train/precalculus/1023.json

{
    "problem": "Let\n\\[\\mathbf{A} = \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & -1 \\\\ 0 & 1 & 0 \\end{pmatrix}.\\]Compute $\\mathbf{A}^{95}.$",
    "level": "Level 3",
    "type": "Precalculus",
    "solution": "Note that\n\\[\\mathbf{A}^2 = \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & -1 \\\\ 0 & 1 & 0 \\end{pmatrix} \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & -1 \\\\ 0 & 1 & 0 \\end{pmatrix} = \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & -1 & 0 \\\\ 0 & 0 & -1 \\end{pmatrix}.\\]Then\n\\[\\mathbf{A}^4 = \\mathbf{A}^2 \\mathbf{A}^2 = \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & -1 & 0 \\\\ 0 & 0 & -1 \\end{pmatrix} \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & -1 & 0 \\\\ 0 & 0 & -1 \\end{pmatrix} = \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & 1 \\end{pmatrix}.\\]Since $\\mathbf{A}^4$ is a diagonal matrix, any power of $\\mathbf{A}^4$ is\n\\begin{align*}\n(\\mathbf{A}^4)^{k} = \\begin{pmatrix} 0^k & 0 & 0 \\\\ 0 & 1^k & 0 \\\\ 0 & 0 & 1^k \\end{pmatrix} = \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & 1 \\end{pmatrix} = \\mathbf{A}^4.\n\\end{align*}Hence,\n\\begin{align*}\n\\mathbf{A}^{95} &= (\\mathbf{A}^4)^{23} \\mathbf{A}^3 = \\mathbf{A}^4 \\mathbf{A} \\mathbf{A}^2 \\\\\n&= \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & 1 \\end{pmatrix} \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & -1 \\\\ 0 & 1 & 0 \\end{pmatrix} \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & -1 & 0 \\\\ 0 & 0 & -1 \\end{pmatrix} \\\\\n&= \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & 1 \\end{pmatrix} \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & 1 \\\\ 0 & -1 & 0 \\end{pmatrix} \\\\\n&= \\boxed{\\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & 1 \\\\ 0 & -1 & 0 \\end{pmatrix}}\n\\end{align*}"
}