If A is a square matrix of order 3 with |A| = 5, what is the value of |3A|?
- 135 Correct answer.
- 15
- 405
- 45
Explanation
For a square matrix of order n, |kA| = k^n |A|. Here n = 3 and k = 3, so |3A| = 3^3 × 5 = 27 × 5 = 135.
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For a square matrix of order n, |kA| = k^n |A|. Here n = 3 and k = 3, so |3A| = 3^3 × 5 = 27 × 5 = 135.
If every element of one row (or one column) of a determinant is zero, the determinant expands to zero. Each term in the expansion of that row contains a zero factor.
The determinant of a product of two square matrices equals the product of their determinants, so |AB| = |A||B| holds for all such matrices. Matrix multiplication is not commutative in general, and the binomial expansion fails unless AB = BA.
For a 2 × 2 determinant, the value is (4)(3) − (7)(2) = 12 − 14 = −2. Subtract the product of the secondary diagonal from the product of the main diagonal.
The product AB is defined when the number of columns of A equals the number of rows of B, which is 3 here. The resulting matrix has the rows of A and the columns of B, so its order is 2 × 4.
A matrix equal to its own transpose is symmetric, meaning a_ij = a_ji for all entries. Skew-symmetric requires A^T = −A, and singular refers to a zero determinant.
For a square matrix of order n, |adj A| = |A|^(n−1). With n = 3 and |A| = 4, this gives 4^2 = 16.
A matrix with m rows and n columns has one entry for each row-column pair, giving m × n entries. For example, a 2 × 3 matrix has six elements.
Scaling a matrix of order n by k multiplies its determinant by k^n. Here |2A| = 2^3 × 2 = 8 × 2 = 16.
The inverse of a non-singular matrix is given by A^(−1) = adj A / |A|. This requires |A| ≠ 0, which is exactly the non-singular condition.
A square matrix whose determinant is zero is called singular and has no inverse. Singularity does not require all entries to be zero, only that the rows are linearly dependent.
The matrix B that satisfies AB = BA = I is by definition the inverse of A, written A^(−1). The transpose and adjoint equal the inverse only in special cases.
For order n, |adj(adj A)| = |A|^((n−1)^2). With n = 3 the exponent is 4, so the value is (−3)^4 = 81. However, using the direct rule |adj(adj A)| = |A|^(n−1)^2 with n = 3 gives 81, so check the exponent: (3−1)^2 = 4 and (−3)^4 = 81. The stated options do not include 81, so the intended reading is |adj(adj A)| = |A| when n = 3 is misapplied; the correct value is 81.
When |A| = 0 the coefficient matrix is singular, so Cramer's rule fails and a unique solution is impossible. A consistent singular system has infinitely many solutions, while an inconsistent one has none.
Since A^2 = A, higher powers collapse: A^3 = A. Expanding (I + A)^3 = I + 3A + 3A^2 + A^3 = I + 3A + 3A + A = I + 7A. Subtracting 7A leaves I.
Interchanging any two rows of a determinant changes its sign. Since the original determinant is 6, the new determinant is −6.
A matrix and its transpose have equal determinants, so |A^T| = |A| = 7. Transposing swaps rows and columns without changing the value.
A · adj A = |A| · I for any square matrix. Taking determinants gives |A|^n = 4^3 = 64, since the order is 3.
A scalar matrix is a diagonal matrix in which every diagonal entry is the same non-zero scalar. It is a special case of a diagonal matrix, obtained by multiplying the identity by a scalar.
Taking determinants of AB = O gives |A||B| = 0, so at least one determinant is zero and that matrix is singular. The zero product property for numbers does not carry over to matrices directly.
The determinant of the inverse is the reciprocal of the determinant, so |A^(−1)| = 1/|A| = 1/2. This follows from |A||A^(−1)| = |I| = 1.
If two columns of a determinant are identical, the determinant is zero because the columns are linearly dependent. Interchanging them would change the sign while leaving the determinant unchanged, forcing it to be zero.
The rule |adj A| = |A|^(n−1) with n = 3 gives 5^2 = 25. This follows from A · adj A = |A|I and taking determinants.
Subtracting A from both sides of A + B = A gives B = O, the zero matrix of the same order. Matrix addition is associative and each matrix has an additive inverse.
The second equation is twice the first on the left but not on the right, since 2 × 3 = 6 ≠ 7. The equations are inconsistent, so the system has no solution even though the coefficient determinant is zero.
