AS June 2018 Paper 1 Q4

OCR ACurrent spec7 marksMatrices

4 The matrix \(\mathbf{A}\) is given by \(\mathbf{A} = \begin{pmatrix} 2 & 1 & 2 \\ 1 & -1 & 1 \\ 2 & 2 & a \end{pmatrix}\).

(i) Show that \(\det\mathbf{A} = 6 - 3a\). [2]

(ii) State the value of \(a\) for which \(\mathbf{A}\) is singular. [1]

(iii) Given that \(\mathbf{A}\) is non-singular find \(\mathbf{A}^{-1}\) in terms of \(a\). [4]

Mark scheme (i)

Scheme	Marks	AO
e.g. \(2(-a - 2) - 1(a - 2) + 2(2 + 2)\) (1^st row) or \(2(-a - 2) - 1(a - 4) + 2(1 + 2)\) (1^st col)	M1	1.1a
\(= -2a - 4 - a + 2 + 8 = 6 - 3a\)	A1 (AG)	1.1
	[2]

Notes

M1: Attempt to expand determinant. Could use any row or column or other method

A1: Must be convincing

Scheme	Marks	AO
2	B1	2.2a
	[1]

Mark scheme (iii)

Scheme	Marks	AO
Matrix of cofactors: \(\begin{pmatrix} -a - 2 & 2 - a & 4 \\ 4 - a & 2a - 4 & -2 \\ 3 & 0 & -3 \end{pmatrix}\)	M1*	1.1a
	A1	1.1
	M1dep*	1.1
\(\mathbf{A}^{-1} = \dfrac{1}{6 - 3a}\begin{pmatrix} -a - 2 & 4 - a & 3 \\ 2 - a & 2a - 4 & 0 \\ 4 & -2 & -3 \end{pmatrix}\)	A1	1.1
	[4]

Notes

M1*: At least 4 co-factors correct, or correct apart from sign. Could be seen in separate calculations or in \(\mathbf{A}^{-1}\). Could be transposed, even if stated as matrix of cofactors. If ambiguity use \(\mathbf{A}^{-1}\). If \(\mathbf{A}^{-1}\) not given then make whichever assumption, transposed or not, which results in most marks.
If not anywhere in matrix form only award M1 if it is clear where the cofactors come from. Cofactor must not be multiplied by anything
Alternative method using cross product also ok.
Matrix of cofactors is given by \((C_2 \times C_3, C_3 \times C_1\ C_1 \times C_2)\)

A1: (1st) 6 cofactors correct
Must include correct sign.

M1dep*: Transposing matrix of cofactors and dividing by determinant

A1: (2nd) cao