QCAA Specialist Mathematics Alternative Sequence Matrices

3 sample questions with marking guides and sample answers

Q1
2020
QCAA
Paper 1
1 mark
Q1
1 mark

The matrix equation [4−231]X=[1−1]\begin{bmatrix} 4 & -2 \\ 3 & 1 \end{bmatrix} \mathbf{X} = \begin{bmatrix} 1 \\ -1 \end{bmatrix} has the solution of

A

X=−12[1−1][12−34]\mathbf{X} = \frac{-1}{2} \begin{bmatrix} 1 \\ -1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ -3 & 4 \end{bmatrix}

B

X=−12[1−1][4−231]\mathbf{X} = \frac{-1}{2} \begin{bmatrix} 1 \\ -1 \end{bmatrix} \begin{bmatrix} 4 & -2 \\ 3 & 1 \end{bmatrix}

C

X=110[12−34][1−1]\mathbf{X} = \frac{1}{10} \begin{bmatrix} 1 & 2 \\ -3 & 4 \end{bmatrix} \begin{bmatrix} 1 \\ -1 \end{bmatrix}

D

X=110[4−231][1−1]\mathbf{X} = \frac{1}{10} \begin{bmatrix} 4 & -2 \\ 3 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ -1 \end{bmatrix}

Reveal Answer
A

X=−12[1−1][12−34]\mathbf{X} = \frac{-1}{2} \begin{bmatrix} 1 \\ -1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ -3 & 4 \end{bmatrix}

This option incorrectly calculates the determinant as −2-2 instead of 1010. Additionally, matrix multiplication is not commutative, so the inverse matrix must be multiplied on the left side of the constant vector, not the right.

B

X=−12[1−1][4−231]\mathbf{X} = \frac{-1}{2} \begin{bmatrix} 1 \\ -1 \end{bmatrix} \begin{bmatrix} 4 & -2 \\ 3 & 1 \end{bmatrix}

This option uses an incorrect determinant of −2-2, fails to find the adjugate matrix (it uses the original matrix instead), and multiplies the matrices in the wrong order.

C

X=110[12−34][1−1]\mathbf{X} = \frac{1}{10} \begin{bmatrix} 1 & 2 \\ -3 & 4 \end{bmatrix} \begin{bmatrix} 1 \\ -1 \end{bmatrix}

Correct Answer

To solve AX=BA\mathbf{X} = B, we multiply both sides by A−1A^{-1} to get X=A−1B\mathbf{X} = A^{-1}B. The inverse of the 2×22 \times 2 matrix is found by dividing by the determinant (4(1)−(−2)(3)=104(1) - (-2)(3) = 10) and finding the adjugate matrix, resulting in 110[12−34]\frac{1}{10} \begin{bmatrix} 1 & 2 \\ -3 & 4 \end{bmatrix}.

D

X=110[4−231][1−1]\mathbf{X} = \frac{1}{10} \begin{bmatrix} 4 & -2 \\ 3 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ -1 \end{bmatrix}

While this option correctly calculates the determinant as 1010, it fails to find the adjugate matrix by swapping the diagonal elements and negating the off-diagonal elements.

Q5
2020
QCAA
Paper 2
1 mark
Q5
1 mark

Solve the matrix equation for X\mathbf{X}.

[0123]X[4567]=[8901]\begin{bmatrix} 0 & 1 \\ 2 & 3 \end{bmatrix} \mathbf{X} \begin{bmatrix} 4 & 5 \\ 6 & 7 \end{bmatrix} = \begin{bmatrix} 8 & 9 \\ 0 & 1 \end{bmatrix}

A

[−9−944]\begin{bmatrix} -9 & -9 \\ 4 & 4 \end{bmatrix}

B

[3−4−12]\begin{bmatrix} 3 & -4 \\ -1 & 2 \end{bmatrix}

C

[13−14−1112]\begin{bmatrix} 13 & -14 \\ -11 & 12 \end{bmatrix}

D

[5456−28−29]\begin{bmatrix} 54 & 56 \\ -28 & -29 \end{bmatrix}

Reveal Answer
A

[−9−944]\begin{bmatrix} -9 & -9 \\ 4 & 4 \end{bmatrix}

This is incorrect. This matrix does not satisfy the equation and likely results from arithmetic errors during matrix inversion or multiplication.

