## Linear Algebra and Its Applications, Exercise 2.4.15

Exercise 2.4.15. For each of the following matrices $A = \begin{bmatrix} 1&1&0 \\ 0&1&1 \end{bmatrix} \qquad M = \begin{bmatrix} 1&0 \\ 1&1 \\ 0&1\end{bmatrix} \qquad T = \begin{bmatrix} a&b \\ 0&a \end{bmatrix}$

find a left inverse and right inverse if they exist.

Answer: We begin with the 2 by 3 echelon matrix $A$. Since $A$ has two pivots its rank $r = 2$. Since $A$ has two rows we also have $r = m$ so that $A$ has a (3 by 2) right inverse. It does not have a left inverse since $r \ne n = 3$.

We now attempt to find a right inverse $C$. We must have $AC = I$ or $\begin{bmatrix} 1&1&0 \\ 0&1&1 \end{bmatrix} \begin{bmatrix} c_{11}&c_{12} \\ c_{21}&c_{22} \\ c_{31}&c_{32} \end{bmatrix} = \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix}$

Multiplying the two rows of $A$ by the first column of $C$ we must have $c_{11} + c_{21} = 1$ and $c_{21} + c_{31} = 0$. One way to achieve this is to set $c_{11} = 1$ and $c_{21} = c_{31} = 0$.

Multiplying the two rows of $A$ by the second column of $C$ we must have $c_{12} + c_{22} = 0$ and $c_{22} + c_{32} = 1$. One way to achieve this is to set $c_{12} = c_{22} = 0$ and $c_{32} = 1$. We then have $C = \begin{bmatrix} 1&0 \\ 0&0 \\ 0&1 \end{bmatrix}$

so that $AC = \begin{bmatrix} 1&1&0 \\ 0&1&1 \end{bmatrix} \begin{bmatrix} 1&0 \\ 0&0 \\ 0&1 \end{bmatrix} = \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix} = I$

We now consider the matrix $M$. The first and third rows of $M$ are linearly independent, and the second row of $M$ is equal to the sum of the first and third rows. The rank of $M$ is therefore $r = 2$. Since $M$ has two columns we also have $r = n$ so that $M$ has a (3 by 2) right inverse. It does not have a right inverse since $r \ne m = 3$.

We now attempt to find a left inverse $B$ of $M$ such that $BM = I$. We could proceed as we did above, but there is a shortcut: Note that $M = A^T$ so that we are looking for $B$ such that $BA^T = I$. But $BA^T = AB^T$ and we already have a matrix $C$ such that $AC = I$. We can therefore choose $B$ such that $B^T = C$ or $B = C^T$: $B = \begin{bmatrix} 1&0&0 \\ 0&0&1 \end{bmatrix}$

We then have $BM = \begin{bmatrix} 1&0&0 \\ 0&0&1 \end{bmatrix} \begin{bmatrix} 1&0 \\ 1&1 \\ 0&1\end{bmatrix} = \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix} = I$

so that $B = C^T$ is a left inverse of $M$.

Finally we consider the 2 by 2 triangular matrix $T$. If $a \ne 0$ then $T$ has two pivots and rank $r = 2$. In this case since $r = m = n$ both a left inverse $B$ and right inverse $C$ exist and are the same: $T$ is invertible with $T^{-1} = B = C$.

From the standard formula for the inverse of a 2 by 2 matrix (see page 42) we have $T^{-1} = 1/(a \cdot a - b \cdot 0) \begin{bmatrix} a&-b \\ -0&a \end{bmatrix}$ $= 1/a^2 \begin{bmatrix} a&-b \\ 0&a \end{bmatrix} = \begin{bmatrix} 1/a&-b/a^2 \\ 0&1/a \end{bmatrix}$

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition , Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books .

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