# 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)

## Hint.

1. For (a), consider the augmented matrix $[A|I]$ and reduce it.
2. Note that given vectors are column vectors of the matrix $A$.
3. 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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