Q.Find the inverse of the following matrix, if it exists: $A = \begin{bmatrix} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{bmatrix}$

Concept: Lower triangular matrix — its inverse (if it exists) is also lower triangular, found by forward substitution.

Step 1: Check determinant. For a triangular matrix, $\det(A) = 1 \cdot 3 \cdot (-1) = -3 \neq 0$, so the inverse exists.

Step 2: Let $A^{-1} = \begin{bmatrix} a & 0 & 0 \\ b & c & 0 \\ d & e & f \end{bmatrix}$. Multiply $A \cdot A^{-1} = I$:

Row 1: $1 \cdot a = 1 \Rightarrow a = 1$.

Row 2: $3a + 3b = 0 \Rightarrow 3(1) + 3b = 0 \Rightarrow b = -1$; and $3c = 1 \Rightarrow c = \frac{1}{3}$.

Row 3: $5a + 2b - d = 0 \Rightarrow 5(1) + 2(-1) - d = 0 \Rightarrow d = 3$; …

Why This Problem Is Almost Solved Before You Start

The matrix $A$ is lower triangular — every entry above the main diagonal is zero. This structure is a gift. For triangular matrices, the inverse (when it exists) is also triangular of the same type. More importantly, the diagonal entries of the inverse are simply the reciprocals of the original diagonal entries. That alone gives us three entries for free.

The diagonal of $A$ is $(1, 3, -1)$. None are zero, so the matrix is invertible. The inverse will have diagonal $(1, \frac13, -1)$.

Now we only need to find the six entries below the diagonal. Because the matrix is $3 \times 3$, we can do this cleanly with forward substitution — solving $A X = I$ column by column.

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Step-by-Step Solution

Let $X = A^{-1}$. Write $X$ as a lower triangular matrix with unknown entries:

We know $x_{11} = 1$, $x_{22} = \frac13$, $x_{33} = -1$ from the diagonal rule. So:

Now solve $A X = I$ column by column.

1. First column of $X$ — solve $A \cdot \text{col}_1(X) = e_1$:

Row 1: $1 \cdot 1 = 1$ — checks out.

Row 2: $3 \cdot 1 + 3 \cdot x_{21} = 0 \implies 3 + 3x_{21} = 0 \implies x_{21} = -1$.

Row 3: $5 \cdot 1 + 2 \cdot x_{21} + (-1) \cdot x_{31} = 0 \implies 5 + 2(-1) - x_{31} = 0 \implies 5 - 2 - x_{31} = 0 \implies x_{31} = 3$.

So the first column is $(1, -1, 3)^T$.

2. Second column of $X$ — solve $A \cdot \text{col}_2(X) = e_2$:

Row 1: $1 \cdot 0 = 0$ — fine.

Row 2: $3 \cdot 0 + 3 \cdot \frac13 = 1$ — checks out.

Row 3: $5 \cdot 0 + 2 \cdot \frac13 + (-1) \cdot x_{32} = 0 \implies \frac23 - x_{32} = 0 \implies x_{32} = \frac23$.

So the second column is $(0, \frac13, \frac23)^T$. …

Method: Finding the Inverse of a Triangular Matrix by Forward Substitution

A shortcut method for triangular matrices that avoids full cofactor/adjoint computation.

Steps

Step 1: Confirm invertibility from the diagonal

For a triangular matrix, $\det(A)$ is the product of the diagonal entries; the inverse exists exactly when none of them is $0$.

Step 2: Use the fact that the inverse is triangular of the same type, with reciprocal diagonal entries

If $A$ is lower triangular with diagonal $(d_1,d_2,d_3)$, then $A^{-1}$ is also lower triangular, with diagonal $(1/d_1,1/d_2,1/d_3)$ — these three entries are known immediately, with no computation.

Step 3: Solve $A\,X = I$ column by column for the remaining unknown entries …

Common Mistakes

Mistake 1: Forgetting that the inverse of a lower triangular matrix is also lower triangular, and trying to fill in the upper-triangular entries

Why it's wrong: this wastes time solving for entries that are guaranteed to be $0$, and risks introducing a spurious nonzero value there by mistake. Correct approach: use the guaranteed-zero upper-triangular entries as free information and only solve for the lower-triangular unknowns.

Mistake 2: Solving the columns of $AX=I$ out of order, or mixing up which equation corresponds to which unknown …