{7} & -\frac{2}{7} & \frac{1}{7} \\ \frac{5}{7} & -\frac{1}{7} & -\frac{3}{7}\end{pmatrix}

    Use the adjugate divided by the determinant.

  7. Pre-multiply both sides by the inverse

    x=A−1b=(−17372737−271757−17−37)(48−3)\mathbf{x}=\mathbf{A}^{-1}\mathbf{b}=\begin{pmatrix}-\frac{1}{7} & \frac{3}{7} & \frac{2}{7} \\ \frac{3}{7} & -\frac{2}{7} & \frac{1}{7} \\ \frac{5}{7} & -\frac{1}{7} & -\frac{3}{7}\end{pmatrix}\begin{pmatrix}4 \\ 8 \\ -3\end{pmatrix}

    This isolates the column vector of unknowns.

  8. Carry out the matrix-vector multiplication

    x=(2−13)\mathbf{x}=\begin{pmatrix}2 \\ -1 \\ 3\end{pmatrix}

    Each entry is a row of the inverse dotted with the right-hand side.

  9. Check equation 1 by substitution

    (1)(2)+(1)(−1)+(1)(3)=4(1)(2)+(1)(-1)+(1)(3)=4

    The solution satisfies the original equation.

  10. Check equation 2 by substitution

    (2)(2)+(−1)(−1)+(1)(3)=8(2)(2)+(-1)(-1)+(1)(3)=8

    The solution satisfies the original equation.

  11. State the values of the unknowns

    x=2, y=−1, z=3x=2,\ y=-1,\ z=3

    These values satisfy every equation simultaneously.

  12. Recall how matrices are added

    (A+B)ij=aij+bij(\mathbf{A}+\mathbf{B})_{ij}=a_{ij}+b_{ij}

    Addition of matrices is carried out entry by entry.

  13. Recall the rule for scalar multiplication

    (λA)ij=λaij(\lambda\mathbf{A})_{ij}=\lambda a_{ij}

    Every entry of the matrix is multiplied by the scalar.

  14. Recall the rule for matrix multiplication

    (AB)ij=∑kaikbkj(\mathbf{A}\mathbf{B})_{ij}=\sum_{k} a_{ik}b_{kj}

    Each entry is a row of the first matrix dotted with a column of the second.

  15. Select the matching option

    x=2, y=−1, z=3x=2,\ y=-1,\ z=3

    This is the unique solution.

Answer
x=2, y=−1, z=3x=2,\ y=-1,\ z=3
Question 2
8 markschallenging
Given A=(23141−2503)\mathbf{A}=\begin{pmatrix}2 & 3 & 1 \\ 4 & 1 & -2 \\ 5 & 0 & 3\end{pmatrix}, which of the following is det⁡(A)\det(\mathbf{A})?
Show worked solution

Worked solution

  1. Identify the quantity required

    det⁡(A)\det(\mathbf{A})

    The expression asks for a single numerical value.

  2. Evaluate it

    det⁡(A)=−65\det(\mathbf{A})=-65

    Apply the determinant rules.

  3. Compare with the given options

    −65-65

    Only one option matches the computed value.

  4. Choose to expand along the first row

    det⁡A=2M11−3M12+1M13\det\mathbf{A}=2 M_{11}-3 M_{12}+1 M_{13}

    Here M1jM_{1j} is the minor obtained by deleting row 1 and column jj.

  5. Evaluate the minor for entry 2

    det⁡(1−203)=(1)(3)−(−2)(0)=3\det\begin{pmatrix}1 & -2 \\ 0 & 3\end{pmatrix}=(1)(3)-(-2)(0)=3

    Delete row 1 and column 1 and take the 2×22\times 2 determinant.

  6. Evaluate the minor for entry 3

    det⁡(4−253)=(4)(3)−(−2)(5)=22\det\begin{pmatrix}4 & -2 \\ 5 & 3\end{pmatrix}=(4)(3)-(-2)(5)=22

    Delete row 1 and column 2 and take the 2×22\times 2 determinant.

  7. Evaluate the minor for entry 1

    det⁡(4150)=(4)(0)−(1)(5)=−5\det\begin{pmatrix}4 & 1 \\ 5 & 0\end{pmatrix}=(4)(0)-(1)(5)=-5

    Delete row 1 and column 3 and take the 2×22\times 2 determinant.

