Further Maths Matrices: arithmetic and determinants Practice Questions

Free Further Maths Matrices: arithmetic and determinants practice questions with full step-by-step worked solutions. Covers matrices, addition, 2x2, subtraction. Practise exam-style problems and check your method.

matricesaddition2x2subtractionscalar-multiplicationmultiplication
Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
Given A=(1234)\mathbf{A}=\begin{pmatrix}1 & 2 \\ 3 & 4\end{pmatrix} and B=(5−102)\mathbf{B}=\begin{pmatrix}5 & -1 \\ 0 & 2\end{pmatrix}, find A+B\mathbf{A}+\mathbf{B}.
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Worked solution

  1. Add corresponding entries

    (1234)+(5−102)\begin{pmatrix}1 & 2 \\ 3 & 4\end{pmatrix}+\begin{pmatrix}5 & -1 \\ 0 & 2\end{pmatrix}

    Matrix addition works entry by entry.

  2. Simplify each entry

    (6136)\begin{pmatrix}6 & 1 \\ 3 & 6\end{pmatrix}

    Add the two numbers in each position.

  3. State the resulting matrix

    A+B=(6136)\mathbf{A}+\mathbf{B}=\begin{pmatrix}6 & 1 \\ 3 & 6\end{pmatrix}

    This is A+B\mathbf{A}+\mathbf{B}.

Answer
(6136)\begin{pmatrix}6 & 1 \\ 3 & 6\end{pmatrix}
Question 2
2 markseasy
Given A=(−3251)\mathbf{A}=\begin{pmatrix}-3 & 2 \\ 5 & 1\end{pmatrix}, which of the following is det⁡(A)\det(\mathbf{A})?
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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)=−13\det(\mathbf{A})=-13

    Apply the determinant rules.

  3. Select the matching option

    −13-13

    This is the correct value.

Answer
−13-13
Question 3
3 marksintermediate
The matrix M=(5k24)\mathbf{M}=\begin{pmatrix}5 & k \\ 2 & 4\end{pmatrix} is singular. Which of the following is the value of kk?
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Worked solution

  1. Set the determinant equal to zero

    det⁡M=0\det\mathbf{M}=0

    This is the condition for a singular matrix.

  2. Expand the determinant

    det⁡M=−2k+20\det\mathbf{M}=-2k+20

    Treat kk as an unknown constant.

  3. Solve for kk

    k=10k=10

    Rearrange the linear equation.

  4. Recall the singularity condition

    det⁡M=0\det\mathbf{M}=0

    A singular matrix has no inverse, so its determinant vanishes.

  5. Expand the determinant in terms of kk

    det⁡M=−2k+20\det\mathbf{M}=-2k+20

    Treat kk as an unknown and expand as usual.

  6. Select the correct value of kk

    1010

    This value makes the determinant zero.

Answer
1010
Question 4
5 markshard
Given A=(10221−1031)\mathbf{A}=\begin{pmatrix}1 & 0 & 2 \\ 2 & 1 & -1 \\ 0 & 3 & 1\end{pmatrix} and B=(2101−13021)\mathbf{B}=\begin{pmatrix}2 & 1 & 0 \\ 1 & -1 & 3 \\ 0 & 2 & 1\end{pmatrix}, which of the following is AB\mathbf{A}\mathbf{B}?
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Worked solution

  1. Identify the required calculation

    AB\mathbf{A}\mathbf{B}

    Decide which matrix operation the expression asks for.

  2. Carry out the calculation

    AB=(2525−123−110)\mathbf{A}\mathbf{B}=\begin{pmatrix}2 & 5 & 2 \\ 5 & -1 & 2 \\ 3 & -1 & 10\end{pmatrix}

    Apply the rules of matrix arithmetic.

  3. Compare with the given options

    (2525−123−110)\begin{pmatrix}2 & 5 & 2 \\ 5 & -1 & 2 \\ 3 & -1 & 10\end{pmatrix}

    Only one option matches the computed matrix.

  4. Check the product is defined

    (3×3)(3×3)→(3×3)(3\times 3)(3\times 3)\to(3\times 3)

    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=13aikbkj(\mathbf{A}\mathbf{B})_{ij}=\sum_{k=1}^{3} 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)+(0)(1)+(2)(0)=2(1)(2)+(0)(1)+(2)(0)=2

    Multiply matching terms and add.

  7. Compute the entry in row 1, column 2

    (1)(1)+(0)(−1)+(2)(2)=5(1)(1)+(0)(-1)+(2)(2)=5

    Multiply matching terms and add.

  8. Compute the entry in row 1, column 3

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

    Multiply matching terms and add.

  9. Compute the entry in row 2, column 1

    (2)(2)+(1)(1)+(−1)(0)=5(2)(2)+(1)(1)+(-1)(0)=5

    Multiply matching terms and add.

  10. Select the matching option

    (2525−123−110)\begin{pmatrix}2 & 5 & 2 \\ 5 & -1 & 2 \\ 3 & -1 & 10\end{pmatrix}

    This is the correct matrix.

Answer
(2525−123−110)\begin{pmatrix}2 & 5 & 2 \\ 5 & -1 & 2 \\ 3 & -1 & 10\end{pmatrix}
Question 5
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?
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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

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