Exam-1 Practice (MATH 2450): Interactive Solutions

Q1. Is \( e^{-3x} + 4y = 5 \) linear in \(x\) and \(y\)?

No.

Linear equations have variables only to the first power and no nonlinear functions (like exponents or products).

The term \(e^{-3x}\) is a nonlinear exponential in \(x\); so the equation is not linear.

Q2. Find the parametric solution of \( -3x + 2y - 5z = 1 \).

Let \(y = s,\, z = t\):
\( x = \frac{2s - 5t - 1}{3},\quad y = s,\quad z = t \).

Solve for \(x\) in terms of free parameters for \(y, z\).

With one equation and three variables, two are free. \( x = \frac{2y - 5z - 1}{3} \).

Q3. Solve \( \begin{cases} 40x_1 + 30x_2 = 21 \\ 20x_1 + 15x_2 = -19 \end{cases} \)

No solution

Multiply row 2 by 2: \(40x_1 + 30x_2 = -38\), which contradicts \(40x_1 + 30x_2 = 21\). No solution.

Q4. Solution set for augmented matrix: \(\begin{bmatrix} 1 & 2 & | & -5 \\ 2 & 1 & | & 5 \end{bmatrix}\)

\(x_1 = 5\), \(x_2 = -5\)

Solve: \(x_1 + 2x_2 = -5\), \(2x_1 + x_2 = 5\). Substitute and solve to get both values.

Q5. Is \(\begin{bmatrix}-1 & 9 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}\) in row-echelon/reduced row-echelon form?

Neither.

REF requires leading entry in each nonzero row to be 1. Here, first row starts with -1. Not in REF or RREF.

Q6. Add: \( \begin{bmatrix}1&3\\8&-2\\9&0\end{bmatrix} + \begin{bmatrix}9&-3&5\\2&0&0\\0&0&0\end{bmatrix} \)

Not possible.

Matrix addition is defined only for matrices of the same size. 3×2 + 3×3 is not defined.

Q7. Solve using Gaussian elimination:
\(\begin{cases}-3x_1-x_2+x_3=-11 \\ 2x_1+4x_2-5x_3=12 \\ x_1-2x_2+3x_3=1\end{cases}\)

\(x_1 = 3\), \(x_2 = 4\), \(x_3 = 2\)

Form the augmented matrix, use row operations to get upper-triangular, then back-substitute.

Start: \( \begin{bmatrix} -3 & -1 & 1 & -11 \\ 2 & 4 & -5 & 12 \\ 1 & -2 & 3 & 1 \end{bmatrix} \)
After row reduction to RREF, we get: \( \begin{bmatrix} 1 & 0 & 0 & 3 \\ 0 & 1 & 0 & 4 \\ 0 & 0 & 1 & 2 \end{bmatrix} \)
This gives the solution \(x_1 = 3\), \(x_2 = 4\), \(x_3 = 2\).

Q8. For \(A = \begin{bmatrix}1&3&-2\\0&1&3\end{bmatrix}\), find \(A^T\), \(A^T A\), \(A A^T\)

\(A^T = \begin{bmatrix}1&0\\3&1\\-2&3\end{bmatrix}\)
\(A^T A = \begin{bmatrix}1&3&-2\\3&10&-3\\-2&-3&13\end{bmatrix}\)
\(A A^T = \begin{bmatrix}14&-3\\-3&10\end{bmatrix}\)

Transpose: swap rows/columns.
\(A^T A\) and \(A A^T\) are found by matrix multiplication.

Q9. Inverse of \(\begin{bmatrix}3&-1\\-27&9\end{bmatrix}\)?

Does not exist (DNE).

Determinant: \(3\times9 - (-1)\times(-27) = 27-27=0\). If determinant is 0, matrix is singular (not invertible).

Q10. Find \(x\) so \(A = \begin{bmatrix}7 & 1 \\ x & -1\end{bmatrix}\) is nonsingular.

\(x \neq -7\)

\(\det(A) = 7\times(-1) - 1\times x = -7-x\). Set \(-7-x \neq 0\) so \(x \neq -7\).

Q11. Inverse of \( \begin{bmatrix}1&0&0\\0&8&0\\0&0&1\end{bmatrix} \)?

\( \begin{bmatrix}1&0&0\\0&\frac{1}{8}&0\\0&0&1\end{bmatrix} \)

Inverse of a diagonal matrix: reciprocate each nonzero diagonal entry.