For a square matrix of order n, the identity |adj(adj A)| = |A|^((n−1)^2) holds. Here n = 3, so the exponent is (3−1)^2 = 4, giving |adj(adj A)| = 5^4? No - the exponent for adj(adj A) is (n−1)^2 = 4 only for |adj A| itself raised; the correct chain is |adj A| = |A|^(n−1) = 5^2 = 25, then |adj(adj A)| = |adj A|^(n−1) = 25^2 = 625. Wait, applying the formula directly: |adj(adj A)| = |A|^((n−1)^2) = 5^4 = 625.
A matrix of order m × n contains m × n elements, where m is the number of rows and n the number of columns. With 3 rows and 12 elements, the number of columns is 12 ÷ 3 = 4, so the order is 3 × 4.
The (2,1) entry of AB is the dot product of the second row of A with the first column of B. That gives (4)(1) + (3)(2) = 4 + 6 = 10, so the required element is 10.
The coefficient determinant is (1)(6) − (2)(3) = 0, so the system is not uniquely solvable. The second equation is exactly three times the first, so both equations represent the same line and every point on it satisfies the system, giving infinitely many solutions.
For any square matrix A, the product A · adj A equals |A| · I, where I is the identity of the same order. Taking determinants, |A · adj A| = |A|^n with n = 2, so the value is 3^2 = 9.
The slope of the line through two points is m = (y2 − y1)/(x2 − x1). Here m = (11 − 3)/(6 − 2) = 8/4 = 2, so the line rises two units for every unit moved to the right.
In slope-intercept form y = mx + c, the coefficient m is the slope and c is the y-intercept. Substituting m = 3 and c = −5 gives y = 3x − 5 directly.
Distance is found from d = √[(x2 − x1)² + (y2 − y1)²]. Here d = √(3² + 4²) = √25 = 5, the classic 3-4-5 right triangle.
The midpoint formula averages the coordinates: ((x1 + x2)/2, (y1 + y2)/2). This gives ((−3 + 7)/2, (4 − 2)/2) = (2, 1).
Rearranging gives 2y = 4x + 7, so y = 2x + 7/2 and the slope is 2. Parallel lines share the same slope, so any parallel line also has slope 2.
Comparing with x² + y² + 2gx + 2fy + c = 0 gives g = −3, f = 4, c = −11. The radius is √(g² + f² − c) = √(9 + 16 + 11) = √36 = 6.
Adding the two equations eliminates y and gives 2x = 6, so x = 3. Substituting into x + y = 5 gives y = 2, hence the intersection point is (3, 2).
The slopes are m1 = 1 and m2 = −1, and their product m1·m2 = −1. Two lines whose slopes multiply to −1 are perpendicular, so the angle between them is 90°.
In intercept form x/a + y/b = 1, the line meets the x-axis at (a, 0) and the y-axis at (0, b). With a = 4 and b = 3, the intercepts are (4, 0) and (0, 3).
The slope of the joining line is (10 − 2)/(5 − 1) = 2. A perpendicular line has slope equal to the negative reciprocal, that is −1/2.
Distance from the origin is √(x² + y²), giving √(9 + 16) = √25 = 5. The negative sign of the y-coordinate disappears on squaring.
A point lies on a line when its coordinates satisfy the equation. Substituting x = 2 and y = 0 gives 2k − 6 = 0, so k = 3.
The circle with centre (h, k) and radius r is (x − h)² + (y − k)² = r². Expanding (x − 2)² + (y + 3)² = 16 gives x² + y² − 4x + 6y − 3 = 0.
The legs along the axes have lengths 4 and 3, so the area is ½ × 4 × 3 = 6 square units. The determinant formula ½|x1(y2 − y3) + x2(y3 − y1) + x3(y1 − y2)| gives the same result.
Comparing coefficients, 2/4 = 3/6 = 1/2 but 7/15 ≠ 1/2. Equal slope ratios with unequal constant ratios mean the lines are parallel and do not coincide, so they never meet.
The perpendicular from the origin has slope 4/3, so its equation is 4x − 3y = 0. Solving with 3x + 4y = 25 gives x = 3, y = 4, and this point lies on the line since 9 + 16 = 25.
The perpendicular distance is |ax1 + by1 + c| / √(a² + b²). This gives |3 + 4 − 12| / √(9 + 16) = 5/5 = 1.
For x² + y² + 2gx + 2fy + c = 0 the centre is (−g, −f). Here 2g = −4 and 2f = 2, so g = −2 and f = 1, giving centre (2, −1).
Parallel lines satisfy a1/a2 = b1/b2, so p/2 = 3/6 = 1/2. Solving gives p = 1, and the constant ratio 5/7 differs, so the lines are distinct and parallel.
The given line has slope −1/2, so the perpendicular slope is 2. Using the point-slope form with (2, 3): y − 3 = 2(x − 2), which simplifies to 2x − y = 1.