Applied Mathematics

Algebra

Hard

applied

If the matrix $A = \begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix}$ satisfies the equation $A^T A = I_3$ , then $x^2 + y^2 + z^2$ is :

✅ Correct Option: 2

$A^T A = I$ implies columns of $A$ are orthonormal. Column norms: $0 + x^2 + x^2 = 2x^2 = 1$ , $4y^2 + y^2 + y^2 = 6y^2 = 1$ , $z^2 + z^2 + z^2 = 3z^2 = 1$ . So $x^2 = \frac{1}{2}$ , $y^2 = \frac{1}{6}$ , $z^2 = \frac{1}{3}$ . Sum $= \frac{3+1+2}{6} = 1$ .

2023: 11 Aug Shift 1

If $\begin{bmatrix} 2x & -7 \\ 5y & 8 \end{bmatrix} = \begin{bmatrix} 6 & -7 \\ -5 & 3x + y \end{bmatrix}$ , then value of $5x - 3y$ is :

2023: 7 June Shift 1

A die is rolled thrice. What is the probability of getting a number greater than 4 in the first and the second throw of dice and a number less than 4 in the third throw?

2023: 11 Aug Shift 1

Solution for the inequality $|3x - 7| \leq 2$ is :

2022: 25 May Shift 1

Match List I with List II

LIST I	LIST II
A. The solution of $3x + 7 > 12$ is	I. $[-1, \infty)$
B. The solution of $\frac{3x+5}{2} \geq 1$ is	II. $\left[\frac{17}{8}, \infty\right)$
C. The solution of $2x + 5 < 7x + 9$ is	III. $\left(\frac{5}{3}, \infty\right)$
D. The solution of $6x - 5 \geq -2x + 12$ is	IV. $\left(-\frac{4}{5}, \infty\right)$

Choose the correct answer from the options given below:

2026: 14 May Shift 2

A and B are square matrices of order 3 each such that $|A|=3, |B|=2$ , then the value of $|4AB|$ will be?

2023: 22 May Shift 3

If A is square matrix of order 3 with $|A| = 3$ , then $|4 \text{ adj} A|$ is equal to :

If the matrix A=[02yzxy−zx−yz]A = \begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix}A=​0xx​2yy−y​z−zz​​ satisfies the equation ATA=I3A^T A = I_3ATA=I3​, then x2+y2+z2x^2 + y^2 + z^2x2+y2+z2 is :

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

If the matrix $A = \begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix}$ satisfies the equation $A^T A = I_3$ , then $x^2 + y^2 + z^2$ is :