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Matrices 1 — chapter review — 9231 FP1

Use cofactors or elimination for a determinant and inverse. Connect a transformation matrix to area and invariant directions.

Notes and key methods

A 3×3 determinant can be expanded along a row or found using determinant-preserving row operations. A row swap reverses its sign; multiplying a row scales it by the same factor.

An inverse exists when the determinant is nonzero. Solve \(\displaystyle AX=I\) by elimination or use cofactors and transpose the cofactor matrix. Check a proposed inverse by multiplying it on both sides.

The columns of a 2×2 transformation matrix are the images of the coordinate basis vectors. Composition acts right to left on column vectors: AB applies B first and A second.

An invariant line through the origin has an eigenvector direction. Solve \(\displaystyle A(1,m)^{T}=\lambda (1,m)^{T}\) for a nonvertical direction and separately check the vertical axis, which cannot be represented by a finite m. Area scales by |det A|.

Check the conditions

Three worked examples

Example 1 · Three-by-three matrices

\(A=\left[\begin{matrix}5 & 1 & 0\\0 & 6 & 1\\1 & 0 & 7\end{matrix}\right]\). For the system question use \(b=\left[\begin{matrix}37\\51\\69\end{matrix}\right]\).

  1. Find det A.
  2. Find the (1,1) entry of A inverse.
  3. Find the first coordinate of the solution of Av=b.
  4. Find trace(\(\displaystyle A^{2}\)).
Show the reasoning

(a)

\(\det A=5(6)(7)+1=211\)

Answer: \(211\).

(b)

\(\displaystyle A^{-1}=\left[\begin{matrix}\frac{42}{211} & - \frac{7}{211} & \frac{1}{211}\\\frac{1}{211} & \frac{35}{211} & - \frac{5}{211}\\- \frac{6}{211} & \frac{1}{211} & \frac{30}{211}\end{matrix}\right]\)

Answer: \(\displaystyle \frac{42}{211}\).

(c)

\(v=\left[\begin{matrix}6\\7\\9\end{matrix}\right]\)

Answer: \(6\).

(d)

\(\operatorname{tr}(A^2)=110\)

Answer: \(110\).

Example 2 · Linear transformations and invariant lines

\(A=\left[\begin{matrix}6 & 7\\0 & 9\end{matrix}\right],\ B=\left[\begin{matrix}1 & 0\\1 & 1\end{matrix}\right]\). Use column vectors; AB means B acts first.

  1. Find the area scale factor of A.
  2. Find the nonzero finite slope of an invariant line of A through the origin.
  3. Find the (1,1) entry of AB.
  4. Find the (2,1) entry of AB.
Show the reasoning

(a)

\(|\det A|=54\)

Answer: \(54\).

(b)

\(\displaystyle 6m+ 7m^{2}= 9m\implies m= \frac{9- 6}{7}= \frac{3}{7}\)

Answer: \(\displaystyle \frac{3}{7}\).

(c)

\(AB=\left[\begin{matrix}13 & 7\\9 & 9\end{matrix}\right]\)

Answer: \(13\).

(d)

\(AB=\left[\begin{matrix}13 & 7\\9 & 9\end{matrix}\right]\)

Answer: \(9\).

Example 3 · Three-by-three matrices

\(A=\left[\begin{matrix}7 & 1 & 0\\0 & 8 & 1\\1 & 0 & 9\end{matrix}\right]\). For the system question use \(b=\left[\begin{matrix}65\\83\\107\end{matrix}\right]\).

  1. Find det A.
  2. Find the (1,1) entry of A inverse.
  3. Find the first coordinate of the solution of Av=b.
  4. Find trace(\(\displaystyle A^{2}\)).
Show the reasoning

(a)

\(\det A=7(8)(9)+1=505\)

Answer: \(505\).

(b)

\(\displaystyle A^{-1}=\left[\begin{matrix}\frac{72}{505} & - \frac{9}{505} & \frac{1}{505}\\\frac{1}{505} & \frac{63}{505} & - \frac{7}{505}\\- \frac{8}{505} & \frac{1}{505} & \frac{56}{505}\end{matrix}\right]\)

Answer: \(\displaystyle \frac{72}{505}\).

(c)

\(v=\left[\begin{matrix}8\\9\\11\end{matrix}\right]\)

Answer: \(8\).

(d)

\(\operatorname{tr}(A^2)=194\)

Answer: \(194\).

Common mistakes

Practise with answer checking

Original exam-style problems

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