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Chapter 1: Problem 17

Let \(A\) be an \(n \times n\) matrix and let \(\mathbf{x}\) and \(\mathbf{y}\) be vectors in \(\mathbb{R}^{n} .\) Show that if \(A \mathbf{x}=A \mathbf{y}\) and \(\mathbf{x} \neq \mathbf{y},\) then the matrix \(A\) must be singular.

Short Answer

Expert verified
To show that the matrix $A$ is singular given $A\mathbf{x}=A\mathbf{y}$ and $\mathbf{x}\neq \mathbf{y}$, we first find the non-zero vector $\mathbf{z} = \mathbf{x} - \mathbf{y}$, and then show that $A\mathbf{z} = 0$. Since there exists a non-zero vector $\mathbf{z}$ such that $A\mathbf{z}=0$, it means that the matrix $A$ is not invertible, which implies the determinant of $A$ is 0. Thus, the matrix $A$ is singular.

Step by step solution

01

Finding the vector z

Since Ax = Ay and x 鈮� y, we can find a vector z such that z = x - y. Notice that since x and y are different, z is a non-zero vector.
02

Show that Az = 0

Multiply both sides of the equation Ax = Ay by A, and using the given vector z, we have: \[A(\mathbf{x} - \mathbf{y}) = A \mathbf{x} - A \mathbf{y}\] Since Ax = Ay, we have: \[A \mathbf{x} - A \mathbf{y} = 0\] So, \[A \mathbf{z} = 0\] where z is a non-zero vector.
03

Conclude that the determinant of A is 0

Since we have found a non-zero vector z such that Az = 0, this means that the matrix A is not invertible. This is because if A were invertible, we could multiply both sides of the equation Az = 0 by the inverse of A, but then we would have z = 0, which contradicts the fact that z is a non-zero vector. Therefore, the determinant of A must be 0 since A is not invertible.
04

Conclude that the matrix A is singular

A matrix is singular if and only if its determinant is 0. In Step 3, we showed that the determinant of A is 0. Therefore, the matrix A is singular.

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Most popular questions from this chapter

Let \(A\) and \(B\) be symmetric \(n \times n\) matrices. Prove that \(A B=B A\) if and only if \(A B\) is also symmetric.

Consider a linear system whose augmented matrix is of the form \\[ \left(\begin{array}{rrr|r} 1 & 2 & 1 & 1 \\ -1 & 4 & 3 & 2 \\ 2 & -2 & a & 3 \end{array}\right) \\] For what values of a will the system have a unique solution?

For each of the systems of equations that follow, use Gaussian elimination to obtain an equivalent system whose coefficient matrix is in row echelon form. Indicate whether the system is consistent. If the system is consistent and involves no free variables, use back substitution to find the unique solution. If the system is consistent and there are free variables, transform it to reduced row echelon form and find all solutions $$\begin{aligned} &\text { (a) } \quad x_{1}-2 x_{2}=3\\\ &2 x_{1}-x_{2}=9 \end{aligned}$$ $$\begin{aligned} &\text { (b) } \quad 2 x_{1}-3 x_{2}=5\\\ &-4 x_{1}+6 x_{2}=8 \end{aligned}$$ $$\begin{aligned} &\text { (c) } \quad x_{1}+x_{2}=0\\\ &\begin{array}{l} 2 x_{1}+3 x_{2}=0 \\ 3 x_{1}-2 x_{2}=0 \end{array} \end{aligned}$$ $$\begin{aligned} &\text { (d) } 3 x_{1}+2 x_{2}-x_{3}=4\\\ &\begin{array}{r} x_{1}-2 x_{2}+2 x_{3}=1 \\ 11 x_{1}+2 x_{2}+x_{3}=14 \end{array} \end{aligned}$$ $$\begin{aligned} &\text { (e) } 2 x_{1}+3 x_{2}+x_{3}=1\\\ &\begin{array}{r} x_{1}+x_{2}+x_{3}=3 \\ 3 x_{1}+4 x_{2}+2 x_{3}=4 \end{array} \end{aligned}$$ $$\begin{aligned} &\text { (f) } \quad x_{1}-x_{2}+2 x_{3}=4\\\ &\begin{array}{l} 2 x_{1}+3 x_{2}-x_{3}=1 \\ 7 x_{1}+3 x_{2}+4 x_{3}=7 \end{array} \end{aligned}$$ $$\begin{aligned} &\text { (g) } x_{1}+x_{2}+x_{3}+x_{4}=0\\\ &\begin{array}{l} 2 x_{1}+3 x_{2}-x_{3}-x_{4}=2 \\ 3 x_{1}+2 x_{2}+x_{3}+x_{4}=5 \\ 3 x_{1}+6 x_{2}-x_{3}-x_{4}=4 \end{array} \end{aligned}$$ $$\begin{aligned} &\text { (h) } \quad x_{1}-2 x_{2}=3\\\ &\begin{aligned} 2 x_{1}+x_{2} &=1 \\ -5 x_{1}+8 x_{2} &=4 \end{aligned} \end{aligned}$$ $$\begin{aligned} &\text { (i) } \quad-x_{1}+2 x_{2}-x_{3}=2\\\ &\begin{aligned} -2 x_{1}+2 x_{2}+x_{3} &=4 \\ 3 x_{1}+2 x_{2}+2 x_{3} &=5 \\ -3 x_{1}+8 x_{2}+5 x_{3} &=17 \end{aligned} \end{aligned}$$ $$\begin{aligned} &\text { (j) } \quad x_{1}+2 x_{2}-3 x_{3}+x_{4}=1\\\ &\begin{array}{r} -x_{1}-x_{2}+4 x_{3}-x_{4}=6 \\ -2 x_{1}-4 x_{2}+7 x_{3}-x_{4}=1 \end{array} \end{aligned}$$ $$\begin{aligned} &\text { (k) } x_{1}+3 x_{2}+x_{3}+x_{4}=3\\\ &\begin{array}{c} 2 x_{1}-2 x_{2}+x_{3}+2 x_{4}=8 \\ x_{1}-5 x_{2}+x_{4}=5 \end{array} \end{aligned}$$ $$\begin{aligned} &\text { (I) } \quad x_{1}-3 x_{2}+\quad x_{3}=1\\\ &\begin{aligned} 2 x_{1}+x_{2}-x_{3} &=2 \\ x_{1}+4 x_{2}-2 x_{3} &=1 \\ 5 x_{1}-8 x_{2}+2 x_{3} &=5 \end{aligned} \end{aligned}$$

Let \(A=\left[\begin{array}{ll}A_{11} & A_{12} \\ A_{21} & A_{22}\end{array}\right] \quad\) and \(\quad A^{T}=\left[\begin{array}{cc}A_{11}^{T} & A_{21}^{T} \\ A_{12}^{T} & A_{22}^{T}\end{array}\right]\) Is it possible to perform the block multiplications of \(A A^{T}\) and \(A^{T} A ?\) Explain.

Find nonzero \(2 \times 2\) matrices \(A\) and \(B\) such that \(A B=O\).
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