  8. Combine the three signed products

    det⁡A=(2)(3)−(3)(22)+(1)(−5)=−65\det\mathbf{A}=(2)(3)-(3)(22)+(1)(-5)=-65

    Apply the +,−,++,-,+ sign pattern along the first row.

  9. State whether the matrix is singular

    det⁡A=−65\det\mathbf{A}=-65

    A zero determinant would mean the matrix is singular.

  10. Recall how matrices are added

    (A+B)ij=aij+bij(\mathbf{A}+\mathbf{B})_{ij}=a_{ij}+b_{ij}

    Addition of matrices is carried out entry by entry.

  11. Recall the rule for scalar multiplication

    (λA)ij=λaij(\lambda\mathbf{A})_{ij}=\lambda a_{ij}

    Every entry of the matrix is multiplied by the scalar.

  12. Recall the rule for matrix multiplication

    (AB)ij=∑kaikbkj(\mathbf{A}\mathbf{B})_{ij}=\sum_{k} a_{ik}b_{kj}

    Each entry is a row of the first matrix dotted with a column of the second.

  13. Note that matrix multiplication is not commutative

    AB≠BA in general\mathbf{A}\mathbf{B}\neq\mathbf{B}\mathbf{A}\ \text{in general}

    The order of the factors changes the product.

  14. Recall the determinant of a 2×22\times 2 matrix

    det⁡(abcd)=ad−bc\det\begin{pmatrix}a & b \\ c & d\end{pmatrix}=ad-bc

    The determinant is the difference of the two diagonal products.

  15. Select the matching option

    −65-65

    This is the correct value.

Answer
−65-65
Question 3
8 markschallenging
Given A=(2111)\mathbf{A}=\begin{pmatrix}2 & 1 \\ 1 & 1\end{pmatrix}, which of the following is A3\mathbf{A}^3?
Show worked solution

Worked solution

  1. Identify the required calculation

    A3\mathbf{A}^3

    Decide which matrix operation the expression asks for.

  2. Carry out the calculation

    A3=(13885)\mathbf{A}^3=\begin{pmatrix}13 & 8 \\ 8 & 5\end{pmatrix}

    Apply the rules of matrix arithmetic.

  3. Compare with the given options

    (13885)\begin{pmatrix}13 & 8 \\ 8 & 5\end{pmatrix}

    Only one option matches the computed matrix.

  4. State the order of each matrix

    A is 2×2, B is 2×2\mathbf{A}\ \text{is }2\times 2,\ \mathbf{B}\ \text{is }2\times 2

    Matrices must have the same order to be added or subtracted.

  5. Write the sum entry by entry

    (2111)+(2111)\begin{pmatrix}2 & 1 \\ 1 & 1\end{pmatrix}+\begin{pmatrix}2 & 1 \\ 1 & 1\end{pmatrix}

    Combine corresponding entries of the two matrices.

  6. Work along row 1

    2+2=4,1+1=22+2=4, 1+1=2

    Each entry of row 1 is obtained independently.

  7. Work along row 2

    1+1=2,1+1=21+1=2, 1+1=2

    Each entry of row 2 is obtained independently.

  8. Assemble the resulting matrix

    (4222)\begin{pmatrix}4 & 2 \\ 2 & 2\end{pmatrix}

    Place each combined entry in its original position.

  9. Check the order of the answer

    order 2×2\text{order }2\times 2

    The result has the same order as the matrices being combined.

  10. Recall how matrices are added

    (A+B)ij=aij+bij(\mathbf{A}+\mathbf{B})_{ij}=a_{ij}+b_{ij}

    Addition of matrices is carried out entry by entry.

  11. Recall the rule for scalar multiplication

    (λA)ij=λaij(\lambda\mathbf{A})_{ij}=\lambda a_{ij}

    Every entry of the matrix is multiplied by the scalar.

  12. Recall the rule for matrix multiplication

    (AB)ij=∑kaikbkj(\mathbf{A}\mathbf{B})_{ij}=\sum_{k} a_{ik}b_{kj}

    Each entry is a row of the first matrix dotted with a column of the second.

  13. Note that matrix multiplication is not commutative

    AB≠BA in general\mathbf{A}\mathbf{B}\neq\mathbf{B}\mathbf{A}\ \text{in general}

    The order of the factors changes the product.

