CUET UG 2025 — Mathematics Algebra

Given $A = \begin{bmatrix} 4 & 3 \\ 2 & -1 \\ 1 & 0 \end{bmatrix}$A=​421​3−10​​ and $B^T = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 1 & -1 \end{bmatrix}$BT=[12​21​3−1​].

Matrix $A$A is a $3 \times 2$3×2 matrix.

Matrix $B^T$BT is a $2 \times 3$2×3 matrix.

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To find $B$B, transpose $B^T$BT by converting rows to columns.

Row 1 of $B^T$BT: $[1, 2, 3]$[1,2,3] becomes Column 1 of $B$B

Row 2 of $B^T$BT: $[2, 1, -1]$[2,1,−1] becomes Column 2 of $B$B

$$B = \begin{bmatrix} 1 & 2 \\ 2 & 1 \\ 3 & -1 \end{bmatrix}$$B=​123​21−1​​

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Both $A$A and $B$B are $3 \times 2$3×2 matrices. Subtract corresponding elements:

$$A - B = \begin{bmatrix} 4 & 3 \\ 2 & -1 \\ 1 & 0 \end{bmatrix} - \begin{bmatrix} 1 & 2 \\ 2 & 1 \\ 3 & -1 \end{bmatrix}$$A−B=​421​3−10​​−​123​21−1​​

$$A - B = \begin{bmatrix} 4-1 & 3-2 \\ 2-2 & -1-1 \\ 1-3 & 0-(-1) \end{bmatrix}$$A−B=​4−12−21−3​3−2−1−10−(−1)​​

$$A - B = \begin{bmatrix} 3 & 1 \\ 0 & -2 \\ -2 & 1 \end{bmatrix}$$A−B=​30−2​1−21​​