B

[3−4−12]\begin{bmatrix} 3 & -4 \\ -1 & 2 \end{bmatrix}

Correct Answer

This is correct. To isolate X\mathbf{X} in the equation AXB=CA\mathbf{X}B = C, you must multiply both sides by A−1A^{-1} on the left and B−1B^{-1} on the right, yielding X=A−1CB−1\mathbf{X} = A^{-1}CB^{-1}.

C

[13−14−1112]\begin{bmatrix} 13 & -14 \\ -11 & 12 \end{bmatrix}

This is incorrect. This is the result of calculating B−1CA−1B^{-1}CA^{-1}, which incorrectly applies the inverses to the wrong sides of CC. Remember that matrix multiplication is not commutative.

D

[5456−28−29]\begin{bmatrix} 54 & 56 \\ -28 & -29 \end{bmatrix}

This is incorrect. This matrix does not satisfy the equation and likely results from incorrect matrix operations, such as multiplying the matrices without taking their inverses.

Q11
2022
QCAA
Paper 1
6 marks
Q11

Consider the simultaneous equations
4x+3y=−24x + 3y = -2
−2x+y=3-2x + y = 3

Q11a
1 mark

Express these equations in the form AX=B\mathbf{AX} = \mathbf{B} where A\mathbf{A} is a 2×22 \times 2 matrix and X\mathbf{X} and B\mathbf{B} are column vectors.

Reveal Answer

[43−21][xy]=[−23]\begin{bmatrix} 4 & 3 \\ -2 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} -2 \\ 3 \end{bmatrix}

Marking Criteria
DescriptorMarks

correctly expresses the simultaneous equations in AX=B\mathbf{AX} = \mathbf{B} form

1
Q11b
1 mark

Calculate A−1\mathbf{A}^{-1}.

Reveal Answer

A−1=110[1−324]\mathbf{A}^{-1} = \frac{1}{10} \begin{bmatrix} 1 & -3 \\ 2 & 4 \end{bmatrix}

Marking Criteria
DescriptorMarks

calculates A−1\mathbf{A}^{-1}

1
Q11c
2 marks

Use matrix algebra to solve the equation formed in Question 11a).

Reveal Answer

[xy]=110[1−324][−23]\begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{10} \begin{bmatrix} 1 & -3 \\ 2 & 4 \end{bmatrix} \begin{bmatrix} -2 \\ 3 \end{bmatrix}
=110[−118]= \frac{1}{10} \begin{bmatrix} -11 \\ 8 \end{bmatrix}
=[−1.10.8]= \begin{bmatrix} -1.1 \\ 0.8 \end{bmatrix}
x=−1.1,y=0.8x = -1.1, y = 0.8

Marking Criteria
DescriptorMarks

demonstrates the use of matrix algebra to determine an expression for X\mathbf{X}

1

determines solution

1
Q11d
2 marks

Evaluate the reasonableness of your solution for Question 11c).

Reveal Answer

Substituting x=−1.1,y=0.8x = -1.1, y = 0.8 into the simultaneous equations
4(−1.1)+3(0.8)=−4.4+2.4=−24(-1.1) + 3(0.8) = -4.4 + 2.4 = -2 (as required)
−2(−1.1)+(0.8)=2.2+0.8=3-2(-1.1) + (0.8) = 2.2 + 0.8 = 3 (as required)

The solution is reasonable.

Marking Criteria
DescriptorMarks

verifies solution by substituting results into one of the given equations

1

verifies solution by substituting results into the remaining equation and communicates that the result is verified

1

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