# Linear Independent Vectors, Invertible Matrix, and Expression of a Vector as a Linear Combinations

## Problem 66

Consider the matrix

\[A=\begin{bmatrix}

1 & 2 & 1 \\

2 &5 &4 \\

1 & 1 & 0

\end{bmatrix}.\]

**(a)**Calculate the inverse matrix $A^{-1}$. If you think the matrix $A$ is not invertible, then explain why.

**(b)**Are the vectors

\[ \mathbf{A}_1=\begin{bmatrix}

1 \\

2 \\

1

\end{bmatrix}, \mathbf{A}_2=\begin{bmatrix}

2 \\

5 \\

1

\end{bmatrix},

\text{ and } \mathbf{A}_3=\begin{bmatrix}

1 \\

4 \\

0

\end{bmatrix}\] linearly independent?

**(c)**Write the vector $\mathbf{b}=\begin{bmatrix}

1 \\

1 \\

1

\end{bmatrix}$ as a linear combination of $\mathbf{A}_1$, $\mathbf{A}_2$, and $\mathbf{A}_3$.

(*The Ohio State University, Linear Algebra Exam*)

Contents

## Hint.

- For (a), consider the augmented matrix $[A|I]$ and reduce it.
- Note that given vectors are column vectors of the matrix $A$.
- Use the inverse matrix $A^{-1}$ to solve a system.

## Solution.

### (a) Calculate the inverse matrix $A^{-1}$

We consider the augmented matrix

\[ \left[\begin{array}{rrr|rrr}

1 & 2 & 1 & 1 &0 & 0 \\

2 & 5 & 4 & 0 & 1 & 0 \\

1 & 1 & 0 & 0 & 0 & 1 \\

\end{array} \right] \]
and reduce this matrix using the elementary row operations as follows.

\begin{align*}

&\left[ \begin{array}{rrr|rrr}

1 & 2 & 1 & 1 &0 & 0 \\

2 & 5 & 4 & 0 & 1 & 0 \\

1 & 1 & 0 & 0 & 0 & 1 \\

\end{array} \right]
\xrightarrow[R_3-R_1]{R_2-2R_1}

\left[\begin{array}{rrr|rrr}

1 & 2 & 1 & 1 &0 & 0 \\

0 & 1 & 2 & -2 & 1 & 0 \\

0 & -1 & -1 & -1 & 0 & 1 \\

\end{array} \right]
\xrightarrow[R_3+R_2]{R_1-2R_2} \\[6pt]
&

\left[\begin{array}{rrr|rrr}

1 & 0 & -3 & 5 &-2 & 0 \\

0 & 1 & 2 & -2 & 1 & 0 \\

0 & 0 & 1 & -3 & 1 & 1 \\

\end{array} \right]
\xrightarrow[R_2-2R_3]{R_1+3R_3} \left[\begin{array}{rrr|rrr}

1 & 0 & 0 & -4 &1 & 3 \\

0 & 1 & 0 & 4 & -1 & -2 \\

0 & 0 & 1 & -3 & 1 & 1 \\

\end{array} \right].

\end{align*}

Since the left $3 \times 3$ part of the last matrix is the identity matrix, the inverse matrix of $A$ is

\[A^{-1}=\begin{bmatrix}

-4 & 1 & 3 \\

4 &-1 &-2 \\

-3 & 1 & 1

\end{bmatrix}.\]

### (b) Are the Vectors Linearly Independent?

To check whether $\mathbf{A}_1, \mathbf{A}_2, \mathbf{A}_3$ are linearly independent, we consider the linear combination

\[x_1\mathbf{A}_1+x_2\mathbf{A}_2+x_3\mathbf{A}_3=\mathbf{0}\]
and if this equation has only zero solution, the vectors are linearly independent.

This equation can be written as the matrix equation

\[A\mathbf{x}=\mathbf{0},\]
where $\mathbf{x}=\begin{bmatrix}

x_1 \\

x_2 \\

x_3

\end{bmatrix}$.

Since by part (a), the inverse matrix $A^{-1}$ exists. Thus multiplying by $A^{-1}$ on the left we get $\mathbf{x}=\mathbf{0}$. Thus the solution is $\mathbf{x}=\mathbf{0}$ and the vectors $\mathbf{A}_1, \mathbf{A}_2, \mathbf{A}_3$ are linearly independent.

#### Another Solution of (b)

Note that $\mathbf{A}_1, \mathbf{A}_2, \mathbf{A}_3$ are columns vectors of the matrix $A$. We proved in part (a) that $A$ is invertible. We know that the column vectors of an invertible matrix are linearly independent. Thus the vectors $\mathbf{A}_1, \mathbf{A}_2, \mathbf{A}_3$ are linearly independent.

### (c) Write the vector $\mathbf{b}$ as a linear combination of $\mathbf{A}_1$, $\mathbf{A}_2$, and $\mathbf{A}_3$

We want to solve the vector equation

\[x_1\mathbf{A}_1+x_2\mathbf{A}_2+x_3\mathbf{A}_3=\mathbf{b}.\]
This can be written as the matrix equation

\[A\mathbf{x}=\mathbf{b}.\]

Since $A$ is invertible, we have

\[\mathbf{x}=A^{-1}\mathbf{b}=\begin{bmatrix}

-4 & 1 & 3 \\

4 &-1 &-2 \\

-3 & 1 & 1

\end{bmatrix}

\begin{bmatrix}

1 \\

1 \\

1

\end{bmatrix}=\begin{bmatrix}

0 \\

1 \\

-1

\end{bmatrix}.\]

Thus $x_1=0$, $x_2=1$, and $x_3=-1$ and the linear combination is

\[\mathbf{b}=\begin{bmatrix}

1 \\

1 \\

1

\end{bmatrix}=\mathbf{A}_2-\mathbf{A}_3.\]

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