Inverse of a matrix (adjoint and row reduction)

At a Glance

• Frequency: 3 sub-parts across 3 of 13 years (2013, 2016, 2024)
• Priority tier: T3
• Marks (count): 10 (1), 15 (1), 6 (1)
• Average solve time: ~7 min
• Difficulty mix: easy 3
• Section: A | Dominant type: computation

Why This Chapter Matters

Matrix inversion by row operations is the most reliably “easy” computation in the linear algebra section — all three appearances are rated easy, and two ask you to use the result to solve a linear system (doubling the payoff for the same inverse computation). The Gauss–Jordan method ($[A|I]\to[I|A^{-1}]$) is a universal technique that never fails when $A$ is invertible. The 2024 variant (operator invertibility) extends naturally: write the matrix, read off the triangular determinant, back-substitute for $T^{-1}$. These are guaranteed full-marks questions once the technique is automatic.

Minimum Theory

Existence. An $n\times n$ matrix $A$ is invertible iff $\det A\neq 0$. Equivalently, $A$ has full rank $n$.

Gauss–Jordan method. Form the augmented matrix $[A\,|\,I_n]$. Apply elementary row operations to reduce the left block to $I_n$. The right block becomes $A^{-1}$: $[A\,|\,I]\;\xrightarrow{\text{row ops}}\;[I\,|\,A^{-1}].$

Adjoint (classical adjugate). $A^{-1}=\dfrac{1}{\det A}\,\operatorname{adj}(A)$, where $\operatorname{adj}(A)_{ij}=(-1)^{i+j}M_{ji}$ (cofactors, transposed). Efficient for $2\times2$ and $3\times3$ matrices; cumbersome for larger ones.

Solving $A\mathbf x=\mathbf b$ via the inverse. Once $A^{-1}$ is known, $\mathbf x=A^{-1}\mathbf b$ — a single matrix–vector multiplication. This is why computing $A^{-1}$ and then solving the system is profitable when the same matrix appears in multiple right-hand-sides.

Triangular matrices. The determinant of a triangular matrix is the product of its diagonal entries. For a lower-triangular matrix, inversion by back-substitution (solving column by column) is faster than full Gauss–Jordan.

Invertibility of a linear operator. For $T:\mathbb R^n\to\mathbb R^n$, invertibility is equivalent to $\det[T]\neq0$. $T^{-1}$ can be found by solving $T(x)=y$ for $x$ in terms of $y$.

Question Archetypes

inverse-by-row-ops (2 question(s); 2013, 2016)

Recognition Cues

• “Find the inverse of the matrix $A$ using elementary row operations.”
• Often followed by “hence solve the system $A\mathbf x=\mathbf b$.”
• A $3\times3$ integer matrix is given.

Solution Template

1. Check invertibility. Compute $\det A$; if $\det A=0$, the inverse does not exist.
2. Form $[A\,|\,I_3]$.
3. Forward elimination. Clear below each pivot using row operations; normalise each pivot to 1.
4. Back elimination. Clear above each pivot.
5. Read off $A^{-1}$ from the right block.
6. Solve $A\mathbf x=\mathbf b$ (if required): $\mathbf x=A^{-1}\mathbf b$.
7. Verify: substitute $\mathbf x$ back into the original system.

Worked Example(s)

2013 Paper 1, 2013-P1-Q1a (10 marks)

Find the inverse of $A=\begin{pmatrix}1&3&1\\2&-1&7\\3&2&-1\end{pmatrix}$ by elementary row operations. Hence solve $x+3y+z=10$, $2x-y+7z=21$, $3x+2y-z=4$.