  14. Recall the determinant of a 2×22\times 2 matrix

    det⁡(abcd)=ad−bc\det\begin{pmatrix}a & b \\ c & d\end{pmatrix}=ad-bc

    The determinant is the difference of the two diagonal products.

  15. Select the matching option

    (13885)\begin{pmatrix}13 & 8 \\ 8 & 5\end{pmatrix}

    This is the correct matrix.

Answer
(13885)\begin{pmatrix}13 & 8 \\ 8 & 5\end{pmatrix}
Question 4
8 markschallenging
Given A=(1123)\mathbf{A}=\begin{pmatrix}1 & 1 \\ 2 & 3\end{pmatrix} and B=(2011)\mathbf{B}=\begin{pmatrix}2 & 0 \\ 1 & 1\end{pmatrix}, which of the following is (AB)−1(\mathbf{A}\mathbf{B})^{-1}?
Show worked solution

Worked solution

  1. Identify the required calculation

    (AB)−1(\mathbf{A}\mathbf{B})^{-1}

    Decide which matrix operation the expression asks for.

  2. Carry out the calculation

    (AB)−1=(32−12−7232)(\mathbf{A}\mathbf{B})^{-1}=\begin{pmatrix}\frac{3}{2} & -\frac{1}{2} \\ -\frac{7}{2} & \frac{3}{2}\end{pmatrix}

    Apply the rules of matrix arithmetic.

  3. Compare with the given options

    (32−12−7232)\begin{pmatrix}\frac{3}{2} & -\frac{1}{2} \\ -\frac{7}{2} & \frac{3}{2}\end{pmatrix}

    Only one option matches the computed matrix.

  4. Check the product is defined

    (2×2)(2×2)→(2×2)(2\times 2)(2\times 2)\to(2\times 2)

    The number of columns of the first matrix matches the rows of the second.

  5. Set up the row-by-column products

    (AB)ij=∑k=12aikbkj(\mathbf{A}\mathbf{B})_{ij}=\sum_{k=1}^{2} a_{ik}b_{kj}

    Every entry is a dot product of a row with a column.

  6. Compute the entry in row 1, column 1

    (1)(2)+(1)(1)=3(1)(2)+(1)(1)=3

    Multiply matching terms and add.

  7. Compute the entry in row 1, column 2

    (1)(0)+(1)(1)=1(1)(0)+(1)(1)=1

    Multiply matching terms and add.

  8. Compute the entry in row 2, column 1

    (2)(2)+(3)(1)=7(2)(2)+(3)(1)=7

    Multiply matching terms and add.

  9. Compute the entry in row 2, column 2

    (2)(0)+(3)(1)=3(2)(0)+(3)(1)=3

    Multiply matching terms and add.

  10. Assemble the product matrix

    (3173)\begin{pmatrix}3 & 1 \\ 7 & 3\end{pmatrix}

    Each computed entry sits in its row and column.

  11. Confirm the order of the product

    order 2×2\text{order }2\times 2

    The product has as many rows as the first matrix and columns as the second.

  12. Recall how matrices are added

    (A+B)ij=aij+bij(\mathbf{A}+\mathbf{B})_{ij}=a_{ij}+b_{ij}

    Addition of matrices is carried out entry by entry.

  13. Recall the rule for scalar multiplication

    (λA)ij=λaij(\lambda\mathbf{A})_{ij}=\lambda a_{ij}

    Every entry of the matrix is multiplied by the scalar.

  14. Recall the rule for matrix multiplication

    (AB)ij=∑kaikbkj(\mathbf{A}\mathbf{B})_{ij}=\sum_{k} a_{ik}b_{kj}

    Each entry is a row of the first matrix dotted with a column of the second.

  15. Select the matching option

    (32−12−7232)\begin{pmatrix}\frac{3}{2} & -\frac{1}{2} \\ -\frac{7}{2} & \frac{3}{2}\end{pmatrix}

    This is the correct matrix.

Answer
(32−12−7232)\begin{pmatrix}\frac{3}{2} & -\frac{1}{2} \\ -\frac{7}{2} & \frac{3}{2}\end{pmatrix}
Question 5
8 markschallenging
Use the inverse matrix method to solve the simultaneous equations 2x+y+z=52x+y+z=5, x−y+2z=7x-y+2z=7 and 3x+2y−z=43x+2y-z=4.
Show worked solution

Worked solution

  1. Write the system in matrix form

    (2111−1232−1)(xyz)=(574)\begin{pmatrix}2 & 1 & 1 \\ 1 & -1 & 2 \\ 3 & 2 & -1\end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix}=\begin{pmatrix}5 \\ 7 \\ 4\end{pmatrix}

    Collect the coefficients, unknowns and constants into matrices.

