Linear Systems and Matrices: math test solutions, College – download the PDF

1 Gaussian elimination / 4 pts

Augmented matrix \(\left(\begin{array}{ccc|c} 2 & 1 & -1 & 1 \\ 1 & -1 & 2 & 5 \\ 3 & 2 & 1 & 10 \end{array}\right)\) (1 pt).

Swap \(R_1 \leftrightarrow R_2\). Then \(R_2 \leftarrow R_2 - 2R_1\) gives \((0, 3, -5 \mid -9)\) and \(R_3 \leftarrow R_3 - 3R_1\) gives \((0, 5, -5 \mid -5)\) (1 pt).

Divide \(R_3\) by \(5\): \((0, 1, -1 \mid -1)\). Swap \(R_2 \leftrightarrow R_3\), then \(R_3 \leftarrow R_3 - 3R_2\) gives \((0, 0, -2 \mid -6)\) (1 pt).

So \(z = 3\), \(y = z - 1 = 2\), \(x = 5 + y - 2z = 1\). Solution \((1, 2, 3)\) (1 pt).

2 Free variable / 3 pts

\(R_2 \leftarrow R_2 - 2R_1\) gives \((0, 0, 1 \mid 2)\), and \(R_3 \leftarrow R_3 - R_1\) gives \((0, 0, -1 \mid -2)\) (1 pt).

Adding these rows leaves \(0 = 0\), so the system is consistent, with \(z = 2\) (1 pt).

Let \(y = t\). From the first equation, \(x = 3 - t - 4 = -1 - t\). Solution set: \((-1 - t,\ t,\ 2)\) for every real \(t\) (1 pt).

3 Matrix operations / 3 pts

\(2A - B = \begin{pmatrix} 2 & -4 \\ 5 & 2 \end{pmatrix}\) (1 pt).

\(AB = \begin{pmatrix} 0 + 1 & 2 - 4 \\ 0 - 3 & 4 + 12 \end{pmatrix} = \begin{pmatrix} 1 & -2 \\ -3 & 16 \end{pmatrix}\) (1 pt).

\(BA = \begin{pmatrix} 0 + 4 & 0 + 6 \\ -1 + 8 & 1 + 12 \end{pmatrix} = \begin{pmatrix} 4 & 6 \\ 7 & 13 \end{pmatrix}\) (1 pt).

4 Inverse and solution / 4 pts

(a) \(\det A = 9 - 8 = 1 \neq 0\), so \(A\) is invertible (1 pt).

(b) \(A^{-1} = \begin{pmatrix} 3 & -2 \\ -4 & 3 \end{pmatrix}\) (1 pt).

(c) \(\begin{pmatrix} x \\ y \end{pmatrix} = A^{-1}\begin{pmatrix} 7 \\ 10 \end{pmatrix} = \begin{pmatrix} 21 - 20 \\ -28 + 30 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}\) (1 pt). Check: \(3 + 4 = 7\) and \(4 + 6 = 10\) (1 pt).

5 A 3 by 3 determinant / 3 pts

Expand along the first row: \(1\cdot(1\cdot 0 - (-1)\cdot 4) - 0 + 2\cdot(3\cdot 4 - 1\cdot 2) = 4 + 20 = 24\) (1 pt).

\(\det A^T = 24\) and \(\det(2A) = 2^3\cdot 24 = 192\) (1 pt).

Since \(\det A \neq 0\), \(A\) is invertible (1 pt).

6 LU factorization / 3 pts

\(R_2 \leftarrow R_2 - 2R_1\) gives \((0, 1, 3)\); \(R_3\) already has a zero in column 1 (multiplier \(0\)); \(R_3 \leftarrow R_3 - 3R_2\) gives \((0, 0, -4)\) (1 pt).

\(U = \begin{pmatrix} 1 & 2 & 0 \\ 0 & 1 & 3 \\ 0 & 0 & -4 \end{pmatrix}\) and \(L = \begin{pmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 0 & 3 & 1 \end{pmatrix}\) (1 pt).

Forward: \(y_1 = 2\), \(y_2 = 1 - 4 = -3\), \(y_3 = -5 + 9 = 4\). Back: \(z = -1\), \(y = -3 + 3 = 0\), \(x = 2\). Solution \((2, 0, -1)\) (1 pt).