Step 1. $\det A=1(1-14)-3(-2-21)+1(4+3)=-13+69+7=63\neq0$. ✓

Step 2. Augment: $\left[\begin{array}{ccc|ccc}1&3&1&1&0&0\\2&-1&7&0&1&0\\3&2&-1&0&0&1\end{array}\right].$

Step 3 — Forward sweep. $R_2\to R_2-2R_1$, $R_3\to R_3-3R_1$: $\left[\begin{array}{ccc|ccc}1&3&1&1&0&0\\0&-7&5&-2&1&0\\0&-7&-4&-3&0&1\end{array}\right].$ $R_3\to R_3-R_2$: $\left[\begin{array}{ccc|ccc}1&3&1&1&0&0\\0&-7&5&-2&1&0\\0&0&-9&-1&-1&1\end{array}\right].$ $R_2\to-\tfrac{1}{7}R_2$, $R_3\to-\tfrac{1}{9}R_3$: $\left[\begin{array}{ccc|ccc}1&3&1&1&0&0\\0&1&-5/7&2/7&-1/7&0\\0&0&1&1/9&1/9&-1/9\end{array}\right].$

Step 4 — Back sweep. $R_2\to R_2+\tfrac{5}{7}R_3$: $R_2=(0,1,0\;|\;23/63,\,-4/63,\,-5/63).$ $R_1\to R_1-R_3$: $(1,3,0\;|\;8/9,\,-1/9,\,1/9)$. Then $R_1\to R_1-3R_2$: $R_1=\bigl(1,0,0\;\big|\;8/9-69/63,\;-1/9+12/63,\;1/9+15/63\bigr)=\bigl(1,0,0\;\big|\;-13/63,\;5/63,\;22/63\bigr).$

Step 5 — Inverse. $\boxed{\;A^{-1}=\frac{1}{63}\begin{pmatrix}-13&5&22\\23&-4&-5\\7&7&-7\end{pmatrix}.\;}$

Step 6 — Solve. $\mathbf x=A^{-1}(10,21,4)^T=\dfrac{1}{63}(-130+105+88,\;230-84-20,\;70+147-28)^T=\dfrac{1}{63}(63,126,189)^T$: $\boxed{\;x=1,\;y=2,\;z=3.\;}$

Verify: $1+6+3=10$ ✓, $2-2+21=21$ ✓, $3+4-3=4$ ✓.

2016 Paper 1, 2016-P1-Q1a-i (6 marks)

Using elementary row operations, find the inverse of $A=\begin{pmatrix}1&2&1\\1&3&2\\1&0&1\end{pmatrix}$.

Step 1. $\det A=1(3-0)-2(1-2)+1(0-3)=3+2-3=2\neq0$. ✓

Step 2–3 — Forward sweep. $R_2\to R_2-R_1$, $R_3\to R_3-R_1$; then $R_1\to R_1-2R_2$, $R_3\to R_3+2R_2$; then $R_3\to\tfrac{1}{2}R_3$ (pivot $=2$): $\left[\begin{array}{ccc|ccc}1&0&-1&3&-2&0\\0&1&1&-1&1&0\\0&0&1&-3/2&1&1/2\end{array}\right].$

Step 4 — Back sweep. $R_1\to R_1+R_3$, $R_2\to R_2-R_3$: $\boxed{\;A^{-1}=\frac{1}{2}\begin{pmatrix}3&-2&1\\1&0&-1\\-3&2&1\end{pmatrix}.\;}$

Verify: $AA^{-1}$: row 1 of $A$ $(1,2,1)$ times col 1 of $A^{-1}$ $(3/2,1/2,-3/2)$: $3/2+1-3/2=1$ ✓.

Common Traps

• The third pivot in the 2013 matrix is $-9$, not $1$. Divide $R_3$ by $-9$ before clearing above; failing to normalise gives off-by-factor errors in $A^{-1}$.
• Carry exact fractions. For the 2013 matrix, denominator $63=7\times 9$ appears throughout. Converting to decimals invites rounding. The final answer can be checked by $\det A\cdot A^{-1}$ being an integer matrix.
• The third pivot in the 2016 matrix is $2$, not $1$. Divide $R_3$ by $2$ before back-clearing.
• Always verify $A\cdot A^{-1}=I$ (one row, one column is sufficient) and, if a system is also being solved, substitute back into the original equations.
• Confirm $\det A\neq0$ before starting — stating the determinant is non-zero is a required mark.

operator-invertibility (1 question(s); 2024)

Recognition Cues

• “Is the linear operator $T$ invertible? If yes, find $T^{-1}$.”
• $T$ is defined by an explicit formula, e.g. $T(x,y,z)=(2x,\ldots)$.
• The matrix of $T$ is typically triangular, making the determinant trivial to read off.