Q12. Find \(\det(A)\), \(A = \begin{bmatrix}-24&0&3\\3&9&-6\\27&-3&6\end{bmatrix}\)

-1620

Use cofactor expansion along first row:
\( -24\begin{vmatrix}9 & -6 \\ -3 & 6\end{vmatrix} + 0 + 3\begin{vmatrix}3 & 9 \\ 27 & -3\end{vmatrix} \)
\( = -24(9\times6 - (-6)\times(-3)) + 3(3\times(-3) - 9\times27) \)
\( = -24(54 - 18) + 3(-9 - 243) = -24\times36 + 3\times(-252) = -864 - 756 = -1620 \)

Q13. Determinant of \( \begin{bmatrix}3&0&0&0\\-9&1&0&0\\6&-1&9&0\\2&3&1&-1\end{bmatrix} \)

-27

Lower-triangular matrix. Determinant = product of diagonals: \(3 \times 1 \times 9 \times -1 = -27\).

Q14. For \(A, B\), check \(|A||B| = |AB|\). \(A = \begin{bmatrix}0&1&2\\5&4&3\\7&6&8\end{bmatrix},\ B = \begin{bmatrix}3&1&3\\1&-1&0\\0&4&-3\end{bmatrix}\)

\(|A| = -15,\quad |B| = 24,\quad |AB| = -360\)

Multiply the determinants: \( -15 \times 24 = -360 = |AB| \). The property holds.

Q15. For \(A = \begin{bmatrix}-3&9\\4&1\end{bmatrix}\), compute \(|A^T|\), \(|A^3|\), \(|A^T A|\), \(|7A|\).

\( |A| = -39 \)
\( |A^T| = -39 \)
\( |A^3| = (-39)^3 = -59,\!319 \)
\( |A^T A| = (-39)^2 = 1,\!521 \)
\( |7A| = 7^2 \times (-39) = -1,\!911 \)

Properties: \( |A^T| = |A| \), \( |A^k| = |A|^k \), \( |kA| = k^n|A| \) for \( n \times n \) matrix.

Q16. If \(A\) is invertible, is \(A^2\) invertible?

Yes.

If \(A\) is invertible, \(\det(A) \neq 0\), so \(\det(A^2) = (\det(A))^2 \neq 0\). Inverse is \((A^{-1})^2\).

Q17. If \(AB = AC\), can we "cancel" \(A\) to say \(B=C\)?

Only if \(A\) is invertible.

If \(A\) is invertible, multiply both sides by \(A^{-1}\) on left: \(A^{-1}AB = A^{-1}AC \implies B = C\). If not invertible, can't cancel.

Q18. Compute adjoint of \( \begin{bmatrix}4&1\\5&2\end{bmatrix} \).

\( \begin{bmatrix}2&-1\\-5&4\end{bmatrix} \)

For a \(2\times2\) matrix \(\begin{bmatrix}a&b\\c&d\end{bmatrix}\), adjoint is \(\begin{bmatrix}d&-b\\-c&a\end{bmatrix}\).

Q19. Find minors and cofactors of \( \begin{bmatrix}4&2\\3&1\end{bmatrix} \).

Minors: \(M_{11}=1,\ M_{12}=3,\ M_{21}=2,\ M_{22}=4\)
Cofactors: \(C_{11}=1,\ C_{12}=-3,\ C_{21}=-2,\ C_{22}=4\)

Cofactor: \(C_{ij} = (-1)^{i+j}M_{ij}\).

Q20. Find all \(x\) so that \( \begin{bmatrix}3&x\\1&2\end{bmatrix} \) is singular.

\(x=6\)

Singular: determinant is 0.
\(3\times2-x\times1=0\), \(6-x=0\) so \(x=6\).

Q21. Find all \(x\) so \( \begin{bmatrix}x&1\\1&x\end{bmatrix} \) is nonsingular.

\(x \neq \pm1\)

Nonsingular: determinant \( \neq 0 \).
\(x^2-1 \neq 0\), so \(x \neq 1, -1\).

Q22. If \(A\) is singular and \(B\) is invertible, is \(AB\) invertible?

No.

\(\det(AB) = \det(A)\det(B) = 0\). So, \(AB\) is singular.

Q23. If \(A\) is invertible, must its transpose be invertible?

Yes.

\(\det(A^T) = \det(A)\), so \(A^T\) is invertible if \(A\) is. Inverse: \((A^T)^{-1} = (A^{-1})^T\).

Exam-1 Practice (MATH 2450)

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Interactive Widgets (With Worked Examples)

Practice Questions