Question

Let T: R3→R3 be a L.T. and defined by T(a,b,c)=(a-b,b-c,c-a).Find the matrix with respect to basis (v1v2v3)where v1=(1,0,1),v2=(0,1,1),v3=(1,1,0) is this matrix invertible if so find inverse corresponding to formula for T-1.

Answer 1

Task. Let $T: \mathbb{R}^3 \to \mathbb{R}^3$ be a linear transformation defined by $T(a, b, c) = (a - b, b - c, c - a)$. Find the matrix with respect to basis $(v_1, v_2, v_3)$, where

$$v_1 = (1, 0, 1), \qquad v_2 = (0, 1, 1), \qquad v_3 = (1, 1, 0).$$

Is this matrix invertible? If so find inverse corresponding to formula for $T^{-1}$.

Solution. The matrix of $T$ has the following form:

$$T = \begin{pmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ -1 & 0 & 1 \end{pmatrix}$$

The transition matrix to the basis $(v_1, v_2, v_3)$ has the form

$$A = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 0 \end{pmatrix}$$

Therefore the matrix of $T$ in the basis $(v_1, v_2, v_3)$ is

$$ATA^{-1}.$$

Let us compute the inverse of $A$.

First find the determinant of $A$:

$$\begin{vmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 0 \end{vmatrix} = 1 \cdot 1 \cdot 0 + 0 \cdot 1 \cdot 1 + 1 \cdot 0 \cdot 1 - 1 \cdot 1 \cdot 1 - 1 \cdot 1 \cdot 1 - 0 \cdot 0 \cdot 0 = -2.$$

Now compute the minores

$$M_{11} = \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} = 1 \cdot 0 - 1 \cdot 1 = -1$$ $$M_{12} = \begin{vmatrix} 0 & 1 \\ 1 & 0 \end{vmatrix} = 0 \cdot 0 - 1 \cdot 1 = -1$$ $$M_{13} = \begin{vmatrix} 0 & 1 \\ 1 & 1 \end{vmatrix} = 0 \cdot 1 - 1 \cdot 1 = -1$$ $$M_{21} = \begin{vmatrix} 0 & 1 \\ 1 & 0 \end{vmatrix} = 0 \cdot 0 - 1 \cdot 1 = -1$$ $$M_{22} = \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} = 1 \cdot 0 - 1 \cdot 1 = -1$$ $$M_{23} = \begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix} = 1 \cdot 1 - 0 \cdot 1 = 1$$ $$M_{31} = \begin{vmatrix} 0 & 1 \\ 1 & 1 \end{vmatrix} = 0 \cdot 1 - 1 \cdot 1 = -1$$ $$M_{32} = \begin{vmatrix} 1 & 1 \\ 0 & 1 \end{vmatrix} = 1 \cdot 1 - 1 \cdot 0 = 1$$ $$M_{33} = \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = 1 \cdot 1 - 0 \cdot 0 = 1$$

Then the inverse of $A$ is given by the formula:

$$A^{-1} = \frac{1}{|A|} \begin{pmatrix} M_{11} & -M_{21} & M_{31} \\ -M_{12} & M_{22} & -M_{32} \\ M_{13} & -M_{23} & M_{33} \end{pmatrix} = \frac{1}{-2} \begin{pmatrix} -1 & 1 & -1 \\ 1 & -1 & -1 \\ -1 & -1 & 1 \end{pmatrix} =$$

2

Now compute

$$\begin{array}{l} AT = \left( \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 0 \end{array} \right) \left( \begin{array}{ccc} 1 & -1 & 0 \\ 0 & 1 & -1 \\ -1 & 0 & 1 \end{array} \right) \\ = \left( \begin{array}{cccc} 1 \cdot 1 + 0 \cdot 0 + 1 \cdot (-1) & 1 \cdot (-1) & + 0 \cdot 1 + 1 \cdot 0 & 1 \cdot 0 + 0 \cdot (-1) + 1 \cdot 1 \\ 0 \cdot 1 + 1 \cdot 0 + 1 \cdot (-1) & 0 \cdot (-1) & + 1 \cdot 1 + 1 \cdot 0 & 0 \cdot 0 + 1 \cdot (-1) + 1 \cdot 1 \\ 1 \cdot 1 + 1 \cdot 0 + 0 \cdot (-1) & 1 \cdot (-1) & + 1 \cdot 1 + 0 \cdot 0 & 1 \cdot 0 + 1 \cdot (-1) + 0 \cdot 1 \end{array} \right) \\ = \left( \begin{array}{ccc} 0 & -1 & 1 \\ -1 & 1 & 0 \\ 1 & 0 & -1 \end{array} \right) \end{array}$$

Hence

$$\begin{array}{l} ATA^{-1} = \left( \begin{array}{ccc} -1 & 1 & -1 \\ 1 & -1 & -1 \\ -1 & -1 & 1 \end{array} \right) \left( \begin{array}{ccc} 0 & -1 & 1 \\ -1 & 1 & 0 \\ 1 & 0 & -1 \end{array} \right) \\ = -\frac{1}{2} \left( \begin{array}{cccc} 0 \cdot (-1) - 1 \cdot 1 + 1 \cdot (-1) & 0 \cdot 1 - 1 \cdot (-1) + 1 \cdot (-1) & 0 \cdot (-1) - 1 \cdot (-1) + 1 \cdot 1 \\ -1 \cdot (-1) + 1 \cdot 1 + 0 \cdot (-1) & -1 \cdot 1 + 1 \cdot (-1) + 0 \cdot (-1) & -1 \cdot (-1) + 1 \cdot (-1) + 0 \cdot 1 \\ 1 \cdot (-1) + 0 \cdot 1 - 1 \cdot (-1) & 1 \cdot 1 + 0 \cdot (-1) - 1 \cdot (-1) & 1 \cdot (-1) + 0 \cdot (-1) - 1 \cdot 1 \end{array} \right) \\ = -\frac{1}{2} \left( \begin{array}{ccc} -2 & 0 & 2 \\ 2 & -2 & 0 \\ 0 & 2 & -2 \end{array} \right) = \left( \begin{array}{ccc} 1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{array} \right) \end{array}$$

Compute the determinant of $T$:

$$|T| = \left| \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 0 \end{array} \right| = 1 \cdot 1 \cdot 0 + 0 \cdot 1 \cdot 1 + 1 \cdot 0 \cdot 1 - 1 \cdot 1 \cdot 1 - 1 \cdot 1 \cdot 1 - 0 \cdot 0 \cdot 0 = 0.$$

It zero, so the transformation $T$ is not invertible.

**Answer.** $T$ is not invertible, so its inverse does not exist.

The matrix of $T$ in the basis $(v_{1}, v_{2}, v_{3})$ is

$$\left( \begin{array}{ccc} 1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{array} \right)$$