Solution Template

1. Write the matrix. Read off $[T]$ from the formula: column $j$ is $T(e_j)$.
2. Compute $\det[T]$. For triangular matrices, this is the product of diagonal entries.
3. Conclude invertibility ($\det[T]\neq0$) or non-invertibility ($\det[T]=0$).
4. Find $T^{-1}$ by solving $T(x,y,z)=(a,b,c)$ for $(x,y,z)$ in terms of $(a,b,c)$ (back-substitution for triangular $[T]$).
5. Verify: compute $T(T^{-1}(a,b,c))=(a,b,c)$.

Worked Example

2024 Paper 1, 2024-P1-Q2a (15 marks)

Consider $T:\mathbb R^3\to\mathbb R^3$ defined by $T(x,y,z)=(2x,\;4x-y,\;2x+3y-z)$. Is $T$ invertible? If yes, find $T^{-1}$.

Step 1 — Matrix. $[T]=\begin{pmatrix}2&0&0\\4&-1&0\\2&3&-1\end{pmatrix}\quad\text{(lower triangular).}$

Step 2 — Determinant. Product of diagonal entries: $\det[T]=2\cdot(-1)\cdot(-1)=2\neq0$.

Step 3 — Invertible. $\det[T]\neq0$, so $T$ is invertible. $\checkmark$

Step 4 — Compute $T^{-1}$. Solve $T(x,y,z)=(a,b,c)$:

• $2x=a\;\Rightarrow\;x=a/2$.
• $4x-y=b\;\Rightarrow\;y=4x-b=2a-b$.
• $2x+3y-z=c\;\Rightarrow\;z=2x+3y-c=a+3(2a-b)-c=7a-3b-c$.

$\boxed{\;T^{-1}(a,b,c)=\Bigl(\frac{a}{2},\;2a-b,\;7a-3b-c\Bigr).\;}$

Verify: $T(a/2,\,2a-b,\,7a-3b-c)=(a,\;b,\;c)$ — direct substitution confirms. ✓

Common Traps

• For a lower-triangular matrix, read the determinant directly from the diagonal. Don’t expand cofactors along a row/column when the triangular structure is obvious.
• Back-substitution order matters. For lower-triangular $[T]$, solve for $x$ from the first equation, then $y$ from the second (using $x$), then $z$ from the third (using $x$ and $y$). Going in reverse order (from $z$ first) is the standard for upper-triangular; here go top to bottom.
• The adjoint formula $T^{-1}=(1/\det)\operatorname{adj}(T)$ works but is slower for triangular matrices. Back-substitution is more efficient.
• State the invertibility conclusion before finding $T^{-1}$: “Since $\det[T]=2\neq0$, $T$ is invertible.” This sentence is worth marks on its own.

Marks-Aware Writing

6-mark question (row-ops only, 2016): Three forward sweeps + one normalisation + two back-clears. Write each row operation label ($R_2\to R_2-R_1$) on the left side of the arrow. One verification row is expected. No system to solve, so the matrix inverse is the complete answer.

10-mark question (inverse + system, 2013): Two parts — finding $A^{-1}$ (roughly 7 marks) and solving the system (3 marks). For the system, the computation is $\mathbf x=A^{-1}\mathbf b$ — one matrix–vector product — plus a three-equation verification. The verification is required and earns marks.

15-mark question (operator invertibility, 2024): Three parts — matrix (2), determinant + invertibility conclusion (4), find $T^{-1}$ by back-substitution (6), verify (3). The verify step is essential at 15 marks: compute $T(T^{-1}(a,b,c))$ explicitly.

Practice Set

• 2024-P1-Q3a (15 m) — — Hint: matrix inverse or row reduction at 15 marks.
• 2014-P1-Q2b-ii (10 m) — — Hint: inverse by row operations on a $3\times3$ matrix.
• 2021-P1-Q1a (10 m) — — Hint: row-reduce $[A|I]$; verify $A^{-1}$.
• 2020-P1-Q4a (15 m) — — Hint: matrix inverse + linear system at 15 marks.
• 2019-P1-Q1d (10 m) — — Hint: $3\times3$ inverse computation.
• 2018-P1-Q1b (10 m) — — Hint: inverse by row operations; check determinant first.