  2. Evaluate the determinant of the coefficient matrix

    det⁡A=6\det\mathbf{A}=6

    The determinant is non-zero so the inverse method applies.

  3. Write down the inverse matrix

    A−1=16(−3337−5−35−1−3)\mathbf{A}^{-1}=\frac{1}{6}\begin{pmatrix}-3 & 3 & 3 \\ 7 & -5 & -3 \\ 5 & -1 & -3\end{pmatrix}

    Divide the adjugate by the determinant.

  4. Apply x=A−1b\mathbf{x}=\mathbf{A}^{-1}\mathbf{b}

    x=(−12121276−56−1256−16−12)(574)=(3−21)\mathbf{x}=\begin{pmatrix}-\frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\ \frac{7}{6} & -\frac{5}{6} & -\frac{1}{2} \\ \frac{5}{6} & -\frac{1}{6} & -\frac{1}{2}\end{pmatrix}\begin{pmatrix}5 \\ 7 \\ 4\end{pmatrix}=\begin{pmatrix}3 \\ -2 \\ 1\end{pmatrix}

    Pre-multiplying by the inverse isolates the unknowns.

  5. Write the system in matrix form

    (2111−1232−1)(xyz)=(574)\begin{pmatrix}2 & 1 & 1 \\ 1 & -1 & 2 \\ 3 & 2 & -1\end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix}=\begin{pmatrix}5 \\ 7 \\ 4\end{pmatrix}

    Collect the coefficients into a matrix and the unknowns into a column vector.

  6. Compute the determinant of the coefficient matrix

    det⁡A=6\det\mathbf{A}=6

    A non-zero determinant guarantees a unique solution.

  7. Find the inverse of the coefficient matrix

    A−1=(−12121276−56−1256−16−12)\mathbf{A}^{-1}=\begin{pmatrix}-\frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\ \frac{7}{6} & -\frac{5}{6} & -\frac{1}{2} \\ \frac{5}{6} & -\frac{1}{6} & -\frac{1}{2}\end{pmatrix}

    Use the adjugate divided by the determinant.

  8. Pre-multiply both sides by the inverse

    x=A−1b=(−12121276−56−1256−16−12)(574)\mathbf{x}=\mathbf{A}^{-1}\mathbf{b}=\begin{pmatrix}-\frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\ \frac{7}{6} & -\frac{5}{6} & -\frac{1}{2} \\ \frac{5}{6} & -\frac{1}{6} & -\frac{1}{2}\end{pmatrix}\begin{pmatrix}5 \\ 7 \\ 4\end{pmatrix}

    This isolates the column vector of unknowns.

  9. Carry out the matrix-vector multiplication

    x=(3−21)\mathbf{x}=\begin{pmatrix}3 \\ -2 \\ 1\end{pmatrix}

    Each entry is a row of the inverse dotted with the right-hand side.

  10. Check equation 1 by substitution

    (2)(3)+(1)(−2)+(1)(1)=5(2)(3)+(1)(-2)+(1)(1)=5

    The solution satisfies the original equation.

  11. Check equation 2 by substitution

    (1)(3)+(−1)(−2)+(2)(1)=7(1)(3)+(-1)(-2)+(2)(1)=7

    The solution satisfies the original equation.

  12. State the values of the unknowns

    x=3, y=−2, z=1x=3,\ y=-2,\ z=1

    These values satisfy every equation simultaneously.

  13. Recall how matrices are added

    (A+B)ij=aij+bij(\mathbf{A}+\mathbf{B})_{ij}=a_{ij}+b_{ij}

    Addition of matrices is carried out entry by entry.

  14. Recall the rule for scalar multiplication

    (λA)ij=λaij(\lambda\mathbf{A})_{ij}=\lambda a_{ij}

    Every entry of the matrix is multiplied by the scalar.

  15. State the solution of the system

    x=3, y=−2, z=1x=3,\ y=-2,\ z=1

    These values satisfy all of the equations.

Answer
x=3, y=−2, z=1x=3,\ y=-2,\ z=1

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